Solve Applications with Systems of Equations

Solve Direct Translation Applications

Systems of linear equations are very useful for solving applications. Some people find setting up word problems with two variables easier than setting them up with just one variable. To solve an application, we’ll first translate the words into a system of linear equations. Then we will decide the most convenient method to use, and then solve the system.

Example. The sum of two numbers is zero. One number is nine less than the other. Find the numbers.

We are looking for two numbers. Let $n=$ the first number and $m=$ the second number.

“The sum of two numbers is zero” translates to one equation, and “one number is nine less than the other” translates to the second:

We solve by substitution. Substitute $m-9$ for $n$ in the first equation:

Substituting $m=\tfrac{9}{2}$ into $n=m-9$ gives $n=\tfrac{9}{2}-9=-\tfrac{9}{2}$.

Check: is the sum zero? $\tfrac{9}{2}+\left(-\tfrac{9}{2}\right)\overset{?}{=}0$, and indeed $0=0$. ✓ Is one number nine less than the other? $\tfrac{9}{2}-9\overset{?}{=}-\tfrac{9}{2}$, and indeed $-\tfrac{9}{2}=-\tfrac{9}{2}$. ✓ The numbers are $\tfrac{9}{2}$ and $-\tfrac{9}{2}$.

Example. Heather has been offered two options for her salary as a trainer at the gym. Option A would pay her $25,000 plus $15 for each training session. Option B would pay her $10,000 plus $40 for each training session. How many training sessions would make the salary options equal?

We are looking for the number of training sessions that would make the pay equal. Let $s=$ Heather’s salary and $n=$ the number of training sessions.

“Option A would pay her $25,000 plus $15 for each training session” translates to $s=25{,}000+15n$. “Option B would pay her $10,000 plus $40 for each training session” translates to $s=10{,}000+40n$:

We solve by substitution. Substitute $25{,}000+15n$ for $s$ in the second equation:

Check: are $600$ training sessions a year reasonable? Yes. Are the two options equal when $n=600$? $25{,}000+15(600)=34{,}000$ and $10{,}000+40(600)=34{,}000$. ✓ The salary options would be equal for $600$ training sessions.

Example. When Jenna spent 10 minutes on the elliptical trainer and then did circuit training for 20 minutes, her fitness app said she burned 278 calories. When she spent 20 minutes on the elliptical trainer and 30 minutes of circuit training she burned 473 calories. How many calories does she burn for each minute on the elliptical trainer? How many calories does she burn for each minute of circuit training?

We are looking for the number of calories burned each minute on the elliptical trainer and each minute of circuit training. Let $e=$ calories burned per minute on the elliptical trainer and $c=$ calories burned per minute of circuit training.

Each workout gives us one equation relating the two rates:

We solve by elimination. Multiply the first equation by $-2$ so the $e$-terms have opposite coefficients, then add the equations:

So $c=8.3$. Substituting $c=8.3$ into the first equation:

Check: $10(11.2)+20(8.3)=112+166=278$ ✓, and $20(11.2)+30(8.3)=224+249=473$ ✓. Jenna burns $11.2$ calories per minute on the elliptical trainer and $8.3$ calories per minute of circuit training.

Solve Geometry Applications

When we learned about geometry applications earlier, we solved them using properties of triangles and rectangles. Now we’ll add some properties of angles. The measures of two complementary angles add to $90$ degrees. The measures of two supplementary angles add to $180$ degrees.

Example. Translate to a system of equations and then solve: the difference of two complementary angles is 26 degrees. Find the measures of the angles.

We are looking for the measure of each angle. Let $x=$ the measure of the first angle and $y=$ the measure of the second angle.

“The angles are complementary” translates to $x+y=90$. “The difference of the two angles is 26 degrees” translates to $x-y=26$:

We solve by elimination — the $y$-terms already have opposite coefficients, so we just add the equations:

So $x=58$. Substituting $x=58$ into $x+y=90$ gives $58+y=90$, so $y=32$.

Check: $58+32\overset{?}{=}90$, and indeed $90=90$. ✓ $58-32\overset{?}{=}26$, and indeed $26=26$. ✓ The angle measures are $58$ degrees and $32$ degrees.

Example. Translate to a system of equations and then solve: two angles are supplementary. The measure of the larger angle is twelve degrees less than five times the measure of the smaller angle. Find the measures of both angles.

Let $x=$ the measure of the smaller angle and $y=$ the measure of the larger angle.

“The angles are supplementary” translates to $x+y=180$. “The larger angle is twelve less than five times the smaller angle” translates to $y=5x-12$:

We solve by substitution. Substitute $5x-12$ for $y$ in the first equation:

Substituting $x=32$ into $y=5x-12$ gives $y=5(32)-12=160-12=148$.

Check: $32+148\overset{?}{=}180$, and indeed $180=180$. ✓ $5(32)-12\overset{?}{=}148$, and indeed $148=148$. ✓ The angle measures are $32$ degrees and $148$ degrees.

Recall that the angles of a triangle add up to $180$ degrees. A right triangle has one angle that measures $90$ degrees. What does that tell us about the other two angles? In the next example we’ll find the measures of the other two angles.

Example. The measure of one of the small angles of a right triangle is ten more than three times the measure of the other small angle. Find the measures of both angles.

We are looking for the measures of the angles. Let $a=$ the measure of the first angle and $b=$ the measure of the second angle.

“One small angle is ten more than three times the other” translates to $a=3b+10$. Since the triangle’s angles sum to $180$ and one angle is the right angle ($90$ degrees), the two small angles satisfy $a+b+90=180$, or $a+b=90$:

We solve by substitution. Substitute $3b+10$ for $a$ in the second equation:

Substituting $b=20$ into $a=3b+10$ gives $a=3(20)+10=70$.

Check: $a+b+90\overset{?}{=}180$: $70+20+90\overset{?}{=}180$, and indeed $180=180$. ✓ $3(20)+10\overset{?}{=}70$, and indeed $70=70$. ✓ The measures of the small angles are $70$ degrees and $20$ degrees.

Often it is helpful to draw a picture to visualize a geometry application.

Example. Translate to a system of equations and then solve: Randall has 125 feet of fencing to enclose the rectangular part of his backyard adjacent to his house. He will only need to fence around three sides, because the fourth side will be the wall of the house. He wants the length of the fenced yard (parallel to the house wall) to be 5 feet more than four times as long as the width. Find the length and the width.

We are looking for the length and width of the fenced yard. Let $L=$ the length of the fenced yard and $W=$ the width of the fenced yard.

Since one length and two widths make up the $125$ feet of fencing, $L+2W=125$. Since “the length will be 5 feet more than four times as long as the width,” $L=4W+5$:

We solve by substitution. Substitute $4W+5$ for $L$ in the first equation:

Substituting $W=20$ into $L=4W+5$ gives $L=4(20)+5=85$.

Check: is one length plus two widths equal to $125$ feet? $85+20+20\overset{?}{=}125$, and indeed $125=125$. ✓ Is the length $5$ more than four times the width? $4(20)+5\overset{?}{=}85$, and indeed $85=85$. ✓ The length is $85$ feet and the width is $20$ feet.

Solve Uniform Motion Applications

We used a table to organize the information in uniform motion problems when we introduced them earlier. We’ll continue using that table here. The basic equation is $D=rt$, where $D$ is the distance traveled, $r$ is the rate, and $t$ is the time. Now that we can use two variables and two equations, we no longer need to write one person’s time or rate in terms of the other’s.

Example. Translate to a system of equations and then solve: Joni left St. Louis on the interstate, driving west towards Denver at a speed of 65 miles per hour. Half an hour later, Kelly left St. Louis on the same route as Joni, driving 78 miles per hour. How long will it take Kelly to catch up to Joni?

We are looking for the length of time Kelly, $k$, and Joni, $j$, will each drive. Since $D=rt$, a table organizes the rates, times, and distances:

Kelly and Joni will have driven the same distance when Kelly catches up, so $65j=78k$. Since Kelly left later, her time will be $\tfrac{1}{2}$ hour less than Joni’s, so $k=j-\tfrac{1}{2}$:

We solve by substitution. Substitute $j-\tfrac{1}{2}$ for $k$ in the first equation:

Substituting $j=3$ into $k=j-\tfrac{1}{2}$ gives $k=3-\tfrac{1}{2}=2\tfrac{1}{2}$.

Check: Joni drives $3$ hours at $65$ mph, covering $195$ miles. Kelly drives $2\tfrac{1}{2}$ hours at $78$ mph, also covering $195$ miles — the same distance. ✓ Kelly will catch up to Joni in $2\tfrac{1}{2}$ hours; by then Joni will have traveled $3$ hours.

Many real-world uniform motion applications arise because of the effect of a current — of water or air — on an object’s actual speed. Picture a boat with a still-water speed of $b$ traveling on a river whose current moves at speed $c$. Going downstream, in the same direction as the current, the current helps push the boat along, so its actual speed is $b+c$. Going upstream, against the current, the current slows the boat down, so its actual speed is $b-c$. Wind currents affect airplane speeds the same way: a tailwind (blowing in the same direction the plane flies) gives an actual speed of $j+w$, and a headwind (blowing against the plane) gives an actual speed of $j-w$, where $j$ is the plane’s speed in still air and $w$ is the wind’s speed.

Example. Translate to a system of equations and then solve: a river cruise ship sailed 60 miles downstream for 4 hours and then took 5 hours sailing upstream to return to the dock. Find the speed of the ship in still water and the speed of the river current.

We are looking for the speed of the ship in still water and the speed of the current. Let $s=$ the rate of the ship in still water and $c=$ the rate of the current. Going downstream, the current helps the ship, so the actual rate is $s+c$. Going upstream, the current slows the ship, so the actual rate is $s-c$.

Since rate times time is distance, we get the system of equations:

We solve by elimination. First distribute, then divide each equation by its common factor to simplify:

Adding the two equations:

So $s=13.5$. Substituting $s=13.5$ into $s+c=15$ gives $c=1.5$.

Check: downstream, the ship’s actual rate is $13.5+1.5=15$ mph, and in $4$ hours it travels $15\cdot4=60$ miles. ✓ Upstream, its actual rate is $13.5-1.5=12$ mph, and in $5$ hours it travels $12\cdot5=60$ miles. ✓ The rate of the ship is $13.5$ mph and the rate of the current is $1.5$ mph.

Example. Translate to a system of equations and then solve: a private jet can fly 1,095 miles in three hours with a tailwind but only 987 miles in three hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Let $j=$ the speed of the jet in still air and $w=$ the speed of the wind. In a tailwind, the wind helps the jet, so the rate is $j+w$. In a headwind, the wind slows the jet, so the rate is $j-w$.

Since rate times time is distance, the system of equations is:

We solve by elimination. Distribute, then add the equations:

So $j=347$. Substituting $j=347$ into $3(j+w)=1{,}095$ gives $3(347+w)=1{,}095$, so $1{,}041+3w=1{,}095$, and $w=18$.

Check: with the tailwind, the jet’s actual rate is $347+18=365$ mph, and in $3$ hours it flies $365\cdot3=1{,}095$ miles. ✓ Into the headwind, its actual rate is $347-18=329$ mph, and in $3$ hours it flies $329\cdot3=987$ miles. ✓ The rate of the jet is $347$ mph and the rate of the wind is $18$ mph.

Key terms

complementary angles — two angles whose measures add to $90$ degrees; each is called the complement of the other. supplementary angles — two angles whose measures add to $180$ degrees; each is called the supplement of the other. tailwind — a wind current blowing in the same direction a plane is flying, so it increases the plane’s actual speed to $j+w$. headwind — a wind current blowing against the direction a plane is flying, so it decreases the plane’s actual speed to $j-w$.

This section is adapted from Intermediate Algebra 2e, Section 4.2: Solve Applications with Systems of Equations by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: recreated the right-triangle and fenced-yard figures as accessible inline graphics and described the river-current and wind-current illustrations in prose; wrote each system of equations and its elimination or substitution steps as display-math blocks; omitted the “Be Prepared” readiness quiz, the Media link, and the Section 4.2 Exercises (“Practice Makes Perfect”) block; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.