Solve Applications with Systems of Equations

Translate to a System of Equations

Many of the problems we solved in earlier applications related two quantities. Here are two examples from Use a Problem-Solving Strategy:

• The sum of two numbers is negative fourteen. One number is four less than the other. Find the numbers.
• A married couple together earns $110,000 a year. The wife earns $16,000 less than twice what her husband earns. What does the husband earn?

In that chapter we translated each situation into one equation using only one variable. Now that we can solve systems of equations, let’s translate these same two problems into a system of two equations in two variables. We’ll focus on Steps 1 through 4 of our problem-solving strategy.

Example. Translate to a system of equations: the sum of two numbers is negative fourteen. One number is four less than the other. Find the numbers.

This is a number problem, and we are looking for two numbers. Let $m =$ one number and $n =$ the second number.

“The sum of the numbers is negative fourteen” translates to one equation, and “one number is four less than the other” translates to the second:

Example. Translate to a system of equations: a married couple together earns $110,000 a year. The wife earns $16,000 less than twice what her husband earns. What does the husband earn?

We are looking for the amount that the husband and wife each earn. Let $h =$ the amount the husband earns and $w =$ the amount the wife earns.

“A married couple together earns $110,000” translates to one equation, and “the wife earns $16,000 less than twice what her husband earns” translates to the second:

Solve Direct Translation Applications

We just set up, but did not solve, the systems of equations in the two examples above. Now we’ll translate a situation to a system of equations and then solve it, using all seven steps of the strategy.

Example. Devon is 26 years older than his son Cooper. The sum of their ages is 50. Find their ages.

We are looking for the ages of Devon and Cooper. Let $d =$ Devon’s age and $c =$ Cooper’s age.

“Devon is 26 years older than Cooper” translates to $d = c + 26$. “The sum of their ages is 50” translates to $d + c = 50$:

We solve by substitution. Substitute $c + 26$ for $d$ in the second equation:

Substituting $c = 12$ into the first equation gives $d = 12 + 26 = 38$.

Check: is Devon’s age 26 more than Cooper’s? $12 + 26 = 38$. ✓ Is the sum of their ages 50? $38 + 12 = 50$. ✓ Devon is 38 years old and Cooper is 12 years old.

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 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 cancel when we add:

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. 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. 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.

Example. 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. 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 $D = r \cdot t$, we can fill in the distance column of the table below.

To translate into a system of equations, we recognize that Kelly and Joni will drive the same distance when Kelly catches up, so $65j = 78k$. Also, since Kelly left later, her time will be $\tfrac{1}{2}$ hour less than Joni’s time, 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. 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 $15 \cdot 4 = 60$ miles. ✓ Upstream, its actual rate is $13.5 - 1.5 = 12$ mph, and in 5 hours it travels $12 \cdot 5 = 60$ miles. ✓ The rate of the ship is 13.5 mph and the rate of the current is 1.5 mph.

Example. 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 $365 \cdot 3 = 1{,}095$ miles. ✓ Into the headwind, its actual rate is $347 - 18 = 329$ mph, and in 3 hours it flies $329 \cdot 3 = 987$ 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 Elementary Algebra 2e, Section 5.4: 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 Randall’s fenced-yard figure as an accessible inline graphic and described the river-current and wind-current figures in prose; wrote each system of equations and its elimination or substitution steps as display-math blocks; corrected an arithmetic slip in the source’s own worked solution for the supplementary-angle example (the check step’s $5(32) - 12$ was printed as 147 instead of 148); omitted the “Be Prepared” readiness quiz, the Media links block, the Section 5.4 Exercises (“Practice Makes Perfect”) block, and the Self Check checklist; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.