Word Problems on Simultaneous Linear Equations



Solving the solution of two variables of system equation that leads for the word problems on simultaneous linear equations is the ordered pair (x, y) which satisfies both the linear equations.



Problems of different problems with the help of linear simultaneous equations:

We have already learnt the steps of forming simultaneous equations from mathematical problems and different methods of solving simultaneous equations. In connection with any problem, when we have to find the values of two unknown quantities, we assume the two unknown quantities as x, y or any two of other algebraic symbols. Then we form the equation according to the given condition or conditions and solve the two simultaneous equations to find the values of the two unknown quantities. Thus, we can work out the problem.


Worked-out examples for the word problems on simultaneous linear equations:

1. The sum of two number is 14 and their difference is 2. Find the numbers.

Solution:

Let the two numbers be x and y.
x + y = 14………. (i)
x – y = 2………. (ii)
Adding equation (i) and (ii), we get 2x = 16
or, 2x/2 = 16/2or, x = 16/2
or, x = 8

Substituting the value x in equation (i), we get
8 + y = 14
or, 8 – 8 + y = 14 – 8
or, y = 14 – 8
or, y = 6

Therefore, x = 8 and y = 6
Hence, the two numbers are 6 and 8.


2. In a two digit number. The units digit is thrice the tens digit. If 36 is added to the number, the digits interchange their place. Find the number.

Solution: Let the digit in the units place is x
And the digit in the tens place be y.
Then x = 3y and the number = 10y + x
The number obtained by reversing the digits is 10x + y.

If 36 is added to the number, digits interchange their places,
Therefore, we have 10y + x + 36 = 10x + y
or, 10y – y + x + 36 = 10x + y – y
or, 9y + x – 10x + 36 = 10x – 10x
or, 9y – 9x + 36 = 0 or, 9x – 9y = 36
or, 9(x – y) = 36
or, 9(x – y)/9 = 36/9
or, x – y = 4 ………. (i)

Substituting the value of x = 3y in equation (i), we get
3y – y = 4
or, 2y = 4
or, y = 4/2
or, y = 2

Substituting the value of y = 2 in equation (i),we get
x – 2 = 4
or, x = 4 + 2
or, x = 6
Therefore, the number becomes 26.


3. If 2 is added to the numerator and denominator it becomes 9/10 and if 3 is subtracted from the numerator and denominator it become 4/5. Find the fractions.

Solution:

Let the fraction be x/y.
If 2 is added to the numerator and denominator fraction becomes 9/10 so, we have
(x + 2)/(y + 2) = 9/10
or, 10(x + 2) = 9(y + 2)
or, 10x + 20 = 9y + 18
or, 10x – 9y + 20 = 9y – 9y + 18
or, 10x – 9x + 20 – 20 = 18 – 20
or, 10x – 9y = -2………. (i)

If 3 is subtracted from numerator and denominator the fraction becomes 4/5 so, we have
(x – 3)/(y – 3) = 4/5
or, 5(x – 3) = 4(y – 3)
or, 5x – 15 = 4y – 12
or, 5x – 4y – 15 = 4y – 4y – 12
or, 5x – 4y – 15 + 15 = – 12 + 15
or, 5x – 4y = 3………. (ii)
So, we have 10x – 9y = –2………. (iii)
and 5x – 4y = 3………. (iv)

Multiplying both sided of equation (iv) by 2, we get
10x – 8y = 6………. (v)
Now, solving equation (iii) and (v) , we get
10x – 9y = -2
10x – 8y = 6
       - y = - 8
y = 8

Substituting the value of y in equation (iv)
5x – 4 × (8) = 3
5x – 32 = 3
5x – 32 + 32 = 3 + 32
5x = 35
x = 35/5
x = 7
Therefore, fraction becomes 7/8.



4. If twice the age of son is added to age of father, the sum is 56. But if twice the age of the father is added to the age of son, the sum is 82. Find the ages of father and son.

Solution:

Let father’s age be x years
Son’s ages = y years
Then 2y + x = 56 …………… (i)
And 2x + y = 82 …………… (ii)

Multiplying equation (i) by 2, (2y + x = 56 …………… × 2)we get

linear equations



or, 3y/3 = 30/3
or, y = 30/3
or, y = 10 (solution (ii) and (iii) by subtraction)

Substituting the value of y in equation (i), we get;
2 × 10 + x = 56
or, 20 + x = 56
or, 20 – 20 + x = 56 – 20
or, x = 56 – 20
x = 36


5. Two pens and one eraser cost Rs. 35 and 3 pencil and four erasers cost Rs. 65. Find the cost of pencil and eraser separately.

Solution:

Let the cost of pen = x and the cost of eraser = y
Then 2x + y = 35 ……………(i)
And 3x + 4y = 65 ……………(ii)

Multiplying equation (i) by 4,

problems on simultaneous equations



Subtracting (iii) and (ii), we get;
5x = 75
or, 5x/5 = 75/5
or, x = 75/5
or, x = 15

Substituting the value of x = 15 in equation (i) 2x + y = 35 we get;
or, 2 × 15 + y = 35
or, 30 + y = 35
or, y = 35 – 30
or, y = 5
Therefore, the cost of 1 pen is Rs. 15 and the cost of 1 eraser is Rs. 5.



Simultaneous Linear Equations

  • Simultaneous Linear Equations
  • Comparison Method
  • Elimination Method
  • Substitution Method
  • Cross-Multiplication Method
  • Solvability of Linear Simultaneous Equations
  • Pairs of Equations
  • Word Problems on Simultaneous Linear Equations
  • Practice Test on Word Problems Involving Simultaneous
    Linear Equations


  • Simultaneous Linear Equations - Worksheets
  • Worksheet on Simultaneous Linear Equations
  • Worksheet on Problems on Simultaneous Linear
    Equations


  • 8th Grade Math Practice

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