Verbal Problems Solving for two unknowns with two equations

Algebra 2 Tutorial

Verbal Problems Solving for two unknowns with two equations

Intro

This Tutorial I wrote presents six problems with increasing difficulty that use Substitution and Elimination methods to obtain both unknowns. I will break it up into two parts.

Sample Problem

Problem 1) In a certain year, a movie and its sequel grossed a combined total of $ 1536 million. If the sequel grossed $ 112 million more than the movie, find how much each movie grossed.

Problem 2) The price of a small popcorn is $ 4 and it is $ 6 for the large size. The total number of popcorns sold is 78 and the total sales revenue was $ 420. Find the number of small and large popcorns.

Problem 3) If you invest three times as much in a 5% interest account than you did in a 4% interest account, how much was invested in each account if the total interest paid was $ 475?

Solution

Problem 1) X= movie income; Y = sequel income
X + Y = 1536 and Y = X + 112
Use Substitution. X + (X + 112) = 1536 so 2X = 1536-112= 1424
The movie grossed 712 million and the sequel grossed 824 (712+112) million

Problem 2) Let X=Number of small popcorns and Let Y=Number of large popcorns
X + Y = 78 and 4X + 6Y = 420
Use Substitution. X = 78 – Y and substitute into the second equation.
4 (78-Y) + 6Y = 420 so 312 – 4Y + 6Y = 420 so 312 + 2Y = 420 so 2Y = 108
Y = Number of large popcorns = 54 and X = Number of small popcorns = 78-54 = 24
Check: $4(24) + $6(54) = 96+324= $ 420

Problem 3) Let X= amount invested in 4% account and Y = amount invested in 5% account
Y = 3X and 0.04X + 0.05Y = 475 Use Substitution.
0.04X + 0.05 (3X) = 475 0.19X = 475
X = 475/0.19 = $ 2500 and Y = 3 (2500) = $ 7500
Check: 4% of 2500 + 5% of 7500 = 100 + 375 = $ 475



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