4.2 Solve Applications with Systems of Equations

Solving Applications with Systems of Equations

Setting up systems of equations from word problems is one of the most practical skills in algebra. The core idea: when a problem has two unknowns, you need two equations to solve it. This section covers how to translate real-world scenarios into systems and solve them.

Translating Word Problems into Systems

The biggest challenge with word problems isn't the algebra itself. It's figuring out what the equations should be. Here's a reliable process:

1. Read the problem carefully and identify what you're solving for. These are your unknowns.
2. Assign a variable to each unknown quantity ($x$ for one, $y$ for the other). Write down what each variable represents so you don't lose track.
3. Find two relationships in the problem and write each one as an equation. Every word problem with two unknowns will give you (at least) two distinct pieces of information.
4. Solve the system using substitution, elimination, or graphing.
5. Check your answer against the original problem. Does it actually make sense? A negative number of tickets or a speed of 900 mph probably means something went wrong.

Example: A theater sold 200 tickets total. Adult tickets cost $8 and child tickets cost $5. Total revenue was $1,340. How many of each type were sold?

• Let $x$ = adult tickets, $y$ = child tickets
• Equation 1 (total tickets): $x + y = 200$
• Equation 2 (total revenue): $8x + 5y = 1340$
• Solving by substitution: $y = 200 - x$, so $8x + 5(200 - x) = 1340$
• $8x + 1000 - 5x = 1340$, which gives $3x = 340$, so $x = \frac{340}{3} \approx 113.3$

That fractional answer signals you'd want to double-check the problem's numbers. In a well-constructed problem, the values come out clean. The process, though, is exactly right.

Geometry Applications

Geometry problems pair nicely with systems because shapes come with built-in formulas that create your equations.

Common setup: You're given information about a rectangle's perimeter and some relationship between its length and width.

• Let $l$ = length and $w$ = width
• The perimeter formula $2l + 2w = P$ gives you one equation
• The second equation comes from whatever additional relationship the problem states

Example: A rectangle's perimeter is 56 cm. Its length is 4 cm more than twice its width. Find the dimensions.

1. Equation 1 (perimeter): $2l + 2w = 56$
2. Equation 2 (length-width relationship): $l = 2w + 4$
3. Substitute equation 2 into equation 1: $2(2w + 4) + 2w = 56$
4. $4w + 8 + 2w = 56$, so $6w = 48$, giving $w = 8$
5. Back-substitute: $l = 2(8) + 4 = 20$

Check: $2(20) + 2(8) = 40 + 16 = 56$ ✓ and $20 = 2(8) + 4$ ✓

Other geometry setups include complementary angles (sum to $90°$), supplementary angles (sum to $180°$), and triangle problems involving angle relationships.

Uniform Motion Problems

Motion problems use the relationship $d = rt$ (distance = rate × time). These often involve two objects moving at different speeds, or the same object traveling at different speeds under different conditions (like with and against a current).

Steps for motion problems:

1. Draw a simple diagram or set up a table with columns for rate, time, and distance. This keeps the information organized.
2. Assign variables to the unknowns (often the two speeds, or a speed and a time).
3. Fill in the table using $d = rt$ and the given information.
4. Write your equations from the table entries and any other conditions the problem provides.
5. Solve and interpret.

Example: Two cars leave the same point traveling in opposite directions. One goes 55 mph and the other goes 65 mph. After how many hours will they be 360 miles apart?

Both cars travel the same amount of time ($t$), and their distances add up:

$55t + 65t = 360$, so $120t = 360$, giving $t = 3$ hours.

For problems with wind or current, you'd set up two unknowns: the object's speed in still conditions and the wind/current speed. Traveling with the current, the effective rate is $r + c$; against it, $r - c$.

Algebraic Modeling

Some applications ask you to set up a system that models a situation with constraints, meaning conditions that limit the possible solutions. Mixture problems and investment problems fall into this category.

• Mixture problems: You're combining two things (solutions, coffees, etc.) with different concentrations or prices. One equation tracks the total quantity; the other tracks the total value or concentration.
• Investment problems: Money is split between two accounts at different interest rates. One equation tracks total money invested; the other tracks total interest earned.

The approach is always the same: identify your two unknowns, find two relationships, write two equations, and solve.