Solving Systems of Equations Word Problems

Word problems can seem tricky, but they become much easier when you translate them into a system of equations. In these problems, you are usually looking for two unknown values. By writing two equations based on the information given, you can solve for both unknowns.

4 Steps to Solve Word Problems

1. Define your variables: Choose two letters to represent the two unknown quantities.
2. Write two equations: Translate the English sentences into math equations (often one for the total quantity and one for the total value/cost).
3. Solve the system: Use either the substitution or elimination method.
4. Check your work: Ensure your answers make sense in the context of the original problem.

Example 1: Ticket Pricing

The Problem: Adult tickets cost $10 and child tickets cost$6. A total of $50$ tickets were sold for $380. How many of each were sold?

Step 1: Define variables Let $a$ = the number of adult tickets Let $c$ = the number of child tickets

Step 2: Write equations Total tickets equation: $a + c = 50$ Total cost equation: $10a + 6c = 380$

Step 3: Solve the system From the first equation, we can express $a$ in terms of $c$: $a = 50 - c$

Substitute this into the second equation: $10(50 - c) + 6c = 380$ $500 - 10c + 6c = 380$ $500 - 4c = 380$ $-4c = -120$ $c = 30$

Now, plug $c = 30$ back into the first equation: $a = 50 - 30 = 20$

Answer: There were $20$ adult tickets and $30$ child tickets sold.

Example 2: Mixture Problem

The Problem: A mix of two types of nuts costs $96 for$10$pounds. Type A is$8/lb and type B is $12/lb. How many pounds of each type are in the mix?

Step 1: Define variables Let $x$ = pounds of Type A Let $y$ = pounds of Type B

Step 2: Write equations Total weight equation: $x + y = 10$ Total cost equation: $8x + 12y = 96$

Step 3: Solve the system Let's use the elimination method. Multiply the entire first equation by $-8$: $-8x - 8y = -80$

Add this new equation to the second equation: $(-8x - 8y) + (8x + 12y) = -80 + 96$ $4y = 16$ $y = 4$

Plug $y = 4$ back into the weight equation: $x + 4 = 10$ $x = 6$

Answer: The mix contains $6$ pounds of Type A and $4$ pounds of Type B.