Systems of Linear Equations and Word Problems

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This section covers:

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

Introduction to Systems

“Systems of equations” just means that we are dealing with more than one equation and variable. So far, we’ve basically just played around with the equation for a line, which is \(y=mx+b\).

But let’s say we have the following situation. 

You’re going to the mall with your friends and you have $200 to spend from your recent birthday money. You discover a store that has all jeans for $25 and all dresses for $50. You really, really want to take home 6 items of clothing because you “need” that many new things.

Wouldn’t it be clever to find out how many pairs of jeans and how many dresses you can buy so you use the whole $200 (tax not included – your parents promised to pay the tax)?

Now, you can always do “guess and check” to see what would work, but you might as well use algebra! It’s much better to learn the algebra way, because even though this problem is fairly simple to solve, the algebra way will let you solve any algebra problem – even the really complicated ones.

The first trick in problems like this is to figure out what we want to know. This will help us decide what variables (unknowns) to use. What we want to know is how many pairs of jeans we want to buy (let’s say “\(j\)”) and how many dresses we want to buy (let’s say “\(d\)”). Always write down what your variables will be:

  Let \(j=\) the number of jeans you will buy

Let \(d=\) the number of dresses you’ll buy

Like we did before, let’s translate word-for-word from math to English:

English

Math

Explanation

“You really, really want to take home 6 items of clothing because you need that many.”

\(j+d=6\)

If you add up the pairs of jeans and dresses, you want to come up with 6 items.
“… you have $200 to spend from your recent birthday money. You discover a store that has all jeans for $25 and all dresses for $50.”

\(25j+50d=200\)

This one’s a little trickier. Use easier numbers if you need to: if you buy 2 pairs of jeans and 1 dress, you spend \(\left( {2\times \$25} \right)+\left( {1\times \$50} \right)\). Now you can put the variables in with their prices, and they have to add up to $200.

Now we have the 2 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 \(d\) so we can remember what they stand for when we get the answers.

This is what we call a system, since we have to solve for more than one variable – we have to solve for 2 here. The cool thing is to solve for 2 variables, you typically need 2 equations, to solve for 3 variables, you need 3 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 we’re solving; this is called solving simultaneous systems, or solving system simultaneously.

There are several ways to solve systems; we’ll talk about graphing first.

Solving Systems by Graphing

Remember that when you graph a line, you see all the different coordinates (or \(x/y\) combinations) that make the equation work. In systems, you have to make both equations work, so the intersection of the two lines shows the point that fits both equations (assuming the lines do in fact intersect; we’ll talk about that later). So the points of intersections satisfy both equations simultaneously. 

We’ll need to put these equations into the \(y=mx+b\) (\(d=mj+b\)) format, by solving for the \(d\) (which is like the \(y\)):

\(\displaystyle j+d=6;\text{ }\,\text{ }\text{solve for }d:\text{ }d=-j+6\text{ }\)

\(\displaystyle 25j+50d=200;\text{ }\,\,\text{solve for }d:\text{ }d=\frac{{200-25j}}{{50}}=-\frac{1}{2}j+4\)

Now let’s graph:

Graph

Explanation

First of all, to graph, we had to either solve for the “\(y\)” value (“\(d\)” in our case) like we did above, or use the cover-up, or intercept method.

 

The easiest way for the second equation would be the intercept method; when we put 0 in for “\(d\)”, we get 8 for the “\(j\)” intercept; when we put 0 in for “\(j\)”, we get 4 for the “\(d\)” intercept.

 

We can do this for the first equation too, or just solve for “\(d\)”.

 

We can see the two graphs intercept at the point \((4,2)\). This means that the numbers that work for both equations is 4 pairs of jeans and 2 dresses!

We can see the two graphs intercept at the point \((4,2)\). This means that the numbers that work for both equations is 4 pairs of jeans and 2 dresses!

We can also use our graphing calculator to solve the systems of equations:

Graphing Calculator Instructions

Screen

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

 

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

 

Push \(Y=\) and enter the two equations in \({{Y}_{1}}=\) and \({{Y}_{2}}=\), respectively. Note that we don’t have to simplify the equations before we have to put them in the calculator.

 

Push GRAPH. You may need to hit “ZOOM 6” (ZoomStandard) and/or “ZOOM 0” (ZoomFit) to make sure you see the lines crossing in the graph.

 

(You can also use the WINDOW button to change the minimum and maximum values of your \(x\) and \(y\) values.)

 

To get the point of intersection, push “2nd TRACE” (CALC), and then either push 5, or move cursor down to intersect. You should see “First curve?” at the bottom.

 

Then push ENTER. Now you should see “Second curve?” and then press ENTER again. Now you should see “Guess?”. Push ENTER one more time, and you will get the point of intersection on the bottom! Pretty cool!

 

(Note that with non-linear equations, there will most likely be more than one intersection; an example of how to get more than one solution via the Graphing Calculator can be found in the Exponents and Radicals in Algebra section.)

Solving Systems with Substitution

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

Here is the problem again:

You’re going to the mall with your friends and you have $200 to spend from your recent birthday money. You discover a store that has all jeans for $25 and all dresses for $50. You really, really want to take home 6 items of clothing because you “need” that many new things. Wouldn’t it be clever to find out how many pairs of jeans and how many dresses you can buy with your $200 (tax not included – your parents promised to pay the tax)?

Below are our two equations, and let’s solve for “\(d\)” in terms of “\(j\)” in the first equation. Then, let’s substitute what we got for “\(d\)” into the next equation.

Even though it doesn’t matter which equation you start with, remember to always pick the “easiest” equation first (one that we can easily solve for a variable) to get a variable by itself.

Steps Using Substitution

Notes

\(\displaystyle \begin{array}{c}\color{#800000}{\begin{array}{c}j+d=\text{ }6\\25j+50d=200\end{array}}\\\\25j+50(-j+6)=200\\25j-50j+300=200\\-25j=-100\,\,\\j=4\,\\d=-j+6=-4+6=2\end{array}\) Solve for \(d\): \(\displaystyle d=-j+6\). Plug this in for \(d\) in the second equation and solve for \(j\).

 

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

 

The solution is \((4,2)\).

We could buy 4 pairs of jeans and 2 dresses. Note that we could have also solved for “\(j\)” first; it really doesn’t matter. You’ll want to pick the variable that’s most easily solved for. Let’s try another substitution problem that’s a little bit different:

Steps Using Substitution

Notes

\(\displaystyle \begin{array}{c}\color{#800000}{\begin{array}{c}37x+4y=124\,\\x=4\,\end{array}}\\\\37(4)+4y=124\\4y=124-148\\4y=-24\\y=-6\end{array}\) This one is actually easier: we already know that \(x=4\).

 

Now plug in 4 for the second equation and solve for \(y\).

 

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

Solving Systems with Linear Combination or Elimination

Probably the most useful way to solve systems is using linear combination, or linear elimination. The reason it’s most useful is that usually in real life we don’t have one variable in terms of another (in other words, a “\(y=\)” situation).

The main purpose of the linear combination method is to add or subtract the equations so that one variable is eliminated. Now let’s see why we can add, subtract, or multiply both sides of equations by the same numbers – let’s use real numbers as shown below. Remember these are because of the Additive Property of Equality, Subtraction Property of Equality, Multiplicative Property of Equality, and Division Property of Equality:

\(\displaystyle \begin{array}{c}\,\,\,3\,\,=\,\,3\\\underline{{+4\,\,=\,\,4}}\\\,\,\,7\,\,=\,\,7\end{array}\) \(\displaystyle \begin{array}{l}\,\,\,12\,=\,12\\\,\underline{{-8\,\,=\,\,\,8}}\\\,\,\,\,\,4\,\,=\,\,4\end{array}\) \(\displaystyle \begin{array}{c}3\,\,=\,\,3\\4\times 3\,\,=\,\,4\times 3\\12\,\,=\,\,12\end{array}\) \(\displaystyle \begin{array}{c}12\,\,=\,\,12\\\frac{{12}}{3}\,\,=\,\,\frac{{12}}{3}\\4\,\,=\,\,4\end{array}\)

So now if we have a set of 2 equations with 2 unknowns, we can manipulate them by adding, multiplying or subtracting (we usually prefer adding) so that we get one equation with one variable. For, example, let’s use our previous problem:

Linear Elimination Steps Notes
\(\displaystyle \begin{array}{c}\color{#800000}{\begin{array}{c}j+d=6\text{ }\\25j+50d=200\end{array}}\\\\\,\left( {-25} \right)\left( {j+d} \right)=\left( {-25} \right)6\text{ }\\\,\,\,\,-25j-25d\,=-150\,\\\,\,\,\,\,\underline{{25j+50d\,=\,200}}\text{ }\\\,\,\,0j+25d=\,50\\\\25d\,=\,50\\d=2\\\\d+j\,\,=\,\,6\\\,2+j=6\\j=4\end{array}\) Since we need to eliminate a variable, we can multiply the first equation by –25. Remember that we need to multiply every term (anything 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 then solve for “\(d\)”.

 

Now that we get \(d=2\), we can plug in that value in the either original equation (use the easiest!) to get the other variable.

 

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

We could buy 4 pairs of jeans and 2 dresses.

Here’s another example:

Linear Elimination Steps Notes

\(\displaystyle \begin{array}{l}\color{#800000}{{2x+5y=-1}}\,\,\,\,\,\,\,\text{multiply by}-3\\\color{#800000}{{7x+3y=11}}\text{ }\,\,\,\,\,\,\,\text{multiply by }5\end{array}\)

 

\(\displaystyle \begin{array}{l}-6x-15y=3\,\\\,\underline{{35x+15y=55}}\text{ }\\\,29x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=58\\\,\,\,\,\,\,\,\,\,\,\,\,\,x=2\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\2(2)+5y=-1\\\,\,\,\,\,\,4+5y=-1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5y=-5\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=-1\end{array}\)

Since we need to eliminate a variable, we can multiply the first equation by –3 and the second one by 5. There are many ways to do this, but we want to make sure that either the \(x\) or \(y\) will be eliminated when adding the two equations.  (We could have also picked multiplying the first by –7 and the second by 2).

 

We then get the second set of equations to add, and the \(y\)’s are eliminated.  Solving for \(x\), we get \(x=2\).

 

Now 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 above, we found one unique solution to the set of equations. Sometimes, however, there are no solutions (when lines are parallel) or an infinite number of solutions (when the two lines are actually the same line, and one is just a “multiple” of the other) to a set of equations.

When there is at least one solution, the equations are consistent equations, since they have a solution. When there is only one solution, the system is called independent, since they cross at only one point.

When equations have infinite solutions, they are the same equation, are consistent, and are called dependent or coincident (think of one just sitting on top of the other). When equations have 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 the graphing calculator:

Inconsistent Equations Dependent Equations Consistent Equations

Systems with Three Equations

Let’s get a little more complicated with systems; in real life, we rarely just have two unknowns with two equations.

Let’s say at the same store, they also had pairs of shoes for $20 and we managed to get $60 more from our parents since our parents are so great!

Now we have a new problem: to spend the even $260, how many pairs of jeans, dresses, and pairs of shoes should we get if want say exactly 10 total items?

Let’s let \(j=\) 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:

\(\displaystyle \begin{array}{c}j+d+s=10\text{ }\\25j+\text{ }50d+\,20s=260\end{array}\)

We’ll need another equation, since for three variables, we need three equations (otherwise, we’d theoretically have infinite ways to solve the problem). In this type of problem, you would also have/need something like this: we want twice as many pairs of jeans as pairs of shoes. Now, since we have the same number of equations as variables, we can potentially get one solution for the system.

So, again, now we have three equations and three unknowns (variables). We’ll learn later how to put these in our calculator to easily solve using matrices (see the Matrices and Solving Systems with Matrices section), but for now we need to first use two of the equations to eliminate one of the variables, and then use two other equations to eliminate the same variable:

 \(\displaystyle \begin{array}{c}j+d+s=10\text{ }\\25j+50d+\,20s=260\\j=2s\end{array}\) Note that when we say “we have twice as many pairs of jeans as pair of shoes”, it doesn’t translate that well into math.

 

We can think in terms of real numbers, such as if we had 8 pairs of jeans, we’d have 4 pairs of shoes. Then it’s easier to put it in terms of the variables.

Now this gets more difficult to solve, but remember that in “real life”, there are computers to do all this work!

The trick to do these problems “by hand” is to keep working on the equations using either substitution or elimination until we get the answers.

Remember again, that if we ever get to a point where we end up with something like this, it means there are an infinite number of solutions:

       \(4=4\)  (variables are gone and a number equals another number and they are the same)

And if we up with something like this, it means there are no solutions:  

        \(5=2\)  (variables are gone and two numbers are left and they don’t equal each other)

Let’s go for it and solve:     \(\displaystyle \begin{array}{c}j+d+s=10\text{ }\\25j+\text{ }50d+20s=260\\j=2s\end{array}\):

Solving Systems Steps

Notes

\(\displaystyle \begin{array}{c}j+d+s=10\text{ }\\25j+50d+20s=260\\j=2s\end{array}\)

 

\(\displaystyle \begin{align}2s+d+s&=10\\25(2s)+50d+\,20s&=260\\70s+50d&=260\end{align}\)

 

\(\displaystyle \begin{array}{l}-150s-50d=-500\\\,\,\,\,\,\underline{{\,\,70s+50d=\,\,\,\,260}}\\\,\,-80s\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-240\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,s=3\\\\3(3)+d=10;\,\,\,\,\,d=1\,\\j=2s=2(3);\,\,\,\,\,\,j=6\end{array}\)

Use substitution since the last equation makes that easier. We’ll substitute \(2s\) for \(j\) in the other two equations and then we’ll have 2 equations and 2 unknowns.

 

We then multiply the first equation by –50 so we can add the two equations to get rid of the \(d\). We could have also used substitution again.

 

First, we get that \(s=3\), so then we can substitute this in one of the 2 equations we’re working with.

 

Now we know that \(d=1\), so we can plug in \(d\) and \(s\) in the original first equation to get \(j=6\).

 

The solution is \((6,1,3)\).

We could buy 6 pairs of jeans, 1 dress, and 3 pairs of shoes.

Here’s one more example of a three-variable system of equations, where we’ll only use linear elimination:

\(\displaystyle \begin{align}5x-6y-\,7z\,&=\,7\\6x-4y+10z&=\,-34\\2x+4y-\,3z\,&=\,29\end{align}\)

Solving Systems Steps

Notes

 

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

 

 

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

 

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

 

\(\displaystyle \begin{array}{l}32x-46(-3)=202\,\,\,\,\,\,\,\,\,\,\,\,\,x=\frac{{202-138}}{{32}}=\frac{{64}}{{32}}=2\\\\5(2)-6y-\,\,7(-3)\,=\,\,7\,\,\,\,\,\,\,\,y=\frac{{-10+-21+7}}{{-6}}=4\end{array}\)

We first pick any 2 equations and eliminate a variable; we’ll use equations 2 and 3 since we can add them to eliminate the \(y\).

 

We then use 2 different equations (one will be the same!) to also eliminate the \(y\); we’ll use equations 1 and 3. To eliminate the \(y\), we’ll have to multiply the first by 4, and the second by 6.

 

Now we use the 2 equations we’ve just created without the \(y\)’s and solve them just like a normal set of systems. We needed to multiply the first by –4 to eliminate the \(x\)’s to get the \(z\).

 

We can then get the \(x\) from the second equation that we just worked with.

 

Since we have the \(x\) and the \(z\), we can use any of the original equations to get the \(y\).

 

So the solution is \((2,4,-3)\).

I know – this is really difficult stuff!  But if you do it step-by-step and keep using the equations you need with the right variables, you can do it. Think of it like a puzzle – you may not know exactly where you’re going, but do what you can in baby steps, and you’ll get there (sort of like life!).

And we’ll learn much easier ways to do these types of problems.

Also – note that equations with three variables are represented by planes, not lines (you’ll learn about this in Geometry). They could have 1 solution (if all the planes crossed in only one point), no solution (if say two of them were parallel), or an infinite number of solutions (say if two or three of them crossed in a line). OK, enough Geometry for now!

Algebra Word Problems with Systems

Let’s do more word problems; you’ll notice that many of these are the same type that we did earlier in the Algebra Word Problems section, but now we can use more than one variable. This will actually make the problems easier! Again, when doing these word problems:

  • If you’re wondering what the variable (or unknown) should be when working on a word problem, look at what the problem is asking. This is usually what your variable is!
  • If you’re not sure how to set up the equations, use regular numbers (simple ones!) and see what you’re doing. Then put the variables back in!

Investment Word Problem:

Suppose Lindsay’s mom invests $10000, part at 3%, and the rest at 2.5%, in interest bearing accounts. The totally yearly investment income (interest) is $283. How much did Lindsay’s mom invest at each rate?

Solution:

We always have to define a variable, and we can look at what they are asking. Since we’ve learned about systems, let’s use two variables: let \(x=\) the amount of money invested at 3%, and \(y=\) the amount of money invested at 2.5%.

Remember that the yearly investment income or interest is the amount that we get from the yearly percentages. (This is the amount of money that the bank gives us for keeping our money there.)  To get the interest, we have to multiply each percentage by the amount invested at that rate. We can add these amounts up to get the total interest.

We have two equations and two unknowns. We know that the total amount (\(x+y\)) must equal 10000, and we also know that the interest (\(.03x+.025y\)) must equal 283:

Steps Notes
 \(\displaystyle \begin{array}{c}\color{#800000}{\begin{array}{c}x\,+\,y=10000\\.03x+.025y=283\end{array}}\\\\y=10000-x\\.03x+.025(10000-x)=283\\\,\,\,.03x\,+\,250\,-.025x=283\\\,.005x=33\\x=6600\\\,\,y=10000-6600=3400\end{array}\) Remember to turn the percentages into decimals: move the decimal point 2 places to the left.

 

Substitution is the easiest way to solve.

 

We see that Lindsay’s mom invested $6600 at 3% and $3400 at 2.5%.

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

  Amount Turn % to decimal Total  
Amount at 3% \(x\) \(.03\) \(.03x\) Multiply across
Amount at 2.5% \(y\) \(.025\) \(.025y\) Multiply across
Total \(10000\) \(283\) Do Nothing Here
Add Down:

\(x+y=10000\)

Do Nothing Here Add Down: \(.03x+.025y = 283\) and solve the system

Mixture Word Problems:

Two types of milk, one 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 low-fat milk, which has 2% butterfat?

Solution:

(Note that we did a similar mixture problem using only one variable here in the Algebra Word Problems section.)

Let’s first define two variables for the number of liters of each type of milk. Let \(x=\) the number of liters of the 1% milk, and \(y=\) the number of liters of the 3.5% milk.

Let’s use a table again:

Amount Turn % to decimal Total
1% Milk \(x\) \(.01\) \(.01x\) Multiply across
3.5% Milk \(y\) \(.035\) \(.035y\) Multiply across
Total \(10\) \(.02\) \(10(.02)=.2\) Multiply across
Add Down:

\(x+y=10\)

Do Nothing Here Add Down: \(.01x+.035y=.2\) and solve the system

We can also set up mixture problems with the type of figure below. We add up the terms inside the box, and then multiply the amounts in the boxes by the percentages above the boxes, and then add across. This will give us the two equations. System of Equations Mixture Problem with Boxes Now let’s do the math!

Steps Notes
 \(\displaystyle \begin{array}{c}\color{#800000}{\begin{array}{c}x+y=10\\.01x+.035y=10(.02)\end{array}}\\\\\,y=10-x\\.01x+.035(10-x)=.2\\.01x\,+\,.35\,-\,.035x=.2\\\,-.025x=-.15\\x=6\\\,y=10-6=4\end{array}\) Again, let’s use substitution.

 

So, we would need 6 liters of the 1% milk, and 4 liters of the 3.5% milk.

 

Also, remember if mixture problems call for a pure solution, use 100% for the percentage.

Mixture Word Problem with Money:

A store sells two different types of coffee beans; the more expensive one sells for $8 per pound, and the cheaper one 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 50 pounds of the mixture?

Solution: Let’s first define two variables for the number of pounds of each type of coffee bean. Let \(x=\) the number of pounds of the $8 coffee, and $y=$ the number of pounds of the $4 coffee.

Let’s use a table again:

Amount Cost Total
$8 Coffee Beans (lbs) \(x\) \(8\) \(8x\) Multiply across
$4 Coffee Beans (lbs) \(y\) \(4\) \(4y\) Multiply across
Total \(50\) \(6.40\) \(50(6.4)=320\) Multiply across
Add Down:

\(x+y=50\)

Do Nothing Here Add Down: \(8x+4y=320\) and solve the system

Now let’s do the math:

Math Notes
\(\displaystyle \begin{array}{c}x+y=50\\8x+4y=50(6.4)\\\,y=50-x\\8x+4(50-x)=320\\8x\,\,+\,\,\,200\,\,\,-\,\,\,4x=320\\4x=120\\x=30\\\,y=50-30=20\end{array}\) Again, let’s use substitution.

 

So, we would need 30 pounds of the $8 coffee bean, and 20 pounds of the $4 coffee bean.

 

See how similar this problem is to the one where we use percentages?

Distance Word Problem:

Lia walks to the mall from her house at 5 mph. 10 minutes later, Lia’s sister Megan starts riding her bike at 15 mph (from the same house) to the mall to meet Lia. They arrive at the mall the same time. How far is the mall from the sisters’ house? How long did it take Megan to get there?

Solution:

OK, this is another tough one. Remember always that  \(\text{distance}=\text{rate}\times \text{time}\). It’s difficult to know how to define the variables, but usually in these types of distance problems, we want to set the variables to time, since we have rates, and we’ll want to set distances equal to each other (the house is always the same distance from the mall).

Let’s let \(L\) equal the how long (in hours) it will take Lia to get to the mall, and \(M\) equal to how long (in hours) it will take Megan to get to the mall. (Sometimes we’ll need to add the distances together instead of setting them equal to each other.)

We must use the distance formula for each of them separately, and then we can set their distances equal, since they are both traveling the same distance (house to mall).

Let’s draw a picture and work the problem: Systems of Equations Distance Problem Note that there’s an example of a Parametric Distance Problem here in the Parametric Equations section.

Which Plumber Problem:

Many word problems you’ll have to solve have to do with an initial charge or setup charge, and a charge or rate per time period. In these cases, the initial charge will be the y-intercept, and the rate will be the slope. Here is an example:

Michaela’s mom is trying to decide between two plumber companies to fix her sink.  The first company charges $50 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 hour of labor. At how many hours will the two companies charge the same amount of money?

In these cases, the money spent depends on the plumber’s set up charge and number of hours, so let \(y=\) total cost of the plumber, and \(x=\) number of hours of labor. And again, set up charges are typical \(y\)-intercepts, and rates per hour are slopes.

Solution: To get the number of hours when the two companies charge the same amount of money, we just put the two \(y\)’s together and solve for \(x\) (substitution, right?):

Plumber Problem

Geometry Word Problem:

Many times we’ll have a geometry problem as an algebra word problem; these might involve perimeter, area, or sometimes angle measurements (so don’t forget these things!). Let’s do one involving angle measurements. Two angles are supplementary.  The measure of one angle is 30 degrees smaller than twice the other. Find the measure of each angle.

Solution:

We have to know that two angles are supplementary if their angle measurements add up to 180 degrees (and remember also that two angles are complementary if their angle measurements add up to 90 degrees, in case you see that).

Let’s define the variables and turn English into Math. Let \(x=\) the first angle, and \(y=\) the second angle; we really don’t need to worry at this point about which angle is bigger; the math will take care of itself.

Then we know that \(x\) plus \(y\) must equal 180 degrees by definition, and also \(x=2y-30\). (Remember the English-to-Math chart?)  Let’s solve:

Math Notes
 \(\displaystyle \begin{array}{c}\color{#800000}{\begin{array}{c}x\,\,+\,\,y=180\\x=2y-30\end{array}}\\\,\\2y-30+y=180\\3y=210\\y=70\\x=2(70)-30=110\end{array}\) Again, let’s use substitution, since we have \(x\) equals something.

 

The larger angle is 110°, and the smaller is 70°.

 

Let’s check our work: The two angles do in fact add up to 180°, and the larger angle (110°) is 30° less than twice the smaller (70°).

See – these are getting easier!   Here’s one that’s a little tricky though:

Work Problem

8 women and 12 girls can paint a large mural in 10 hours. 6 women and 8 girls can paint it in 14 hours. Find the time to paint the mural, by 1 woman alone, and 1 girl alone.

Solution:

(This is a “work problem” that is typically seen when studying Rational Equations – fraction with variables in them –  and can be found here in the Rational Functions, Equations and Inequalities section.)  But let’s solve it with using systems.  (There’s also a simpler version of this problem here in the Direct, Inverse, Joint and Combined Variation section).

Let’s let \(w=\) the part of the job by 1 woman in 1 hour, and \(g=\) the part of the job by 1 girl in 1 hour. Let’s set up and solve:

Work Problems With Systems

Let’s do one more with three equations and three unknowns:

Three Variable Word Problem:

A florist is making 5 identical bridesmaid bouquets for a wedding. She has $610 to spend (including tax) and wants 24 flowers for each bouquet. Roses cost $6 each, tulips cost $4 each, and lilies cost $3 each. She wants to have twice as many roses as the other 2 flowers combined in each bouquet. How many roses, tulips, and lilies are in each bouquet?

Solution:

Let’s look at the question that is being asked and define our variables: Let \(r=\) the number of roses, \(t=\) the number of tulips, and \(l=\) the number of lilies. Let’s put the money terms together, and also the counting terms together:

\(\begin{array}{l}6r+4t+3l=610\text{ }\,\,\,\text{ (price of each flower times number of each flower = total price)}\\\,\,\,\,\,\,\,\,\,r=2(t+l)\text{ }\,\,\,\,\,\,\,\,\,\,\text{ (two times the sum of the other two flowers = number of roses)}\\\,\,\,\,\,r+t+l=5(24)\text{ }\,\,\,\text{ (total flowers = 5 bouquets, each with 24 flowers)}\end{array}\)

Now let’s do the math:

Systems of Equations Three Equations

The “Candy” Problem

Sometimes we get lucky and can solve a system of equations where we have more unknowns (variables) then equations. (Actually, I think it’s not so much luck, but having good problem writers!)  Here’s one like that:

Sarah buys 2 lb of jelly beans and 4 lb of chocolates for $4.00. She then buys 1 lb of jelly beans and 4 lbs of caramels for $3.00. She also buys 1 lb of jelly beans, 3 lbs of licorice and 1 lb of caramels for $1.50. How much will it cost to buy 1 lb of each of the four candies?

Solution:

Let’s look at the question that is being asked and define our variables: Let \(j=\)  the cost of 1 lb of jelly beans, \(o=\) the cost of 1 lb of chocolates, \(l=\) the cost of 1 lb of licorice, and \(c=\) the cost of 1 lb of caramels. We have this system of equations:

 \(\begin{array}{c}2j+4o=4\\j+4c=3\\j+3l+1c=1.5\\\text{Want: }\,\,j+o+c+l\end{array}\) Wait! Something’s not right since we have 4 variables and 3 equations.  But note that they are not asking for the cost of each candy, but the cost to buy all 4! So maybe the problem will just “work out” so we can solve it; we just have to try it to see.

Now let’s try to do the math:

Math

Notes

\(\begin{array}{c}2j+4o=4\\j+4c=3\\j+3l+1c=1.5\\\text{Want: }j+o+c+l\end{array}\)

 

\(\displaystyle \begin{align}o=\frac{{4-2j}}{4}=\frac{{2-j}}{2}\,\,\,\,\,\,\,\,\,c=\frac{{3-j}}{4}\,\\j+3l+1\left( {\frac{{3-j}}{4}} \right)=1.5\\4j+12l+3-j=6\\\,l=\frac{{6-3-3j}}{{12}}=\frac{{3-3j}}{{12}}=\frac{{1-j}}{4}\end{align}\)

 

\(\require{cancel} \displaystyle \begin{align}j+o+c+l=j+\frac{{2-j}}{2}+\frac{{3-j}}{4}+\frac{{1-j}}{4}\\=\cancel{j}+1-\cancel{{\frac{1}{2}j}}+\frac{3}{4}\cancel{{-\frac{j}{4}}}+\frac{1}{4}\cancel{{-\frac{j}{4}}}=2\end{align}\)

Using substitution, let’s try to eliminate all variables but one (\(j\)).

 

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

 

When we substitute in \(\text{ }j+o+c+l\), our \(j\)s actually cancel out, which is very unusual!

 

We can’t really solve for all the variables, since we don’t know what \(j\) is. But we can see that the cost to buy 1 lb of each of the candies is $2.

You can find a Right Triangle Trigonometry systems problem here in the Right Triangle Trigonometry section.

Understand these problems, and practice, practice, practice!


For Practice: Use the Mathway widget below to try a Systems of Equations problem. Click on Submit (the blue arrow to the right of the problem) and click on Solve by Substitution or Solve by Addition/Elimination to see the answer.

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On to Algebraic Functions, including Domain and Range – you’re ready!