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

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

In this lesson we present some typical word problems and show how to solve them using linear systems of two equations in 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 x be the unknown number of gift wrap in solid colors and y be the unknown number of print gift wraps.
Since the total number of rolls sold was 480, the first equation is
x+%2B+y+=+480

.
Now, the total cost of x gift wraps in solid colors sold for $4.00 per roll is equal to 4%2Ax

dollars,
while the total cost of y print gift sold for $6.00 per roll is equal to 6%2Ay

dollars.
Since the total amount of money collected was $2340, the second equation is
4x+%2B+6y+=+2340

.

So, we reduced our problem to the solution of the linear system of two equations in two unknowns
system%28x+%2B++y+=+480%2C%0D%0A+++++++++4x+%2B+6y+=+2340%0D%0A%29

.

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

Express

from the first equation:
y+=+480-x

.
Substitute this to the second equation:
4%2Ax+%2B+6%2A%28480-x%29+=+2340

.
Open parentheses, collect common terms, and step-by-step simplify:
4X+%2B+6%2A480+-6x+=+2340

.
Substitute the found value of x=270

to the first equation and calculate

.

As a result, you get x=270

as the potential solution.

Check
Substitute these values of x=270

and

into the first and the second equations.
You have
x%2By+=+270%2B210+=+480

for the left side of the first equation, and
4x%2B6y+=+4%2A270%2B6%2A210+=+1080+%2B+1260+=+2340

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

Solution

Let

be the unknown price for one apple as cents and

be the unknown price for one orange as cents.
From the first condition you have the equation
4x+%2B+2y+=+100

,
while from the second problem condition you have another equation
2x+%2B+3y+=+70

.

So, our problem is reduced to the solution of the linear system of two equations in two unknowns
system%284x+%2B+2y+=+100%2C%0D%0A++++++++++2x+%2B+3y+=++70%0D%0A%29

.

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

Keep the first equation as is and multiply the second equation by two:
system%284x+%2B+2y+=+100%2C%0D%0A++++++++++4x+%2B+6y+=+140%0D%0A%29

system%284x+%2B+2y+=+100%2C%0D%0A++++++++++4x+%2B+6y+=+140%0D%0A%29

,

Subtract the second equation from the first one:
0x+-+4y+=+-40

.

Now, substitute the found value of y+=+10

into the first equation and calculate

.

So, you get x=20

as the potential solution.

Check
Substitute these values of x=20

and

into the first and the second equations.
You have
4x+%2B+2y+=+4%2A20+%2B+2%2A10+=+80+%2B+20+=+100

for the first equation left side, and
2x+%2B+3y+=+2%2A20+%2B+3%2A10+=+40+%2B+30+=++70

for the second equation left side.

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

Problem 3. Tickets

Tickets are sold at $4.00 for adults and $2.50 for students. If 100 tickets were sold for $355.00,
how many tickets were adult tickets?

Solution

Let "x" be the numbers of adult tickets, and
let "y" be the numbers of student tickets.

The system is:

  x +     y = 100,           (1)         ("100 tickets were sold")      and
4x + 2.5y = 355.         (2)         ("100 tickets were sold for $355.00")

To solve the system, express "x" from (1):   x = 100 - y, and substitute it into (2). You will get a single equation for y:

4*(100 - y) + 2.5y = 355.

Simplify and solve it:

400 - 4y + 2.5y = 355,     or

-1.5y = 355 - 400,

-1.5y = -45,

y = %28-45%29%2F%28-1.5%29

= 30.

So, we just found y, the number of student tickets.

Then x = 100 - y = 100 - 30 = 70. It is the number of adult tickets.

Check: 70*4 + 30*2.5 = 280 + 75 = 355. Correct !

Answer. 70 adult and 30 student tickets were sold.

Problem 4. Alloys

A piece of metal is 1 ft. long, 6 in. wide, and 4 in. thick, and weighs 189.8125 pounds.
It is composed of an alloy of gold and copper. Determine percentage of gold.
Gold density is 0.70 lbs. per cu. in.
Copper density is 0.32 lbs. per cu. in.

Solution

When they ask "Determine percentage of gold", they mean the ratio of the gold contents by WEIGHT to the total WEIGHT of the alloy;
the ratio, expressed as percentage.

The volume of a piece of metal is  (12 in x 6 in x 4 in) = 288 cubic inches.


The density of the alloy is   = 0.659 lb/in^3.


Now, having the densities expressed in consistent units, you can write the equation.


Let  0 <= x <= 1 be the fraction (by the weight) of gold in the alloy (i.e. the percentage is 100x).

So the weight fraction of copper is (1-x), obviously.


Then the "fraction" equation is

     +  = ,   


where D stands for density, or


    0.70*x + 0.32*(1-x) = 0.659.


Simplify and solve for x:

    0.70x + 0.32 - 0.32x = 0.659  ====>  0.38x = 0.659 - 0.32 = 0.339  ====>  x =  = 0.8921.


Thus the gold fraction by the weight is  0.8921,  or  89.21%.

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