# SOLVING SYSTEM OF LINEAR EQUATIONS WORD PROBLEMS WORKSHEET

Problem 1 :

The denominator of a fraction is 4 more than twice the numerator. When both the numerator and denominator are decreased by 6, then the denominator becomes 12 times the numerator and determine the fraction.

Problem 2 :

The sum of a two digit number and the number formed by interchanging the digit is 132. If 12 is added to the number, then new number becomes 5 times the sum of the digits. Find the number.

Problem 3 :

Ten years ago, father was twelve times as old as his son and ten years hence, he will  be twice as old as his son will be. Find their present ages.

Problem 4 :

A man starts his job with a certain monthly salary and earns a fixed increment every year. If his monthly salary be \$ 1500 after 4 years of service and \$ 1800 after 10 years of service, what is his starting salary and what is his annual increment.

Problem 5 :

Hertz Car Rental rents cars for x dollars per day plus y dollars for each mile driven. John rented a car for 4 days, drove it 160 miles, and spent \$120. David rented a car for 1 day, drove it 240 miles, and spent \$80. Write equations to represent John expenses and David expenses. Then solve the system and tell what each number represents.

Problem 6 :

Adult tickets to Space City amusement park cost dollars. Children’s tickets cost y dollars. The Henson family bought 3 adult and 1 child tickets for \$163. The Garcia family bought 2 adult and 3 child tickets for \$174. Write equations to represent the Hensons’ cost and the Garcias’ cost. Solve the system and interpret the solution in the original context. Let x be the numerator of the fraction.

Then the denominator is (2x + 4).

So, the required fraction is

=  x / (2x + 4)  -----(1)

Given : If the numerator and denominator are decreased by 6, then the denominator becomes 12 times the numerator and determine the fraction.

Then, we have

(x - 6) / (2x + 4 - 6)  =  x / 12x

(x - 6) / (2x - 2)  =  1 / 12

Simplify.

12(x - 6)  =  2x - 2

12x - 72  =  2x - 2

10x  =  70

x  =  7

2x + 4  =  2(7) + 4

2x + 4  =  18

(1)----->  x / (2x + 4)  =  7 / 18

So, the required fraction is 7/18.

Let xy be the two digit number.

Given : The sum of two digit number and the number formed by interchanging the digit is 132.

xy + yx  =  132

10x + y + 10y + x  =  132

11x + 11y  =  132

Divide each side by 11.

x + y  =  12

Subtract x from each side.

y  =  12 - x -----(1)

Given : If 12 is added to the number, then new number becomes 5 times the sum of the digits.

xy + 12  =  5(x + y)

10x + y + 12  =  5x + 5y

5x - 4y + 12  =  0

Substitute (12 - x) for y.

5x - 4(12 - x) + 12  =  0

5x - 48 + 4x + 12  =  0

9x - 36  =  0

9x  =  36

x  =  4

Substitute 4 for x in (1).

(1)-----> y  =  12 - 4

y  =  8

xy  =  48

So, the required two digit number is 48.

Let f and s be the present ages of father and son.

Given : Ten years ago, father was twelve times as old as his son.

Then, we have

f - 10  =  12(s - 10)

Simplify.

f - 10  =  12s - 120

f  =  12s - 110 -----(1)

Given : Ten years hence, father will be twice as old as his son will be

Then, we have

f + 10  =  2(s + 10)

Simplify.

f + 10  =  2s + 20

f  =  2s + 10 -----(2)

From (1) and (2), we get

12s - 110  =  2s + 10

10s  =  120

s  =  12

Substitute 12 for s in (2).

f  =  2(12) + 10

f  =  24 + 10

f  =  34

So, the present ages of father and son are 34 years and 12 years respectively.

The man gets fixed increment every year.

So, his monthly salary each year is growing linearly.

Then, we can have the following linear equation for the information given above.

y  =  mx + b -----(1)

y -----> Monthly salary

x -----> Number of years

m -----> Yearly increment

b -----> Starting monthly salary

So, we have to solve for m and b.

Given : Monthly salary is \$ 1500 after 4 years of service.

Substitute 1500 for y and 4 for x in (1).

(1)----->  1500  =  4m + b

or

4m + b  =  1500 -----(2)

Given : Monthly salary is \$ 1800 after 10 years of service.

Substitute 1800 for y and 10 for x in (2).

(1)----->  1800  =  10m + b

or

10m + b  =  1800 -----(3)

Solving (2) and (3), we get

m  =  50

b  =  1300

So, his starting monthly salary is \$1300 and his annual increment is \$50.

Given : John rented a car for 4 days, drove it 160 miles, and spent \$120.

4x + 160y = 120 ----(1)

Given : David rented a car for 1 day, drove it 240 miles, and spent \$80.

x + 240y = 80

Solve the equation above for x in terms of y.

x = 80 - 240y ----(2)

Substitute x = 80 - 240y into (1).

4(80 - 240y) + 160y = 120

320 - 960y + 160y = 120

320 - 800y = 120

Subtract 320 from both sides.

-800y = -200

Divide both sides by -800.

y = 0.25

Substitute y = 0.25 into (2).

x = 80 - 240(0.25)

x = 80 - 60

x = 20

The soultion of the system is

(x, y) = (20, 0.25)

Interpret the solution in the original context.

Therefore, Hertz Car Rental rents cars for \$20 per day plus \$0.25 for each mile driven.

Given : The Henson family bought 3 adult and 1 child tickets for \$163.

Then, Henson's cost :

3x + y = 163 ----(1)

Given : The Garcia family bought 2 adult and 3 child tickets for \$174.

Then, Garcias' cost :

2x + 3y  =  174 ----(2)

Solve the system :

Solve (1) for y in terms of x

3x + y = 163

y = 163 - 3x ----(3)

Substitute y = 163- 3x in (2).

2x + 3(163 - 3x) = 174

2x + 489 - 9x = 174

489 - 7x = 174

Subtract 489 from each side.

-7x = -315

Divide each side by -7.

x = 45

Substitute x = 45 into (3).

(3)----> y = 163 - 3(45)

y = 163 - 135

y = 28

The soultion of the system is

(x , y) = (45, 28)

Interpret the solution in the original context :

Therefore, adult ticket price is \$45 and child ticket price is \$28.

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