Lesson Solving word problems using linear systems of two equations with two unknowns

Solving word problems using linear systems of two equations with two unknowns

In this lesson we present some typical word problems and show how to solve those using linear systems of two equations with two unknowns.

Problem 1. The Madison Local High School band

The Madison Local High School marching band sold gift wrap to earn money for a band trip to Orlando, Florida.
The gift wrap in solid colors sold for $4.00 per roll, and the print gift wrap sold for $6.00 per roll.
The total number of rolls sold was 480, and the total amount of money collected was $2340.
How many rolls of each kind of gift wrap were sold?

Solution
Let us denote the unknown number of gift wrap in solid colors as x and the unknown number of print gift wrap as y.
Since the total number of rolls sold was 480, the first equation is
.
Now, the total cost of x gift wraps in solid colors sold for $4.00 per roll is equal to dollars,
while the total cost of y print gift sold for $6.00 per roll is equal to dollars.
Since the total amount of money collected was $2340, the second equation is
.

So, we reduced our problem to the solution of the linear system of two equations with two unknowns
.

Let us apply the substitution method (see the lesson Solution of the linear system of two equations with two unknowns by the Substitution method).

Express from the first equation:
.
Substitute this to the second equation:
.
Open brackets, collect common terms, and step-by-step simplify:
,
,
,

.
Substitute the found value of to the first equation and calculate :
,
.

As a result, you get , as the potential solution.

Check
Substitute these values of and into the first and the second equations.
You have
for left side of the first equation, and
for the left side of the second equation.

Answer. 1080 gift wrap in solid colors and 1260 print gift were sold.

Problem 2. Apples and oranges

If 4 apples and 2 oranges equals $1 and 2 apples and 3 orange equals $0.70, how much does each apple and each orange cost?
There are no quantity discounts.

Solution
Let us denote the unknown price for one apple as cents and the unknown price for one orange as cents.
From the first condition we have the equation
,
while from the second problem condition we have another equation
.

So, our problem is reduced to the solution of the linear system of two equations with two unknowns
.

Let us apply the elimination method, which is multiplication and addition/subtraction method
(see the lesson Solution of the linear system of two equations with two unknowns by the Elimination method).

Keep the first equation as is, multiply the second equation by two: and subtract the second equation from the first one:
,

Subtract the second equation from the first one:
,
.

Now, substitute the found value of into the first equation and calculate :
,
,
,
.

So, you get , as the potential solution.

Check
Substitute these values of and into the first and the second equations.
You have
for the first equation left side, and
for the second equation left side.

Answer. Each apple costs 20 cents and each orange costs 10 cents.

Problem 3. Alloy

A piece of alloy composed of copper and zinc weights 81400 N (Newtons).
and is 1000 in volume. How much copper and how much zinc is there in the alloy?
The density of copper is 8.92 , the density of zinc is 7.14 .

    Note that this problem was just solved in the lesson Using linear equations to solve word problems by using the reduction to the linear equation with one variable.
    In this lesson, we use the reduction to the system of two linear equations with two unknowns.

Solution
Let us denote the unknown volume of copper in the alloy as x and the unknown volume of zinc in the alloy as y .
The weight of the copper in the alloy is

981 * 8.92 * x = 8750.52 * x = 8750.52 * x ,

where 981 is the gravity acceleration.
The weight of the zinc in alloy is

981 * 7.14 * y = 7004.34 * y .

The total weight of alloy is 81400 Newtons, which gives us an equation

8750.52 * x + 7004.34 * y = 81400 N = 81400 = 8140000 .

(Note that in the previous line we converted Newtons (that are ) to , because we use grams, centimeters and seconds as units uniformly in the solution of this problem).

This is the first equation.
The second equation express the total volume of 1000 as the sum
.

So, we reduced the original problem to the solution of the linear system of two equations with two unknowns


Let us apply the determinant method (see the lesson Solution of the linear system of two equations with two unknowns using determinant).

First, calculate determinant of the matrix .
.

Next, calculate determinant of the matrix , which is the numerator for calculation the unknown :
,
and then calculate determinant of the matrix , which is the numerator for calculation the unknown :
.

Now, calculate and


So, the volume of copper is 650.368 , and mass of copper in alloy is 8.92 * 650.368 = 5801 .
Volume of zinc in alloy is 349.632 , and mass of zinc in alloy is 7.14 * 349.632 = 2496 .

Answer. Mass of copper in alloy is 5801 g; mass of zinc in alloy is 2496 g.