Algebra Hard
⏱ 25 min 📊 Hard ⭐ Premium

Word Problems with Systems of Equations

Translate real-world scenarios into systems of two equations and solve them.

Before you read
Answer first — see what you already know.

A farmer has chickens and cows. There are 30 animals total and 80 legs. How many chickens are there? (Chickens have 2 legs, cows have 4.)

Check your answer
Answer:

C

Before you read
Answer first — see what you already know.

A cashier has bills and bills totaling . There are 25 bills in all. How many bills are there?

Check your answer
Answer:

B

Theory

Setting Up Systems from Word Problems

Most system word problems on the SAT follow a pattern:

Two unknowns \to define two variables
Two relationships \to write two equations
Solve \to use substitution or elimination

Common types:
1. Quantity + Value: items and costs, tickets and prices
2. Mixture: combining solutions, blending ingredients
3. Distance/Rate: two objects moving
4. Age: relationships between people's ages
5. Comparison: one quantity related to another

Ticket problemPriceQtyTotalAdult12a12aChild8c8cTotal1000
Setting up a system from a word problem
Theory

The Quantity-Value Template

The most common SAT system word problem:

Equation 1 (Quantity): total number of items
x+y=total itemsx + y = \text{total items}

Equation 2 (Value): total value/cost
ax+by=total valuea \cdot x + b \cdot y = \text{total value}

where aa and bb are the per-item values.

Example 1

A theater sells adult tickets for $12\$12 and student tickets for $7\$7. If 200 tickets were sold for a total of $1,900\$1{,}900, how many of each type were sold?

Let aa = adult tickets, ss = student tickets.

Quantity: a+s=200a + s = 200

Value: 12a+7s=190012a + 7s = 1900

From eq1: s=200as = 200 - a.

Substitute: 12a+7(200a)=190012a + 7(200 - a) = 1900

12a+14007a=190012a + 1400 - 7a = 1900

5a=5005a = 500 \to a=100a = 100.

s=200100=100s = 200 - 100 = 100.

100 adult tickets and 100 student tickets
12(100)+7(100)=1200+700=190012(100) + 7(100) = 1200 + 700 = 1900
Example 2

A store sells two types of coffee. Blend A costs $8\$8/lb and Blend B costs $12\$12/lb. A customer buys 5 lbs total for $48\$48. How many pounds of Blend A?

Let aa = lbs of A, bb = lbs of B.

a+b=5a + b = 5 and 8a+12b=488a + 12b = 48.

a=5ba = 5 - b. Substitute: 8(5b)+12b=488(5-b) + 12b = 48.

408b+12b=4840 - 8b + 12b = 48 \to 4b=84b = 8 \to b=2b = 2.

a=52=3a = 5 - 2 = 3.

3 lbs of Blend A

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