# SOLVING SYSTEM OF LINEAR EQUATIONS WORD PROBLEMS WORKSHEET

Solving System of Linear Equations Word Problems Worksheet :

Worksheet given in this section will be much useful to the students who would like to practice solving word problems on system of linear equations.

## Solving System of Linear Equations Word Problems Worksheet - Questions

Question 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.

Question 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.

Question 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.

Question 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. ## Solving System of Linear Equations Word Problems Worksheet - Solutions

Question 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.

Solution :

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.

Question 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.

Solution :

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.

Question 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.

Solution :

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.

Question 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.

Solution :

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. After having gone through the stuff given above, we hope that the students would have understood, how to solve word problems on system of Linear equations.

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