# Word Problems on Simultaneous Linear Equations | Simultaneous Linear Equations Questions

By solving the system of linear equations in two variables, you will get an ordered pair having x coordinate and y coordinate values (x, y) that satisfies both equations. Here those simultaneous linear equations are in the form of word problems. So, you can solve different word problems with the help of linear equations.

We have already learned some steps and methods to solve the simultaneous linear equations in two variables. Assume the unknown quantities in the question as x, y variables and represent them in the form of a linear equation according to the condition mentioned in the question. And follow the methods to solve the formed system of linear equations to get the values of unknown quantities. We have also provided simultaneous equations problems with solutions that help you to grasp the concept.

## Simultaneous Equations Word Problems

Example 1.

One number is greater than thrice the other number by 6. If 4 times the smaller number exceeds the greater by 7, find the numbers?

Solution:

Let x, y be the two numbers

So that x > y

Given that, one number is greater than thrice the other number by 6.

Then, we can write it as

x = 3y + 6 ——— (i)

According to the question,

4 times the smaller number exceeds the greater by 7.

4y – x = 7 ——— (ii)

Substitute (3y + 6) in equation (ii).

4y – (3y + 6) = 7

4y – 3y – 6 = 7

y – 6 = 7

y = 7 + 6

y = 13

Substitute y = 13 in equation (i)

x = 3(13) + 6

x = 39 + 6

x = 45

So, the two numbers are 45, 13.

Example 2.

The sum of two numbers is 25 and their difference is 5. Find those two numbers?

Solution:

Let the two numbers be x and y.

According to the question,

The sum of two numbers is 25

x + y = 25 —— (i)

The difference of numbers is 5.

x – y = 5 —– (ii)

Add both the equations (i) & (ii)

x + y + x – y = 25 + 5

2x = 30

x = 30/2

x = 15

Putting x = 15 in equation (i)

15 + y = 25

y = 25 – 15

y = 10

So, the two numbers are 15, 10.

Example 3.

The sum of two numbers is 50. If the larger is doubled and the smaller is tripled, the difference is 25. Find the two numbers.

Solution:

Let the two numbers be x and y

According to the given question,

The sum of two numbers is 50.

x + y = 50 —— (i)

The larger number is doubled and the smaller number is tripled, the difference is 25.

2x – 3y = 25 —— (ii)

Multiply the first equation by 3.

3(x + y) = 3 x 50

3x + 3y = 150 —— (iii)

Add equation (ii) and equation (iii)

2x – 3y + 3x + 3y = 25 + 150

5x = 175

x = 175/5

x = 35

Substituting the value of x in equation (i)

35 + y = 50

y = 50 – 35

y = 15

Hence the two numbers are 35, 15.

Example 4.

The class IX students of a certain public school wanted to give a farewell party to the outgoing students of class X. They decided to purchase two kinds of sweets, one costing Rs. 70 per kg and the other costing Rs. 85 per kg. They estimated that 34 kg of sweets were needed. If the total money spent on sweets was Rs. 2500, find how much sweets of each kind they purchased?

Solution:

Let the quantity of sweets purchased be x kg which cost Rs. 70 per kg and sweets purchased y kg which cost Rs. 85 per kg.

According to the question,

x + y = 34 ——- (i)

70x + 85y = 2500 ——– (ii)

Multiplying the equation (i) by 70, we get

70(x + y) = 34 x 70

70x + 70y = 2380 —– (iii)

Subtracting equation (iii) from equation (ii), we get

70x + 70y – (70x + 85y) = 2380 – 2500

70x + 70y – 70x – 85y = -120

-15y = -120

y = 120/75

y = 1.6

Substitute y = 1.6 in equation (i)

x + 1.6 = 34

x = 34 – 1.6

x = 32.4

Hence, the sweets purchased 1 kg 600 grams which cost Rs. 85 per kg and 32 kg 400 grams which cost Rs. 70 per Kg.

Example 5.

A two-digit number is eight times the sum of its digits. The number formed by reversing the digits is 18 less than the given number. Find the given number?

Solution:

Let the two-digit number be xy.

Given that, the two digit number is eight times the sum of its digits.

xy = 8(x + y) —– (i)

In the two digit number xy, x is in the tens position and y is in ones position.

Then,

xy = 10 . x + 1 . y

xy = 10x + y ——- (ii)

Substitute equation (ii) in equation (i)

10x + y = 8(x + y)

10x + y = 8x + 8y

10x – 8x = 8y – y

2x = 7y

2x – 7y = 0 —– (iii)

According to the question,

The number formed by reversing the digits 18 less than the given number.

xy – yx = 18

(10 . x + 1 . y) – (10 . y + 1 . x) = 18

(10x + y) – (10y + x) = 18

10x + y – 10y – x = 18

9x – 9y = 18

9(x – y) = 18

(x – y) = 18/9

x – y = 2 —— (iv)

Multiply equation (iv) by 2.

2(x – y) = 2 x 2

2x – 2y = 4 —— (v)

Subtract equation (v) from equation (iii)

(2x – 2y) – (2x – 7y) = 4 – 0

2x – 2y – 2x + 7y = 4

5y = 4

y = 4/5

Substitute y = 4/5 in equation (iv)

x – 4/5 = 2

x = 2 + 4/5

x = (10 + 4)/5

x = 14/5

So, the required two digit number is 144/5.

Example 6.

If the numerator of a certain fraction is increased by 2 and the denominator by 1, the fraction becomes equal to ⅗ and if the numerator and denominator are each diminished by 1, the fraction becomes equal to ⅔, find the fraction.

Solution:

Let the fraction be x/y.

According the given question,

(x + 2) / (y + 1) = ⅗

5(x + 2) = 3(y + 1)

5x + 10 = 3y + 3

5x – 3y = 3 – 10

5x – 3y = -7 —— (i)

(x – 1) / (y – 1) = ⅔

3(x – 1) = 2(y – 1)

3x – 3 = 2y – 2

3x – 2y = -2 + 3

3x – 2y = 1 —— (ii)

Multiply the equation (i) by 2, equation (ii) by 3.

2(5x – 3y) = 2 x -7 ⇒ 10x – 6y = -14

3(3x – 2y) = 3 x 1 ⇒ 9x – 6y = 3

(10x – 6y) – (9x – 6y) = -14 – 3

10x – 6y – 9x + 6y = -17

x = -17

Substitute x = -17 in equation (i)

5(-17) – 3y = -7

-85 – 3y = -7

-3y = -7 + 85

-3y = 78

y = -78 / 3

y = -26

Therefore, the fraction is 17/26.

Example 7.

If four times the age of the son is added to the age of the father, the sum is 64. But if twice the age of the father is added to the age of the son, the sum is 82. Find the ages of father and son?

Solution:

Let the father’s age be x years, the son’s age be y years.

According to the question,

4y + x = 64 —– (i)

2x + y = 82 —— (ii)

Multiply the equation (i) by 2

2(x + 4y) = 64 x 2

2x + 8y = 128

Subtract 2x + 8y = 128 from equation (ii)

(2x + 8y) – (2x + y) = 128 – 82

2x + 8y – 2x – y = 46

7y = 46

y = 46/7

y = 6 years 7 months

Substituting y = 46/7 in equation (ii)

2x + 46/7 = 82

2x = 82 – 46/7

2x = (574 – 64)/7

2x = 510/7

x = 510/14

x = 36 years 4 months

Therefore, the father’s age is 36 years 4 months, the son’s age is 6 years 7 months.

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