How To Set Up Linear Equations From Word Problems

Introduction to Systems	Altitude Give-and-take Trouble
Solving Systems by Graphing	Which Plumber Trouble
Solving Systems with Exchange	Geometry Give-and-take Trouble
Solving Systems with Linear Combination or Elimination	Work Problem
Types of Equations	Three Variable Word Problem
Systems with Three Equations	The "Processed" Problem
Algebra Word Problems with Systems:	Right Triangle Trigonometry Systems Problem
Investment Word Problem	Inequality Word Problem (in Linear Programming section)
Mixture Word Bug	More Do

Note that we solve Algebra Word Problems without Systems here, and nosotros solve systems using matrices in the Matrices and Solving Systems with Matrices section here.

Introduction to Systems

"Systems of equations" just ways that nosotros are dealing with more than than one equation and variable. So far, nosotros've basically just played around with the equation for a line, which is $y=mx+b$. Let's say nosotros have the following situation:

Yous're going to the mall with your friends and you accept $200 to spend from your recent birthday money. You discover a store that has all jeans for $25 and all dresses for $50 . You actually, really want to take dwelling vi items of clothing because you "need" that many new things. How many pairs of jeans and how many dresses you can purchase and so you use the whole $200 (tax not included)?

Now, yous can always exercise "guess and cheque" to come across what would work, merely you might also use algebra! It'due south much better to learn the algebra way, because even though this problem is fairly simple to solve, the algebra fashion will let you solve whatsoever algebra problem – even the really complicated ones.

The first trick in problems like this is to figure out what we want to know. This volition aid us decide what variables (unknowns) to use. What we desire to know is how many pairs of jeans we want to buy (let'south say "$j$") and how many dresses we desire to purchase (let's say "$d$"). Always write down what your variables will be:

Allow $j=$ the number of jeans you will buy
Let $d=$ the number of dresses y'all'll purchase

Similar we did earlier, let's translate discussion-for-give-and-take from math to English:

English	Math	Caption
"You lot really, really desire to take domicile 6 items of wearable because y'all need that many."	\(j+d=6\) (Number of Items)	If you add together up the pairs of jeans and dresses, yous want to come up with 6 items.
"… y'all take $200 to spend from your contempo birthday coin. You detect a store that has all jeans for $25 and all dresses for $50 ."	\(25j+50d=200\) (Coin)	This one's a footling trickier. Utilise easier numbers if you need to: if y'all buy 2 pairs of jeans and 1 dress, you spend \(\left( {2\times \$25} \right)+\left( {1\times \$l} \right)\). Now you can put the variables in with their prices, and they take to add together upward to $200 .

English

Math

Caption

"You lot really, really desire to take domicile 6 items of wearable because y'all need that many."

$j+d=6$

(Number of Items)

If you add together up the pairs of jeans and dresses, yous want to come up with 6 items.

"… y'all take $200 to spend from your contempo birthday coin. You detect a store that has all jeans for $25 and all dresses for $50 ."

$25j+50d=200$

(Coin)

This one's a footling trickier. Utilise easier numbers if you need to: if y'all buy 2 pairs of jeans and 1 dress, you spend $\left( {2\times \$25} \right)+\left( {1\times \$l} \right)$. Now you can put the variables in with their prices, and they take to add together upward to $200 .

Now we have the ii equations as shown below. Notice that the $j$ variable is just like the $x$ variable and the $d$ variable is just like the $y$. It's easier to put in $j$and $d$ and then we can call up what they stand for when we go the answers.

This is what we call a system, since we take to solve for more than one variable – nosotros have to solve for 2 hither. The cool matter is to solve for ii variables, you typically demand 2 equations, to solve for three variables, yous demand iii equations, and so on. That's easy to remember, right?

We need to get an answer that works in both equations; this is what we're doing when nosotros're solving; this is called solving simultaneous systems, or solving system simultaneously. In that location are several ways to solve systems; we'll talk virtually graphing get-go.

Solving Systems by Graphing

Remember that when you graph a line, you see all the different coordinates (or $x/y$ combinations) that brand the equation work. In systems, you have to make both equations work, and so the intersection of the two lines shows the signal that fits both equations (assuming the lines practise in fact intersect; we'll talk near that later).The points of intersections satisfy both equations simultaneously.

Put these equations into the $y=mx+b$ ($d=mj+b$) format, by solving for the $d$ (which is like the $y$):

$\displaystyle j+d=vi;\text{ }\,\text{ }\text{solve for }d:\text{ }d=-j+six\text{ }$

$\displaystyle 25j+50d=200;\text{ }\,\,\text{solve for }d:\text{ }d=\frac{{200-25j}}{{fifty}}=-\frac{1}{2}j+iv$

At present graph both lines:

Solving Systems using Graph	Explanation
	To graph, solve for the "\(y\)" value ("\(d\)" in our case) to use the slope-intercept method, or keep the equations equally is and use the cover-up, or intercept method. The easiest mode to graph the second equation is the intercept method; when we put 0 in for "\(d\)", we get 8 for the "\(j\)" intercept; when nosotros put 0 in for "\(j\)", we go iv for the "\(d\)" intercept. We tin can exercise this for the first equation also, or just solve for "\(d\)" to run across that the slope is \(-1\) and the \(y\)-intercept is \(six\). The two graphs intercept at the signal \((4,2)\). This ways that the numbers that work for both equations are four pairs of jeans and 2 dresses!

Solving Systems using Graph

Explanation

To graph, solve for the "$y$" value ("$d$" in our case) to use the slope-intercept method, or keep the equations equally is and use the cover-up, or intercept method.

The easiest mode to graph the second equation is the intercept method; when we put 0 in for "$d$", we get 8 for the "$j$" intercept; when nosotros put 0 in for "$j$", we go iv for the "$d$" intercept. We tin can exercise this for the first equation also, or just solve for "$d$" to run across that the slope is $-1$ and the $y$-intercept is $six$.

The two graphs intercept at the signal $(4,2)$. This ways that the numbers that work for both equations are four pairs of jeans and 2 dresses!

Nosotros tin too employ our graphing figurer to solve the systems of equations:

Graphing Calculator Instructions	Screens
\(\displaystyle \begin{array}{c}j+d=6\text{ }\\25j+50d=200\end{array}\) Solve for \(y\,\left( d \correct)\) in both equations. Push \(Y=\) and enter the two equations in \({{Y}_{1}}=\)and \({{Y}_{2}}=\), respectively. Annotation that we don't take to simplify the equations before we take to put them in the figurer. Button GRAPH. You may demand to hit "ZOOM 6" (ZoomStandard) and/or "ZOOM 0" (ZoomFit) to make sure you see the lines crossing in the graph. (You lot tin also apply the WINDOW button to alter the minimum and maximum values of your \(x\)- and \(y\)-values.)
To get the point of intersection, push "2^nd TRACE" (CALC), and so either push 5, or move cursor down to intersect. You should come across "First curve?" at the bottom. Then button ENTER. Now you lot should run across "2d curve?" and and then press ENTER again. Now y'all should see "Estimate?". Push ENTER one more time, and yous will get the betoken of intersection on the bottom! Pretty cool!

Graphing Calculator Instructions

Screens

$\displaystyle \begin{array}{c}j+d=6\text{ }\\25j+50d=200\end{array}$

Solve for $y\,\left( d \correct)$ in both equations.

Push $Y=$ and enter the two equations in ${{Y}_{1}}=$and ${{Y}_{2}}=$, respectively. Annotation that we don't take to simplify the equations before we take to put them in the figurer.

Button GRAPH. You may demand to hit "ZOOM 6" (ZoomStandard) and/or "ZOOM 0" (ZoomFit) to make sure you see the lines crossing in the graph.

(You lot tin also apply the WINDOW button to alter the minimum and maximum values of your $x$- and $y$-values.)

To get the point of intersection, push "2^nd TRACE" (CALC), and so either push 5, or move cursor down to intersect. You should come across "First curve?" at the bottom.

Then button ENTER. Now you lot should run across "2d curve?" and and then press ENTER again. Now y'all should see "Estimate?". Push ENTER one more time, and yous will get the betoken of intersection on the bottom! Pretty cool!

Note that with non-linear equations, at that place will well-nigh likely be more than one intersection; an example of how to get more than 1 solution via the Graphing Estimator tin be plant in the Exponents and Radicals in Algebra section. Also, there are some examples of systems of inequality here in the Linear Inequalities section.

Solving Systems with Substitution

Substitution is the favorite way to solve for many students! It involves exactly what information technology says: substituting ane variable in another equation so that you only have one variable in that equation. Here is the same trouble:

You lot're going to the mall with your friends and you have $200 to spend from your recent birthday money. You observe a shop that has all jeans for $25 and all dresses for $50 . You actually, really want to have habitation 6 items of article of clothing because you "demand" that many new things. How many pairs of jeans and how many dresses yous tin purchase then you utilise the whole $200 (tax not included)?

Below are our two equations, and let's solve for "$d$" in terms of "$j$" in the first equation. Then, allow's substitute what nosotros got for "$d$" into the next equation. Even though it doesn't matter which equation you start with, remember to ever pick the "easiest" equation starting time (one that we can easily solve for a variable) to get a variable by itself.

Steps Using Exchange	Notes
\(\displaystyle \brainstorm{array}{c}j+d=\text{ }half dozen;\,\,\,\,d=6-j\\25j+50d=200\end{array}\) \(\displaystyle \brainstorm{array}{c}25j+50(half-dozen-j)=200\\25j+300-50j=200\\-25j=-100\,\,\\j=four\,\\d=6-j=half-dozen-4=two\stop{array}\)	Solve for \(d\): \(\displaystyle d=6-j\). Plug this in for \(d\) in the 2d equation and solve for \(j\). When you become the answer for \(j\), plug this back in the easier equation to get \(d\): \(\displaystyle d=6-(4)=2\). The solution is \((4,2)\).

Steps Using Exchange

Notes

$\displaystyle \brainstorm{array}{c}j+d=\text{ }half dozen;\,\,\,\,d=6-j\\25j+50d=200\end{array}$

$\displaystyle \brainstorm{array}{c}25j+50(half-dozen-j)=200\\25j+300-50j=200\\-25j=-100\,\,\\j=four\,\\d=6-j=half-dozen-4=two\stop{array}$

Solve for $d$: $\displaystyle d=6-j$. Plug this in for $d$ in the 2d equation and solve for $j$.

When you become the answer for $j$, plug this back in the easier equation to get $d$: $\displaystyle d=6-(4)=2$.

The solution is $(4,2)$.

Nosotros could purchase iv pairs of jeans and ii dresses. Note that we could have also solved for "$j$" start; information technology really doesn't thing. Yous'll want to pick the variable that'due south most easily solved for. Permit's endeavour some other exchange trouble that's a trivial bit different:

Steps Using Exchange	Notes
\(\displaystyle \begin{array}{c}\color{#800000}{\begin{array}{c}37x+4y=124\,\\x=4\,\end{array}}\\\\37(four)+4y=124\\4y=124-148\\4y=-24\\y=-6\stop{array}\)	This one is actually easier: nosotros already know that \(x=4\). Now plug in 4 for the 2d equation and solve for \(y\). The solution is \((4,-6)\).

Steps Using Exchange

Notes

$\displaystyle \begin{array}{c}\color{#800000}{\begin{array}{c}37x+4y=124\,\\x=4\,\end{array}}\\\\37(four)+4y=124\\4y=124-148\\4y=-24\\y=-6\stop{array}$

This one is actually easier: nosotros already know that $x=4$.

Now plug in 4 for the 2d equation and solve for $y$.

The solution is $(4,-6)$.

Solving Systems with Linear Combination or Elimination

Probably the nigh useful way to solve systems is using linear combination, or linear elimination. The reason information technology'due south about useful is that usually in real life nosotros don't take one variable in terms of another (in other words, a "$y=$" state of affairs).

The main purpose of the linear combination method is to add or subtract the equations so that ane variable is eliminated. We can add, subtract, or multiply both sides of equations past the same numbers – let'due south use existent numbers as shown below. We are using the Additive Property of Equality, Subtraction Holding of Equality, Multiplicative Property of Equality, and/or Sectionalisation Property of Equality that we saw here in the Types of Numbers and Algebraic Properties section:

$\displaystyle \begin{assortment}{c}\,\,\,3\,\,=\,\,three\\\underline{{+four\,\,=\,\,4}}\\\,\,\,7\,\,=\,\,vii\end{array}$

$\displaystyle \brainstorm{array}{l}\,\,\,12\,=\,12\\\,\underline{{-8\,\,=\,\,\,8}}\\\,\,\,\,\,four\,\,=\,\,4\cease{array}$

$\displaystyle \begin{array}{c}3\,\,=\,\,iii\\iv\times 3\,\,=\,\,4\times 3\\12\,\,=\,\,12\end{array}$

$\displaystyle \begin{assortment}{c}12\,\,=\,\,12\\\frac{{12}}{iii}\,\,=\,\,\frac{{12}}{3}\\iv\,\,=\,\,iv\end{array}$

If nosotros accept a set of ii equations with 2 unknowns, for example, nosotros tin can manipulate them past adding, multiplying or subtracting (nosotros usually adopt calculation) so that we get 1 equation with one variable. Let's use our previous problem:

Linear Elimination Steps	Notes
\(\displaystyle \begin{assortment}{c}\color{#800000}{\begin{array}{c}j+d=6\text{ }\\25j+50d=200\end{array}}\\\\\,\left( {-25} \right)\left( {j+d} \right)=\left( {-25} \correct)6\text{ }\\\,\,\,\,-25j-25d\,=-150\,\\\,\,\,\,\,\underline{{25j+50d\,=\,200}}\text{ }\\\,\,\,0j+25d=\,50\\\\25d\,=\,50\\d=2\\\\d+j\,\,=\,\,vi\\\,ii+j=half dozen\\j=iv\end{assortment}\)	Since we need to eliminate a variable, nosotros can multiply the first equation by –25 . Think that we need to multiply every term (annihilation separated by a plus, minus, or \(=\) sign) by the –25 . Then we add the two equations to get "\(0j\)" and eliminate the "\(j\)" variable (thus, the name "linear elimination"). We and so solve for "\(d\)". At present that we become \(d=two\), nosotros can plug in that value in the either original equation (utilise the easiest!) to go the other variable. The solution is \((4,2)\): \(j=4\) and \(d=2\).

Linear Elimination Steps

Notes

$\displaystyle \begin{assortment}{c}\color{#800000}{\begin{array}{c}j+d=6\text{ }\\25j+50d=200\end{array}}\\\\\,\left( {-25} \right)\left( {j+d} \right)=\left( {-25} \correct)6\text{ }\\\,\,\,\,-25j-25d\,=-150\,\\\,\,\,\,\,\underline{{25j+50d\,=\,200}}\text{ }\\\,\,\,0j+25d=\,50\\\\25d\,=\,50\\d=2\\\\d+j\,\,=\,\,vi\\\,ii+j=half dozen\\j=iv\end{assortment}$

Since we need to eliminate a variable, nosotros can multiply the first equation by –25 . Think that we need to multiply every term (annihilation separated by a plus, minus, or $=$ sign) by the –25 .

Then we add the two equations to get "$0j$" and eliminate the "$j$" variable (thus, the name "linear elimination"). We and so solve for "$d$".

At present that we become $d=two$, nosotros can plug in that value in the either original equation (utilise the easiest!) to go the other variable.

The solution is $(4,2)$: $j=4$ and $d=2$.

We could purchase 4 pairs of jeans and ii dresses.

Here's another example:

Linear Elimination Steps	Notes
\(\displaystyle \begin{array}{50}\color{#800000}{{2x+5y=-i}}\,\,\,\,\,\,\,\text{multiply by –}3\\\color{#800000}{{7x+3y=eleven}}\text{ }\,\,\,\,\,\,\,\text{multiply by }5\end{array}\) \(\displaystyle \begin{array}{50}-6x-15y=3\,\\\,\underline{{35x+15y=55}}\text{ }\\\,29x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=58\\\,\,\,\,\,\,\,\,\,\,\,\,\,x=2\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\2(2)+5y=-ane\\\,\,\,\,\,\,four+5y=-1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5y=-5\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=-ane\end{array}\)	Since nosotros demand to eliminate a variable, nosotros can multiply the offset equation by –3 and the second ane by 5 . There are many ways to practise this, but we want to brand sure that either the \(10\) or \(y\) will be eliminated when calculation the 2 equations. (We could take also picked multiplying the first past –7 and the 2nd past 2 ). We so get the second set of equations to add, and the \(y\)'s are eliminated. Solving for \(ten\), we get \(x=two\). At present we can plug in that value in either original equation (use the easiest!) to get the other variable. The solution is \((2,-1)\).

Linear Elimination Steps

Notes

$\displaystyle \begin{array}{50}\color{#800000}{{2x+5y=-i}}\,\,\,\,\,\,\,\text{multiply by –}3\\\color{#800000}{{7x+3y=eleven}}\text{ }\,\,\,\,\,\,\,\text{multiply by }5\end{array}$

$\displaystyle \begin{array}{50}-6x-15y=3\,\\\,\underline{{35x+15y=55}}\text{ }\\\,29x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=58\\\,\,\,\,\,\,\,\,\,\,\,\,\,x=2\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\2(2)+5y=-ane\\\,\,\,\,\,\,four+5y=-1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5y=-5\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=-ane\end{array}$

Since nosotros demand to eliminate a variable, nosotros can multiply the offset equation by –3 and the second ane by 5 . There are many ways to practise this, but we want to brand sure that either the $10$ or $y$ will be eliminated when calculation the 2 equations. (We could take also picked multiplying the first past –7 and the 2nd past 2 ).

We so get the second set of equations to add, and the $y$'s are eliminated. Solving for $ten$, we get $x=two$.

At present we can plug in that value in either original equation (use the easiest!) to get the other variable.

The solution is $(2,-1)$.

Types of equations

In the example to a higher place, we establish one unique solution to the set up of equations. Sometimes, all the same, for a set of equations, there are no solutions (when lines are parallel) or an infinite number of solutions or infinitely many solutions (when the 2 lines are actually the aforementioned line, and 1 is merely a "multiple" of the other).

When there is at least one solution, the equations are consistent equations, since they have a solution. When at that place is only ane solution, the system is called independent, since they cross at simply one point. When equations have infinite solutions, they are the aforementioned equation, are consistent, and are called dependent or ancillary (think of one merely sitting on top of the other).

When equations accept no solutions, they are called inconsistent equations, since we can never get a solution.

Here are graphs of inconsistent and dependent equations that were created on a graphing reckoner:

Systems of Equations Calculator Screens	Notes
	\(\displaystyle \begin{array}{l}y=-x+4\\y=-x-2\end{array}\) Notice that the slope of these two equations is the same, merely the \(y\)-intercepts are different. In this state of affairs, the lines are parallel, as we can come across from the graph. These types of equations are called inconsistent, since there are no solutions. If nosotros were to "solve" the two equations, nosotros'd end up with "\(4=-2\)"; no affair what values we give to \(x\) or \(y\), \(4\) can never equal \(-2\). Thus, there are no solutions. The symbol \(\emptyset \) is sometimes used for no solutions; it is called the "empty fix".
	\(\displaystyle x+y=6\,\,\,\,\,\,\,\text{or}\,\,\,\,\,\,\,y=-x+six\) \(\displaystyle 2x+2y=12\,\,\,\,\,\,\,\text{or}\,\,\,\,\,\,\,y=\frac{{-2x+12}}{ii}=-ten+6\) Sometimes we accept a situation where the system contains the same equations even though it may non be obvious. Meet how we may not know unless we actually graph, or simplify them? These types of equations are called dependent or coincident since they are ane and the aforementioned equation and they have an infinite number of solutions, since one "sits on top of" the other. Since they accept at least i solution, they are also consistent. If we were to "solve" the ii equations, we'd cease up with "\(six=half-dozen\)", and no matter what values nosotros give to \(x\) or \(y\), \(vi\) ever equals \(six\). Thus, there are an infinite number of solutions (infinitely many), but \(y\) ever has to be equal to \(-x+half dozen\). We can also write the solution as \((x,-x+6)\).

Systems of Equations Calculator Screens

Notes

$\displaystyle \begin{array}{l}y=-x+4\\y=-x-2\end{array}$

Notice that the slope of these two equations is the same, merely the $y$-intercepts are different. In this state of affairs, the lines are parallel, as we can come across from the graph.

These types of equations are called inconsistent, since there are no solutions.

If nosotros were to "solve" the two equations, nosotros'd end up with "$4=-2$"; no affair what values we give to $x$ or $y$, $4$ can never equal $-2$. Thus, there are no solutions. The symbol $\emptyset $ is sometimes used for no solutions; it is called the "empty fix".

$\displaystyle x+y=6\,\,\,\,\,\,\,\text{or}\,\,\,\,\,\,\,y=-x+six$

$\displaystyle 2x+2y=12\,\,\,\,\,\,\,\text{or}\,\,\,\,\,\,\,y=\frac{{-2x+12}}{ii}=-ten+6$

Sometimes we accept a situation where the system contains the same equations even though it may non be obvious. Meet how we may not know unless we actually graph, or simplify them?

These types of equations are called dependent or coincident since they are ane and the aforementioned equation and they have an infinite number of solutions, since one "sits on top of" the other. Since they accept at least i solution, they are also consistent.

If we were to "solve" the ii equations, we'd cease up with "$six=half-dozen$", and no matter what values nosotros give to $x$ or $y$, $vi$ ever equals $six$. Thus, there are an infinite number of solutions (infinitely many), but $y$ ever has to be equal to $-x+half dozen$. We can also write the solution as $(x,-x+6)$.

Systems with Three Equations

Permit's get a lilliputian more complicated with systems; in real life, we rarely simply have two unknowns to solve for.

Let's say at the aforementioned shop, they also had pairs of shoes for $twenty and we managed to get $lx more than to spend! Now we accept a new trouble. To spend the even $260 , how many pairs of jeans, dresses, and pairs of shoes should we go if desire, for instance, exactly ten total items (Remember that jeans cost $25 each and dresses cost $50 each).

Let'southward permit $j=$ the number of pair of jeans, $d=$ the number of dresses, and $s=$ the number of pairs of shoes we should buy. So far, we'll have the following equations:

$\displaystyle \brainstorm{assortment}{c}j+d+s=10\text{ }\\25j+\text{ }50d+\,20s=260\end{assortment}$

We'll demand another equation, since for 3 variables, we demand iii equations (otherwise, we theoretically might have space ways to solve the problem). In this blazon of problem, you would as well need something like this: Nosotros want twice every bit many pairs of jeans equally pairs of shoes . At present, since nosotros have the same number of equations every bit variables, we can potentially get one solution for the organization of equations. Here are the three equations:

$\displaystyle \begin{array}{c}j+d+due south=10\text{ }\\25j+50d+\,20s=260\\j=2s\end{array}$

Notation that when nosotros say "we have twice every bit many pairs of jeans every bit pair of shoes", information technology doesn't translate that well into math.

Nosotros can think in terms of real numbers, such every bit if we had eight pairs of jeans, we'd have 4 pairs of shoes. And then it's easier to put it in terms of the variables.

We'll learn after how to put these in our calculator to easily solve using matrices (see theMatrices and Solving Systems with Matrices section). For now, we can use ii equations at a time to eliminate a variable (using exchange and/or elimination), and keep doing this until we've solved for all variables. These can become really difficult to solve, but remember that in "real life", in that location are computers to do all this piece of work!

Remember again, that if we ever go to a point where we end upwards with something similar this, it means there are an infinite number of solutions: $iv=4$ (variables are gone and a number equals another number and they are the same). And if nosotros upwardly with something similar this, information technology ways at that place are no solutions: $v=ii$ (variables are gone and 2 numbers are left and they don't equal each other).

And another note: equations with three variables are represented past planes, non lines (you lot'll larn about this in Geometry). If all the planes crossed in merely i point, there is one solution, and if, for example, any two were parallel, we'd have no solution, and if, for example, 2 or three of them crossed in a line, nosotros'd have an infinite number of solutions.

Let's solve our system:$\displaystyle \brainstorm{assortment}{c}j+d+s=10\text{ }\\25j+\text{ }50d+20s=260\\j=2s\terminate{array}$:

Solving Systems Steps	Notes
\(\displaystyle \brainstorm{array}{c}j+d+s=x\text{ }\\25j+50d+20s=260\\j=2s\end{array}\) \(\displaystyle \begin{array}{c}2s+d+s=10\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,3s+d=x\\25(2s)+50d+\,20s=260\,\,\,\,\,\,\Rightarrow \,\,\,\,70s+50d=260\end{array}\) \(\displaystyle \brainstorm{array}{l}-150s-50d=-500\\\,\,\,\,\,\underline{{\,\,70s+50d=\,\,\,\,260}}\\\,\,-80s\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-240\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,s=three\\\\3(3)+d=x;\,\,\,\,\,d=one\,\\j=2s=two(3);\,\,\,\,\,\,j=half dozen\end{array}\)	Utilize substitution since the final equation makes that easier. Nosotros'll substitute \(2s\) for \(j\) in the other two equations and then we'll have 2 equations and 2 unknowns. We then multiply the showtime equation by –50 so nosotros can add together the two equations to get rid of the \(d\). We could accept also used substitution again. First, we get that \(s=3\), so then we can substitute this in one of the two equations we're working with. At present nosotros know that \(d=1\), so we can plug in \(d\) and \(south\) in the original showtime equation to get \(j=6\). The solution is \((half dozen,1,3)\).

Solving Systems Steps

Notes

$\displaystyle \brainstorm{array}{c}j+d+s=x\text{ }\\25j+50d+20s=260\\j=2s\end{array}$

$\displaystyle \begin{array}{c}2s+d+s=10\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,3s+d=x\\25(2s)+50d+\,20s=260\,\,\,\,\,\,\Rightarrow \,\,\,\,70s+50d=260\end{array}$

$\displaystyle \brainstorm{array}{l}-150s-50d=-500\\\,\,\,\,\,\underline{{\,\,70s+50d=\,\,\,\,260}}\\\,\,-80s\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-240\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,s=three\\\\3(3)+d=x;\,\,\,\,\,d=one\,\\j=2s=two(3);\,\,\,\,\,\,j=half dozen\end{array}$

Utilize substitution since the final equation makes that easier. Nosotros'll substitute $2s$ for $j$ in the other two equations and then we'll have 2 equations and 2 unknowns.

We then multiply the showtime equation by –50 so nosotros can add together the two equations to get rid of the $d$. We could accept also used substitution again.

First, we get that $s=3$, so then we can substitute this in one of the two equations we're working with.

At present nosotros know that $d=1$, so we can plug in $d$ and $south$ in the original showtime equation to get $j=6$.

The solution is $(half dozen,1,3)$.

We could buy six pairs of jeans, 1 clothes, and 3 pairs of shoes.

Here'due south i more than case of a 3-variable system of equations, where nosotros'll merely use linear emptying:

$\displaystyle \begin{marshal}5x-6y-\,7z\,&=\,7\\6x-4y+10z&=\,-34\\2x+4y-\,3z\,&=\,29\finish{align}$

Solving Systems Steps	Notes
\(\displaystyle \begin{assortment}{fifty}5x-6y-\,7z\,=\,\,7\\6x-4y+10z=\,-34\\2x+4y-\,3z\,=\,29\,\end{array}\)\(\displaystyle \begin{array}{l}6x-4y+10z=-34\\\underline{{2x+4y-\,3z\,=\,29}}\\8x\,\,\,\,\,\,\,\,\,\,\,\,\,+7z=-five\end{assortment}\) \(\require{cancel} \displaystyle \begin{array}{l}\cancel{{5x-6y-7z=seven}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,20x-24y-28z\,=\,28\,\\\cancel{{2x+4y-\,3z\,=29\,\,}}\,\,\,\,\,\,\,\,\underline{{12x+24y-18z=174}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,32x\,\,\,\,\,\,\,\,\,\,\,\,\,\,-46z=202\stop{array}\) \(\displaystyle \begin{array}{l}\,\,\,\cancel{{8x\,\,\,+7z=\,-5}}\,\,\,\,\,-32x\,-28z=\,20\\32x\,-46z=202\,\,\,\,\,\,\,\,\,\,\,\,\underline{{\,\,32x\,-46z=202}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-74z=222\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,z=-3\finish{array}\) \(\displaystyle \brainstorm{array}{l}32x-46(-3)=202\,\,\,\,\,\,\,\,\,\,\,\,\,ten=\frac{{202-138}}{{32}}=\frac{{64}}{{32}}=2\\\\v(two)-6y-seven(-3)=7\,\,\,\,\,\,\,\,y=\frac{{-x+-21+7}}{{-vi}}=4\end{array}\)	We first pick whatsoever 2 equations and eliminate a variable; we'll use equations 2 and 3 since nosotros tin can add them to eliminate the \(y\). We then use 2 dissimilar equations (one will be the same!) to besides eliminate the \(y\); we'll apply equations 1 and three . To eliminate the \(y\), we tin multiply the showtime past four , and the second by six . At present we use the 2 equations we've just created without the \(y\)'due south and solve them merely like a normal set of systems. We tin can multiply the first past –4 to eliminate the \(ten\)'s to get the \(z\), which is –3 . Nosotros can then get the \(x\) from either of the equations we just worked with. Since we accept the \(10\) and the \(z\), we can use whatsoever of the original equations to get the \(y\). The solution is \((2,4,-3)\).

Solving Systems Steps

Notes

$\displaystyle \begin{assortment}{fifty}5x-6y-\,7z\,=\,\,7\\6x-4y+10z=\,-34\\2x+4y-\,3z\,=\,29\,\end{array}$$\displaystyle \begin{array}{l}6x-4y+10z=-34\\\underline{{2x+4y-\,3z\,=\,29}}\\8x\,\,\,\,\,\,\,\,\,\,\,\,\,+7z=-five\end{assortment}$

$\require{cancel} \displaystyle \begin{array}{l}\cancel{{5x-6y-7z=seven}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,20x-24y-28z\,=\,28\,\\\cancel{{2x+4y-\,3z\,=29\,\,}}\,\,\,\,\,\,\,\,\underline{{12x+24y-18z=174}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,32x\,\,\,\,\,\,\,\,\,\,\,\,\,\,-46z=202\stop{array}$

$\displaystyle \begin{array}{l}\,\,\,\cancel{{8x\,\,\,+7z=\,-5}}\,\,\,\,\,-32x\,-28z=\,20\\32x\,-46z=202\,\,\,\,\,\,\,\,\,\,\,\,\underline{{\,\,32x\,-46z=202}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-74z=222\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,z=-3\finish{array}$

$\displaystyle \brainstorm{array}{l}32x-46(-3)=202\,\,\,\,\,\,\,\,\,\,\,\,\,ten=\frac{{202-138}}{{32}}=\frac{{64}}{{32}}=2\\\\v(two)-6y-seven(-3)=7\,\,\,\,\,\,\,\,y=\frac{{-x+-21+7}}{{-vi}}=4\end{array}$

We first pick whatsoever 2 equations and eliminate a variable; we'll use equations 2 and 3 since nosotros tin can add them to eliminate the $y$.

We then use 2 dissimilar equations (one will be the same!) to besides eliminate the $y$; we'll apply equations 1 and three . To eliminate the $y$, we tin multiply the showtime past four , and the second by six .

At present we use the 2 equations we've just created without the $y$'due south and solve them merely like a normal set of systems. We tin can multiply the first past –4 to eliminate the $ten$'s to get the $z$, which is –3 .

Nosotros can then get the $x$ from either of the equations we just worked with.

Since we accept the $10$ and the $z$, we can use whatsoever of the original equations to get the $y$.

The solution is $(2,4,-3)$.

I know – this is really difficult stuff! Merely if you practise information technology step-by-pace and keep using the equations you lot demand with the right variables, you tin can practice it. Think of it like a puzzle – yous may not know exactly where y'all're going, simply practice what you lot can in infant steps, and you'll get there (sort of like life sometimes, right?!). And we'll learn much easier ways to do these types of bug.

Algebra Discussion Problems with Systems

Let'south do more word problems; you lot'll notice that many of these are the same blazon that we did before in the Algebra Discussion Problems department, but at present we can utilise more than one variable. This will really make the problems easier! Again, when doing these word problems:

If you lot're wondering what the variables (or unknowns) should be when working on a word trouble, wait at what the problem is request. These are ordinarily (but not ever) what your variables are!
If you lot're not sure how to ready the equations, apply regular numbers (uncomplicated ones!) and see what you're doing. Then put the variables back in!

Here are some problems:

Investment Word Problem

Investment Word Trouble	Solution
Suppose Lindsay's mom invests $10,000 , role at three% , and the rest at ii.5% , in interest bearing accounts. The totally yearly investment income (interest) is $283 . How much did Lindsay's mom invest at each rate?	Define a variable, and look at what the problem is asking. Use ii variables: allow \(10=\) the amount of coin invested at 3% , and \(y=\) the corporeality of money invested at 2.5% . The yearly investment income or interest is the amount that nosotros get from the yearly percentages. (This is the amount of money that the depository financial institution gives u.s.a. for keeping our money there.) To go the interest, multiply each per centum by the amount invested at that rate. Add these amounts up to go the total interest. We take two equations and two unknowns. The total amount \((10+y)\) must equal $10000 , and the interest \((.03x+.025y)\) must equal $283 : \(\displaystyle \begin{array}{c}x\,+\,y=10000\\.03x+.025y=283\stop{array}\) \(\displaystyle \begin{assortment}{c}y=10000-x\\.03x+.025(10000-x)=283\\\,\,\,.03x\,+\,250\,-.025x=283\\\,.005x=33;\,\,\,\,x=6600\,\,\\\,\,y=10000-6600=3400\terminate{array}\) Plow the percentages into decimals: movement the decimal bespeak two places to the left. Substitution is the easiest manner to solve. Lindsay's mom invested $6600 at three% and $3400 at 2.v% .

Investment Word Trouble

Solution

Suppose Lindsay's mom invests $10,000 , role at three% , and the rest at ii.5% , in interest bearing accounts.

The totally yearly investment income (interest) is $283 .

How much did Lindsay's mom invest at each rate?

Define a variable, and look at what the problem is asking. Use ii variables: allow $10=$ the amount of coin invested at 3% , and $y=$ the corporeality of money invested at 2.5% .

The yearly investment income or interest is the amount that nosotros get from the yearly percentages. (This is the amount of money that the depository financial institution gives u.s.a. for keeping our money there.) To go the interest, multiply each per centum by the amount invested at that rate. Add these amounts up to go the total interest.

We take two equations and two unknowns. The total amount $(10+y)$ must equal $10000 , and the interest $(.03x+.025y)$ must equal $283 :

$\displaystyle \begin{array}{c}x\,+\,y=10000\\.03x+.025y=283\stop{array}$ $\displaystyle \begin{assortment}{c}y=10000-x\\.03x+.025(10000-x)=283\\\,\,\,.03x\,+\,250\,-.025x=283\\\,.005x=33;\,\,\,\,x=6600\,\,\\\,\,y=10000-6600=3400\terminate{array}$

Plow the percentages into decimals: movement the decimal bespeak two places to the left. Substitution is the easiest manner to solve. Lindsay's mom invested $6600 at three% and $3400 at 2.v% .

We besides could have set up this problem with a table:

	Corporeality	Turn % to decimal	Total
Corporeality at 3%	$x$	$.03$	$.03x$	Multiply across
Amount at ii.5%	$y$	$.025$	$.025y$	Multiply across
Full	$10000$		$283$	Do Cipher Here
	Add Down: $x+y=10000$	Practise Nothing Here	Add together Down: $.03x+.025y = 283$ and solve the organization

Mixture Give-and-take Issues

Here's a mixture word trouble. With mixture problems, remember if the problem calls for a pure solution or concentrate, utilise 100% (if the percent is that solution) or 0% (if the percentage is some other solution).

Mixture Discussion Trouble

Solution

Two types of milk, i that has 1% butterfat, and the other that has 3.5% of butterfat, are mixed.

How many liters of these two different kinds of milk are to be mixed together to produce 10 liters of depression-fatty milk, which has 2% butterfat?

(Note that nosotros did a similar mixture trouble using just one variable here in the Algebra Word Problems section.)

First define variables for the number of liters of each type of milk. Permit $x=$ the number of liters of the 1% milk, and $y=$ the number of liters of the 3.5% milk. Apply a table again:

	Amount	Turn % to decimal	Total
1% Milk	$x$	$.01$	$.01x$	Multiply across
3.5% Milk	$y$	$.035$	$.035y$	Multiply across
Total	$10$	$.02$	$x(.02)=.two$	Multiply across
	Add Down: $ten+y=ten$	Do Zippo Here	Add Down: $.01x+.035y=.2$ and solve the arrangement

Nosotros tin can also set up mixture problems with the type of effigy below. We add up the terms within the box, and then multiply the amounts in the boxes past the percentages to a higher place the boxes, and then add together across. This volition give us the two equations.

Let'due south do the math (utilize substitution)!

$\displaystyle \brainstorm{array}{c}x\,\,+\,\,y=x\\.01x+.035y=10(.02)\end{array}$ $\displaystyle \brainstorm{array}{c}\,y=10-x\\.01x+.035(10-x)=.2\\.01x\,+\,.35\,\,-\,.035x=.2\\\,-.025x=-.fifteen;\,\,\,\,\,x=vi\\\,y=x-6=4\end{array}$

We would need 6 liters of the i% milk, and 4 liters of the iii.5% milk.

Hither's another mixture trouble:

Mixture Discussion Trouble with Money

Solution

A shop sells ii different types of coffee beans; the more expensive one sells for $8 per pound, and the cheaper i sells for $4 per pound.

The beans are mixed to provide a mixture of 50 pounds that sells for $6.40 per pound.

How much of each type of coffee bean should be used to create l pounds of the mixture?

First define variables for the number of pounds of each type of coffee bean. Permit $10=$ the number of pounds of the $8 java, and $y=$ the number of pounds of the $4 coffee.

Employ a table again:

	Amount	Cost	Total
$8 Java Beans (lbs)	$ten$	$8$	$8x$	Multiply across
$iv Java Beans (lbs)	$y$	$4$	$4y$	Multiply across
Total	$50$	$half dozen.40$	$50(half dozen.4)=320$	Multiply across
	Add Down: $10+y=50$	Do Nothing Here	Add Down: $8x+4y=320$ and solve the arrangement

Let'south do the math (use substitution)!

$\displaystyle \begin{array}{c}10+y=50\\8x+4y=l\left( {half-dozen.4} \right)\end{array}$ $\displaystyle \begin{array}{c}y=50-x\\8x+4\left( {fifty-x} \correct)=320\\8x+200-4x=320\\4x=120\,;\,\,\,\,10=thirty\\y=50-30=20\\8x+4y=50(6.four)\cease{array}$

We would demand thirty pounds of the $8 coffee bean, and 20 pounds of the $4 coffee bean. Meet how similar this problem is to the i where we use percentages?

Distance Word Problem:

Hither'south a distance word trouble using systems; distance problems accept to do with an object's speed, time, and distance. Note that, likewise every bit the altitude word problem here in the Algebra Discussion Problems section, at that place's an example of a Parametric Distance Problem hither in the Parametric Equations section.

Distance Word Problem	Solution
Lia walks to the mall from her house at 5 mph. 10 minutes later, Lia's sister Megan starts riding her bicycle at xv mph (from the same house) to the mall to meet Lia. They get in at the mall the same time. How far is the mall from the sisters' house? How long did information technology take Megan to get there?	Retrieve always that \(\text{distance}=\text{rate}\times \text{time}\). It'southward difficult to know how to define the variables, merely usually in these types of distance issues, we want to set the variables to time, since we have rates, and we'll want to set distances equal to each other in this case (the house is always the same distance from the mall). (Sometimes we'll need to add together the distances together instead of setting them equal to each other.) Let \(L\) equal the how long (in hours) it will have Lia to go to the mall, and \(M\) equal to how long (in hours) it will take Megan to get to the mall. The rates of the Lia and Megan are 5 mph and fifteen mph respectively. (Usually a rate is "something per something"). Lia's time is Megan'due south timeplus \(\displaystyle \frac{{ten}}{{60}}=\frac{1}{6}\,\,\,\,(50=M+\frac{i}{6})\), since Lia left 10 minutes before than Megan (convert minutes to hours by dividing past threescore – try existent numbers to meet this). Employ the altitude formula for each of them separately, and then fix their distances equal, since they are both traveling the same distance (house to mall). Then use commutation to solve the system for Megan'southward time: after dividing both sides by 5 , multiply both sides by 6 to get rid of the fractions. \(\begin{assortment}{c}L=K+\frac{ane}{half-dozen};\,\,\,\,\,5L=15M\\5\left( {M+\frac{1}{half dozen}} \right)=15M;\,\,\,Chiliad+\frac{1}{half dozen}=3M\,\,\\6M+1=18M;\,\,\,12M=ane;\,\,M\,\,=\frac{1}{{12}}\,\,\text{hr}\text{.}\\D=15\left( {\frac{one}{{12}}} \right)=1.25\,\,\text{miles}\end{array}\) Megan's time is \(\displaystyle \frac{ane}{{12}}\) of any 60 minutes, which is 5 minutes. The distance to the mall is rate times time, which is ane.25 miles.

Distance Word Problem

Solution

Lia walks to the mall from her house at 5 mph. 10 minutes later, Lia's sister Megan starts riding her bicycle at xv mph (from the same house) to the mall to meet Lia. They get in at the mall the same time.

How far is the mall from the sisters' house?

How long did information technology take Megan to get there?

Retrieve always that $\text{distance}=\text{rate}\times \text{time}$. It'southward difficult to know how to define the variables, merely usually in these types of distance issues, we want to set the variables to time, since we have rates, and we'll want to set distances equal to each other in this case (the house is always the same distance from the mall). (Sometimes we'll need to add together the distances together instead of setting them equal to each other.)

Let $L$ equal the how long (in hours) it will have Lia to go to the mall, and $M$ equal to how long (in hours) it will take Megan to get to the mall. The rates of the Lia and Megan are 5 mph and fifteen mph respectively. (Usually a rate is "something per something"). Lia's time is Megan'due south timeplus $\displaystyle \frac{{ten}}{{60}}=\frac{1}{6}\,\,\,\,(50=M+\frac{i}{6})$, since Lia left 10 minutes before than Megan (convert minutes to hours by dividing past threescore – try existent numbers to meet this).

Employ the altitude formula for each of them separately, and then fix their distances equal, since they are both traveling the same distance (house to mall). Then use commutation to solve the system for Megan'southward time: after dividing both sides by 5 , multiply both sides by 6 to get rid of the fractions.

$\begin{assortment}{c}L=K+\frac{ane}{half-dozen};\,\,\,\,\,5L=15M\\5\left( {M+\frac{1}{half dozen}} \right)=15M;\,\,\,Chiliad+\frac{1}{half dozen}=3M\,\,\\6M+1=18M;\,\,\,12M=ane;\,\,M\,\,=\frac{1}{{12}}\,\,\text{hr}\text{.}\\D=15\left( {\frac{one}{{12}}} \right)=1.25\,\,\text{miles}\end{array}$

Megan's time is $\displaystyle \frac{ane}{{12}}$ of any 60 minutes, which is 5 minutes. The distance to the mall is rate times time, which is ane.25 miles.

Which Plumber Problem

Many word issues yous'll take to solve have to do with an initial charge or setup charge, and a charge or rate per fourth dimension period. In these cases, the initial charge will be the $\boldsymbol {y}$-intercept, and therate will be the gradient. Here is an example:

"Which Plumber" Systems Word Problem	Solution
Michaela's mom is trying to decide between ii plumber companies to ready her sink. The get-go company charges $l for a service call, plus an additional $36 per hour for labor. The second company charges $35 for a service call, plus an additional $39 per 60 minutes of labor. At how many hours will the 2 companies charge the same amount of money?	The money spent depends on the plumber'south set upwardly charge and number of hours, and then permit \(y=\) the full cost of the plumber, and \(x=\) the number of hours of labor. Again, set upwardly charges are typical \(\boldsymbol {y}\)-intercepts, and rates per 60 minutes areslopes. The total price of the plumber'due south firm call will be the initial or setup charge, plus the number of hours (\(10\)) at the house times the toll per hour for labor. To become the number of hours when the two companies charge the aforementioned amount of money, we just put the two \(y\)'s together and solve for \(x\)(substitution, correct?): First plumber's full price: \(\displaystyle y=50+36x\) Second plumber's total price: \(\displaystyle y=35+39x\) \(\displaystyle 50+36x=35+39x;\,\,\,\,\,\,ten=five\) Here's what a graph would look like: At v hours, the ii plumbers will charge the same. At this time, the \(y\)-value is 230 , so the total price is $230 . Note that, in the graph, before 5 hours, the kickoff plumber will be more expensive (because of the higher setup charge), only after the first five hours, the second plumber will exist more than expensive. Thus, the plumber would be called based on how many hours Michaela'south mom thinks the plumber will exist in that location.

"Which Plumber" Systems Word Problem

Solution

Michaela's mom is trying to decide between ii plumber companies to ready her sink.

The get-go company charges $l for a service call, plus an additional $36 per hour for labor. The second company charges $35 for a service call, plus an additional $39 per 60 minutes of labor.

At how many hours will the 2 companies charge the same amount of money?

The money spent depends on the plumber'south set upwardly charge and number of hours, and then permit $y=$ the full cost of the plumber, and $x=$ the number of hours of labor. Again, set upwardly charges are typical $\boldsymbol {y}$-intercepts, and rates per 60 minutes areslopes. The total price of the plumber'due south firm call will be the initial or setup charge, plus the number of hours ($10$) at the house times the toll per hour for labor.

To become the number of hours when the two companies charge the aforementioned amount of money, we just put the two $y$'s together and solve for $x$(substitution, correct?):

First plumber's full price: $\displaystyle y=50+36x$

Second plumber's total price: $\displaystyle y=35+39x$

$\displaystyle 50+36x=35+39x;\,\,\,\,\,\,ten=five$

Here's what a graph would look like:

At v hours, the ii plumbers will charge the same. At this time, the $y$-value is 230 , so the total price is $230 . Note that, in the graph, before 5 hours, the kickoff plumber will be more expensive (because of the higher setup charge), only after the first five hours, the second plumber will exist more than expensive. Thus, the plumber would be called based on how many hours Michaela'south mom thinks the plumber will exist in that location.

Geometry Give-and-take Problem:

Many times, we'll accept a geometry problem as an algebra word problem; these might involve perimeter, expanse, or sometimes angle measurements (and then don't forget these things!). Let'southward do i involving angle measurements.

Geometry Systems Discussion Problem	Solution
Two angles are supplementary. The mensurate of 1 angle is 30 degrees smaller than twice the other. Discover the measure of each angle.	From Geometry, we know that ii angles are supplementary if their bending measurements add together up to 180 degrees (and remember likewise that two angles are complementary if their angle measurements add up to 90 degrees). Define the variables and turn English language into Math. Let \(x=\) the first bending, and \(y=\) the second bending. We actually don't need to worry at this point about which angle is bigger; the math will have care of itself. \(ten\) plus \(y\) must equal 180 degrees past definition, and also \(x=2y-xxx\) (Remember the English-to-Math nautical chart?) Solve, using exchange: \(\displaystyle \begin{array}{c}ten+y=180\\x=2y-30\finish{assortment}\) \(\displaystyle \begin{array}{c}2y-30+y=180\\3y=210;\,\,\,\,\,\,\,\,y=70\\x=ii\left( {lxx} \right)-30=110\stop{array}\) The larger angle is 110 ° , and the smaller is seventy ° . Let's check our work: The two angles exercise in fact add up to 180° , and the larger angle ( 110° ) is xxx° less than twice the smaller ( 70° ).

Geometry Systems Discussion Problem

Solution

Two angles are supplementary. The mensurate of 1 angle is 30 degrees smaller than twice the other.

Discover the measure of each angle.

From Geometry, we know that ii angles are supplementary if their bending measurements add together up to 180 degrees (and remember likewise that two angles are complementary if their angle measurements add up to 90 degrees).

Define the variables and turn English language into Math. Let $x=$ the first bending, and $y=$ the second bending. We actually don't need to worry at this point about which angle is bigger; the math will have care of itself.

$ten$ plus $y$ must equal 180 degrees past definition, and also $x=2y-xxx$ (Remember the English-to-Math nautical chart?) Solve, using exchange:

$\displaystyle \begin{array}{c}ten+y=180\\x=2y-30\finish{assortment}$

$\displaystyle \begin{array}{c}2y-30+y=180\\3y=210;\,\,\,\,\,\,\,\,y=70\\x=ii\left( {lxx} \right)-30=110\stop{array}$

The larger angle is 110 ° , and the smaller is seventy ° . Let's check our work: The two angles exercise in fact add up to 180° , and the larger angle ( 110° ) is xxx° less than twice the smaller ( 70° ).

Run into – these are getting easier! Here'south one that'south a picayune tricky though:

Work Problem :

Let'southward do a "work trouble" that is typically seen when studying Rational Equations (fraction with variables in them) and can be plant here in the Rational Functions, Equations and Inequalities section.

Note that there's also a simpler version of this problem hither in the Direct, Inverse, Joint and Combined Variation section.

Work Word Trouble (Systems)	Solution
eight women and 12 girls can paint a large landscape in 10 hours. half dozen women and eight girls can paint it in 14 hours. Observe the time to paint the landscape, past 1 woman alone, and ane girl alone.	Permit's let \(w=\) the role of the task past 1 adult female in 1 hour, and \(g=\) the function of the job by ane girl in i hour. We have 10 hours with 8 women and 12 girls that paint the mural (practice ane job), and fourteen hours with 6 women and 8 girls that paint the mural (practise 1 job). Since \(due west=\) the part of the job that is completed by 1 woman in 1 hour, then \(8w=\) the amount of the chore that is completed by 8 women in i hour. Too, if \(8w=\) the amount of the task that is completed past 8 women in 1 hr, \(10\times 8w\) is the amount of the job that is completed by 8 women in 10 hours. Similarly, \(10\times 12g\) is the corporeality of the job that is completed past 12 girls in x hours. Add these two amounts and we get \(\displaystyle 10\left( {8w+12g} \right)\), which will be the whole job. Use the same logic for the vi women and viii girls to pigment the mural in 14 hours. The whole job is 1 (this is typical in work problems), and we can set up two equations that equal 1 to solve the organisation. Use linear elimination to solve the equations; it gets a footling messy with the fractions, only nosotros tin get it! \(\displaystyle \begin{assortment}{c}ten\left( {8w+12g} \right)=1,\text{ or }8w+12g=\frac{1}{{x}}\\\,fourteen\left( {6w+8g} \correct)=one,\text{ or }\,6w+8g=\frac{ane}{{fourteen}}\terminate{array}\) \(\displaystyle \begin{array}{c}\text{Apply emptying:}\\\left( {-half dozen} \right)\left( {8w+12g} \right)=\frac{ane}{{10}}\left( {-half dozen} \right)\\\left( 8 \right)\left( {6w+8g} \right)=\frac{1}{{14}}\left( 8 \right)\\\cancel{{-48w}}-72g=-\frac{iii}{5}\\\abolish{{48w}}+64g=\frac{4}{7}\,\\\,-8g=-\frac{1}{{35}};\,\,\,\,\,m=\frac{1}{{280}}\cease{array}\) \(\begin{assortment}{c}\text{Substitute in start equation to become }w:\\\,10\left( {8w+12\cdot \frac{one}{{280}}} \correct)=1\\\,80w+\frac{{120}}{{280}}=one;\,\,\,\,\,\,w=\frac{i}{{140}}\\g=\frac{ane}{{280}};\,\,\,\,\,\,\,\,\,\,\,west=\frac{1}{{140}}\end{array}\) The answers we become is the part of the job that is completed by 1 woman or daughter in 1 hr, so to get how long it would accept them to do a whole job, we have to accept the reciprocal. (Recollect near it; if we could consummate \(\displaystyle \frac{1}{iii}\) of a job in an hour, we could complete the whole job in 3 hours). Thus, it would take 1 of the women 140 hours to paint the mural by herself, and ane of the girls 280 hours to pigment the mural by herself.

Work Word Trouble

(Systems)

Solution

eight women and 12 girls can paint a large landscape in 10 hours.

half dozen women and eight girls can paint it in 14 hours.

Observe the time to paint the landscape, past 1 woman alone, and ane girl alone.

Permit's let $w=$ the role of the task past 1 adult female in 1 hour, and $g=$ the function of the job by ane girl in i hour. We have 10 hours with 8 women and 12 girls that paint the mural (practice ane job), and fourteen hours with 6 women and 8 girls that paint the mural (practise 1 job).

Since $due west=$ the part of the job that is completed by 1 woman in 1 hour, then $8w=$ the amount of the chore that is completed by 8 women in i hour. Too, if $8w=$ the amount of the task that is completed past 8 women in 1 hr, $10\times 8w$ is the amount of the job that is completed by 8 women in 10 hours.

Similarly, $10\times 12g$ is the corporeality of the job that is completed past 12 girls in x hours. Add these two amounts and we get $\displaystyle 10\left( {8w+12g} \right)$, which will be the whole job. Use the same logic for the vi women and viii girls to pigment the mural in 14 hours.

The whole job is 1 (this is typical in work problems), and we can set up two equations that equal 1 to solve the organisation. Use linear elimination to solve the equations; it gets a footling messy with the fractions, only nosotros tin get it!

$\displaystyle \begin{assortment}{c}ten\left( {8w+12g} \right)=1,\text{ or }8w+12g=\frac{1}{{x}}\\\,fourteen\left( {6w+8g} \correct)=one,\text{ or }\,6w+8g=\frac{ane}{{fourteen}}\terminate{array}$

$\displaystyle \begin{array}{c}\text{Apply emptying:}\\\left( {-half dozen} \right)\left( {8w+12g} \right)=\frac{ane}{{10}}\left( {-half dozen} \right)\\\left( 8 \right)\left( {6w+8g} \right)=\frac{1}{{14}}\left( 8 \right)\\\cancel{{-48w}}-72g=-\frac{iii}{5}\\\abolish{{48w}}+64g=\frac{4}{7}\,\\\,-8g=-\frac{1}{{35}};\,\,\,\,\,m=\frac{1}{{280}}\cease{array}$ $\begin{assortment}{c}\text{Substitute in start equation to become }w:\\\,10\left( {8w+12\cdot \frac{one}{{280}}} \correct)=1\\\,80w+\frac{{120}}{{280}}=one;\,\,\,\,\,\,w=\frac{i}{{140}}\\g=\frac{ane}{{280}};\,\,\,\,\,\,\,\,\,\,\,west=\frac{1}{{140}}\end{array}$

The answers we become is the part of the job that is completed by 1 woman or daughter in 1 hr, so to get how long it would accept them to do a whole job, we have to accept the reciprocal. (Recollect near it; if we could consummate $\displaystyle \frac{1}{iii}$ of a job in an hour, we could complete the whole job in 3 hours).

Thus, it would take 1 of the women 140 hours to paint the mural by herself, and ane of the girls 280 hours to pigment the mural by herself.

Iii Variable Word Problem:

Let'south do ane more with three equations and 3 unknowns:

Iii Variable Word Problem	Solution
A florist is making five identical bridesmaid bouquets for a wedding. She has $610 to spend (including taxation) and wants 24 flowers for each bouquet. Roses cost $6 each, tulips price $iv each, and lilies cost $3 each. She wants to take twice every bit many roses every bit the other 2 flowers combined in each boutonniere. How many roses, tulips, and lilies are in each bouquet? The trick is to put real numbers in to make certain you're doing the problem correctly, and likewise make sure you're answering what the question is asking!	Expect at the question being asked to define our variables: Let \(r=\) the number of roses, \(t=\) the number of tulips, and \(l=\) the number of lilies in each bouquet. Put the money terms together, and besides the counting terms together: \(\begin{array}{50}five\left( {6r+4t+3l} \right)=610\,\,\,\text{(price of each flower times number of flowers x }5\text{ bouquets= total price)}\\\,\,\,\,\,\,\,\,\,r=2(t+l)\text{ }\,\,\,\,\,\,\,\,\,\,\text{ (two times the sum of the other two flowers = number of roses)}\\\,\,\,\,\,\,r+t+l=24\text{ }\,\,\,\,\,\,\,\,\,\text{(full number of flowers in each bouquet is }24\text{)}\cease{array}\) Utilise substitution and put \(r\) from the center equation in the other equations. And then, use linear emptying to put those ii equations together: multiply the second by –5 to eliminate the \(50\). Nosotros typically accept to use two separate pairs of equations to get the three variables down to 2! \(\begin{array}{c}6r+4t+3l=122\\r=2\left( {t+l} \correct)\\\,r+t+l=24\\\\6\left( {2t+2l} \right)+4t+3l=122\\\,12t+12l+4t+3l=122\\16t+15l=122\\\\\left( {2t+2l} \right)+t+l=24\\3t+3l=24\stop{assortment}\) \(\displaystyle \begin{array}{c}\,16t+15l=122\\\,\,\,\,\,\,\,\,\,\cancel{{3t+3l=24}}\\\,\,\,\,\underline{{-15t-15l=-120}}\\\,\,\,\,\,t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=ii\\16\left( ii \correct)+15l=122;\,\,\,\,50=6\\\\r=2\left( {2+6} \correct)=16\\\,\,\,\,\,\,\,\,\,\,r=16,\,\,\,t=2,\,\,\,l=6\cease{array}\) Nosotros get \(t=2\). Solve for \(l\) in this aforementioned arrangement, and \(r\) by using the value we got for \(t\) and \(l\) (virtually hands in the 2nd equation at the meridian). Thus, for one bouquet, we'll have 16 roses, 2 tulips, and half-dozen lilies. If we had solved for the total number of flowers, we would accept had to carve up each number by 5 .

Iii Variable Word Problem

Solution

A florist is making five identical bridesmaid bouquets for a wedding.

She has $610 to spend (including taxation) and wants 24 flowers for each bouquet. Roses cost $6 each, tulips price $iv each, and lilies cost $3 each. She wants to take twice every bit many roses every bit the other 2 flowers combined in each boutonniere.

How many roses, tulips, and lilies are in each bouquet?

The trick is to put real numbers in to make certain you're doing the problem correctly, and likewise make sure you're answering what the question is asking!

Expect at the question being asked to define our variables: Let $r=$ the number of roses, $t=$ the number of tulips, and $l=$ the number of lilies in each bouquet. Put the money terms together, and besides the counting terms together:

$\begin{array}{50}five\left( {6r+4t+3l} \right)=610\,\,\,\text{(price of each flower times number of flowers x }5\text{ bouquets= total price)}\\\,\,\,\,\,\,\,\,\,r=2(t+l)\text{ }\,\,\,\,\,\,\,\,\,\,\text{ (two times the sum of the other two flowers = number of roses)}\\\,\,\,\,\,\,r+t+l=24\text{ }\,\,\,\,\,\,\,\,\,\text{(full number of flowers in each bouquet is }24\text{)}\cease{array}$

Utilise substitution and put $r$ from the center equation in the other equations. And then, use linear emptying to put those ii equations together: multiply the second by –5 to eliminate the $50$. Nosotros typically accept to use two separate pairs of equations to get the three variables down to 2!

$\begin{array}{c}6r+4t+3l=122\\r=2\left( {t+l} \correct)\\\,r+t+l=24\\\\6\left( {2t+2l} \right)+4t+3l=122\\\,12t+12l+4t+3l=122\\16t+15l=122\\\\\left( {2t+2l} \right)+t+l=24\\3t+3l=24\stop{assortment}$ $\displaystyle \begin{array}{c}\,16t+15l=122\\\,\,\,\,\,\,\,\,\,\cancel{{3t+3l=24}}\\\,\,\,\,\underline{{-15t-15l=-120}}\\\,\,\,\,\,t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=ii\\16\left( ii \correct)+15l=122;\,\,\,\,50=6\\\\r=2\left( {2+6} \correct)=16\\\,\,\,\,\,\,\,\,\,\,r=16,\,\,\,t=2,\,\,\,l=6\cease{array}$

Nosotros get $t=2$. Solve for $l$ in this aforementioned arrangement, and $r$ by using the value we got for $t$ and $l$ (virtually hands in the 2nd equation at the meridian). Thus, for one bouquet, we'll have 16 roses, 2 tulips, and half-dozen lilies. If we had solved for the total number of flowers, we would accept had to carve up each number by 5 .

The "Candy" Trouble

Sometimes nosotros get lucky and can solve a system of equations where nosotros have more unknowns (variables) then equations. (Actually, I recall it's not then much luck, just having good problem writers!) Hither'south i like that:

More Unknowns Than Variables Problem	Solution
Sarah buys 2 pounds of jelly beans and iv pounds of chocolates for $four.00 . She and so buys 1 pound of jelly beans and 4 pounds of caramels for $three.00 . She besides buys 1 pound of jelly beans, 3 pounds of licorice and ane pound of caramels for $1.50 . How much will it cost to buy 1 pound of each of the four candies?	Await at the question being asked to ascertain our variables: Let \(j=\) the cost of 1 pound of jelly beans, \(o=\) the cost of i pound of chocolates, \(c=\) the cost of 1 pound of caramels, and \(l=\) the toll of ane pound of licorice. Hither is the organisation of equations: \(\brainstorm{array}{c}2j+4o=four\\j+4c=three\\j+3l+1c=1.5\\\text{Want: }j+o+c+l\finish{assortment}\) Await! Something's not correct since nosotros have 4 variables and iii equations. But annotation that they are not asking for the cost of each candy, only the toll to buy all 4 ! Maybe the problem will just "piece of work out" and so nosotros tin solve it; allow's try and see. From our three equations above (using substitution), we get values for\(o\), \(c\) and \(l\) in terms of \(j\). \(\displaystyle \begin{marshal}o=\frac{{four-2j}}{four}=\frac{{2-j}}{two}\,\,\,\,\,\,\,\,\,c=\frac{{iii-j}}{4}\,\\j+3l+one\left( {\frac{{iii-j}}{4}} \right)=1.5\\4j+12l+3-j=vi\\\,l=\frac{{half dozen-3-3j}}{{12}}=\frac{{iii-3j}}{{12}}=\frac{{1-j}}{4}\end{align}\) \(\crave{cancel} \displaystyle \begin{align}j+o+c+fifty&=j+\frac{{2-j}}{2}+\frac{{iii-j}}{4}+\frac{{1-j}}{4}\\&=\cancel{j}+ane-\cancel{{\frac{i}{two}j}}+\frac{3}{iv}\cancel{{-\frac{j}{iv}}}+\frac{1}{four}\abolish{{-\frac{j}{4}}}=2\stop{align}\) When nosotros substitute back in the sum \(\text{ }j+o+c+fifty\), all in terms of \(j\), our \(j\) ' s actually cancel out, which is very unusual! We tin can't really solve for all the variables, since we don't know what \(j\) is. But we can meet that the total cost to buy ane pound of each of the candies is $2 . Pretty cool!

More Unknowns Than Variables Problem

Solution

Sarah buys 2 pounds of jelly beans and iv pounds of chocolates for $four.00 .

She and so buys 1 pound of jelly beans and 4 pounds of caramels for $three.00 .

She besides buys 1 pound of jelly beans, 3 pounds of licorice and ane pound of caramels for $1.50 .

How much will it cost to buy 1 pound of each of the four candies?

Await at the question being asked to ascertain our variables: Let $j=$ the cost of 1 pound of jelly beans, $o=$ the cost of i pound of chocolates, $c=$ the cost of 1 pound of caramels, and $l=$ the toll of ane pound of licorice. Hither is the organisation of equations:

$\brainstorm{array}{c}2j+4o=four\\j+4c=three\\j+3l+1c=1.5\\\text{Want: }j+o+c+l\finish{assortment}$

Await! Something's not correct since nosotros have 4 variables and iii equations. But annotation that they are not asking for the cost of each candy, only the toll to buy all 4 ! Maybe the problem will just "piece of work out" and so nosotros tin solve it; allow's try and see.

From our three equations above (using substitution), we get values for$o$, $c$ and $l$ in terms of $j$.

$\displaystyle \begin{marshal}o=\frac{{four-2j}}{four}=\frac{{2-j}}{two}\,\,\,\,\,\,\,\,\,c=\frac{{iii-j}}{4}\,\\j+3l+one\left( {\frac{{iii-j}}{4}} \right)=1.5\\4j+12l+3-j=vi\\\,l=\frac{{half dozen-3-3j}}{{12}}=\frac{{iii-3j}}{{12}}=\frac{{1-j}}{4}\end{align}$ $\crave{cancel} \displaystyle \begin{align}j+o+c+fifty&=j+\frac{{2-j}}{2}+\frac{{iii-j}}{4}+\frac{{1-j}}{4}\\&=\cancel{j}+ane-\cancel{{\frac{i}{two}j}}+\frac{3}{iv}\cancel{{-\frac{j}{iv}}}+\frac{1}{four}\abolish{{-\frac{j}{4}}}=2\stop{align}$

When nosotros substitute back in the sum $\text{ }j+o+c+fifty$, all in terms of $j$, our $j$ ' s actually cancel out, which is very unusual! We tin can't really solve for all the variables, since we don't know what $j$ is. But we can meet that the total cost to buy ane pound of each of the candies is $2 . Pretty cool!

There are more Systems Word Problems in theMatrices and Solving Systems with Matrices section, Linear Programming section, and Right Triangle Trigonometry section.

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