Test: Elimination Method - Year 10 MCQ

Given pair of linear equations:

47x+31y = 63  ---(1)

31x+47y = 15  ---(2)

multiply equation (1) by 31 and equation (2) by 47

substract (2) from (1) we get

(961 - 2209)y = 63x31 - 15x47

-1248 y = 1953 - 705

-1248 y = 1248

Therefore, y = -1.

Substitute y = -1 in equation (1) we get

47x = 94

So, x = 2.
Hence, The values of x and y are 2, -1.

In the elimination method we equate either of the coefficient so that the other is eliminated ,substituting the value of one variable into the other equation and then we are left with the linear equation which is solvable and values of the variables are obtained.

x = (2c + 1) / (m + n)

mx + ny = c.............(1)

nx - ny = c + 1.........(2) add the eqs (1) & (2).

(m + n) x = 2c + 1

x = (2c + 1) / (m + n)

Let us assume x and y are the two digits of the number
Therefore, two-digit number is = 10x + y and the reversed number = 10y + x
Given:
x + y = 12
y = 12 – x  (1)
Also given:
10y + x - 10x – y = 18
9y – 9x = 18
y – x = 2    (2)
Substitute the value of y from Equation 1 in Equation 2
12 – x – x = 2
12 – 2x = 2
2x = 10
x = 5
Therefore, y = 12 – x = 12 – 5 = 7
Therefore, the two-digit number is 10x + y = (10*5) + 7 = 57

Given, x=a and y=b.
Given equations, x-y=2( assume it as equation 1)
x+y=4( assume it as equation 2)

Equation 1+2,
x-y=2
x+y=4
- - - - - - -
2x=6( y gets cancelled as there is a negative y and a positive y implies y-y=0)
x=6/2
x=3
But x=a implies a=3
substitute a in equation 1 (it can be substituted even in equation 2)
x-y=2
3-y=2
y=3-2
y=1
But y=b implies b=1
Therefore, a=3 and b=1