Question

Solve the following pair of linear equations by elimination method and substitution method: a) x + y = 5 and 2x – 3y = 4 b) 3x + 4y = 10 and 2x – 2y = 2

Answer 1

By Elimination Method:
2x+2y=10
2x-3y=4
-    +    -
_______
5y=14
y=14/5

x=5-y
x=11/5

Answer 2

: Required Answer

$\longrightarrow$ (i) x + y = 5 and 2x – 3y = 4

By the method of elimination.

x + y = 5 ……………………………….. (i)

2x – 3y = 4 ……………………………..(ii)

When the equation (i) is multiplied by 2, we get

2x + 2y = 10 ……………………………(iii)

When the equation (ii) is subtracted from (iii) we get,

5y = 6

y = 6/5 ………………………………………(iv)

Substituting the value of y in eq. (i) we get,

x=5−6/5 = 19/5

➝ x = 19/5 , y = 6/5

By the method of substitution.

From the equation (i), we get:

x = 5 – y………………………………….. (v)

When the value is put in equation (ii) we get,

2(5 – y) – 3y = 4

-5y = -6

y = 6/5

When the values are substituted in equation (v), we get:

x =5− 6/5 = 19/5

➝ x = 19/5 ,y = 6/5

(ii) 3x + 4y = 10 and 2x – 2y = 2

➝ By the method of elimination.

3x + 4y = 10……………………….(i)

2x – 2y = 2 ………………………. (ii)

When the equation (i) and (ii) is multiplied by 2, we get:

4x – 4y = 4 ………………………..(iii)

When the Equation (i) and (iii) are added, we get:

7x = 14

x = 2 ……………………………….(iv)

Substituting equation (iv) in (i) we get,

6 + 4y = 10

4y = 4

y = 1

Hence, x = 2 and y = 1

By the method of Substitution

From equation (ii) we get,

➝ x = 1 + y……………………………… (v)

➝ Substituting equation (v) in equation (i) we get,

➝ 3(1 + y) + 4y = 10

➝ 7y = 7

➝ y = 1

When y = 1 is substituted in equation (v) we get,

➝ A = 1 + 1 = 2