Part 7: Algebraic Thinking

# 7.2: Algebra word problems

Notes

To solve algebra words problems, label the unknown quantities as variables (e.g., y), create equations to represent the information given in the problem, and solve the equations for the unknown variable(s).

##### Word problems involving a single variable
• Example 1: Last week Cedric bought some thingamajigs from Uncle Vernon’s Bazaar for \$8 per item plus a \$8 fixed delivery charge. This week he bought some thingamajigs from Aunt Petunia’s Emporium for \$7 per item plus a \$14 fixed delivery charge. If Cedric spent the same amount of money each week, how many thingamajigs did he buy each week and how much did he spend each week?
• Method: Let y = the number of thingamajigs that Cedric buys every week. Form two algebraic equations for the money spent each week, set the equations equal, and solve for y.
• Last week’s money spent = 8y + 8.
• This week’s money spent = 7y + 14.
• Set the equations equal: 8y + 8 = 7y + 14.
• Move expressions involving y to one side and expressions involving just numbers to the other side: 8y – 7y = 14 – 8.
• Simplify: y = 6 thingamajigs.
• Check: Last week’s money spent = 8(6) + 8 = 48 + 8 = \$56.
• This week’s money spent = 7(6) + 14 = 42 + 14 = \$56.
• Example 2: Suppose Uncle Vernon changes his prices to \$9 per item plus a \$5 fixed delivery charge and Aunt Petunia keeps her item charge at \$7 per item but changes her delivery charge to \$15. Solve the problem with the new prices.
• Method: Last week’s money spent = 9y + 5.
• This week’s money spent = 7y + 15.
• Therefore, 9y + 5 = 7y + 15.
• Therefore, 9y – 7y = 15 – 5.
• Simplify: 2y = 10, so y = 5 thingamajigs.
• Check: Last week’s money spent = 9(5) + 5 = 45 + 5 = \$50.
• This week’s money spent = 7(5) + 15 = 35 + 15 = \$50.
##### Word problems involving two variables
• Example 1: Suppose Bailey spends \$7 to buy 6 doodads and gizmos in total. If doodads cost \$2 each and gizmos cost \$1 each, how many each of doodads and gizmos did Bailey buy?
• Method: Let x = the number of doodads that Bailey buys and let y = the number of gizmos that she buys. Form an algebraic equation to represent the total number of doodads and gizmos and another equation to represent how much Bailey spent and solve the equations for x and y.
• Total number of doodads and gizmos: x + y = 6.
• Money spent: 2x + y = 7.
• Rearrange first equation to isolate x: x = 6 – y.
• Substitute this expression for x into the second equation: 2(6 – y) + y = 7.
• Move expressions involving y to one side and expressions involving just numbers to the other side: 2(6) – 7 = 2yy.
• Simplify: y = 5 gizmos.
• Substitute this value for y into the first equation: x = 6 – 5 = 1 doodad.
• Check: Total number of doodads and gizmos: x + y = 1 + 5 = 6.
• Money spent: 2x + y = 2(1) + 5 = 7.
• Example 2: Suppose doodads now cost \$3 each and gizmos cost \$2 each, and Bailey spends \$14 to buy 6 doodads and gizmos in total. How many each of doodads and gizmos did Bailey buy?
• Method: Total number of doodads and gizmos: x + y = 6.
• Money spent: 3x + 2y = 14.
• Rearrange first equation: x = 6 – y.
• Substitute into the second equation: 3(6 – y) + 2y = 14.
• Therefore, 3(6) – 14 = 3y – 2y.
• Simplify: y = 4 gizmos.
• Substitute into the first equation: x = 6 – 4 = 2 doodads.
• Check: Total number of doodads and gizmos: x + y = 2 + 4 = 6.
• Money spent: 3x + 2y = 3(2) + 2(4) = 14.
##### Word problems involving percentage changes
• Example 1: Suppose Kamala spent \$69, including tax, on a pair of jeans. If the tax rate was 15%, what was the (before tax) price on the label?
• Method: Let y = (before tax) price on the label. Form an algebraic equation for the after tax cost to Kamala and solve for y.
• After tax cost: y + 0.15y = 69.
• Simplify: 1.15y = 69.
• Rearrange and solve: y = 69 ÷ 1.15 = \$60.
• Check: After tax cost: 1.15(60) = 69.
• Note that the answer is not 0.85(69) = \$58.65.
• Example 1: Suppose Gregory worked out for 56 minutes today, which was 20% less than yesterday. How many minutes did he work out for yesterday?
• Method: Let y = yesterday’s work-out minutes. Form an algebraic equation for today’s work-out minutes and solve for y.
• Today’s work-out minutes: y – 0.2y = 56.
• Simplify: 0.8y = 56.
• Rearrange and solve: y = 56 ÷ 0.8 = 70.
• Check: Today’s work-out minutes: 0.8(70) = 56.
• Note that the answer is not 1.2(56) = \$67.20.

The videos below work through two of the above examples of algebra word problems.

Practice Exercises

Do the following exercises to practice solving algebra word problems. 