Question

 

Solve each of the following system of equations by eliminating x (by substitution) :

(i) x + y = 7 

2x – 3y = 11

(ii) x + y = 7 

12x + 5y = 7

(iii) 2x – 7y = 1

 4x + 3y = 15

(iv) 3x – 5y = 1

5x + 2y = 19

(v) 5x + 8y = 9

2x + 3y = 4

Answer 1

(i) We have,

x + y = 7 ….(1)

2x – 3y = 11 ….(2)

We shall eliminate x by substituting its value from one equation into the other. from equaton (1), we get

x + y = 7

⇒ x = 7 – y

Substituting the value of x in equation (2), we get

2 × (7 – y) – 3y = 11

⇒ 14 – 2y – 3y = 11

⇒ –5y = – 3 or, y = 3/5

Now, substituting the value of y in equation (1), we get

x + 3/5 = 7 ⇒ x = 32/5.

Hence, x = 32/5 and y = 3/5

(ii) We have,

x + y = 7 ….(1)

12x + 5y = 7 ….(2)

From equation (1), we have

x + y = 7 ⇒ x = 7 – y

Substituting the value of y in equation (2), we get

⇒ 12(7 – y) + 5y = 7

⇒ 84 – 12y + 5y = 7

⇒ –7y = – 77

⇒ y = 11

Now, Substituting the value of y in equation (1), we get

x + 11 = 7 ⇒ x = – 4

Hence, x = – 4, y = 11.

(iii) We have,

2x – 7y = 1 ….(1)

4x + 3y = 15 ….(2)

From equation (1), we get

`2x – 7y = 1 ⇒ x = \frac{7y+1}{2}`

Substituting the value of x in equation (2), we get ;

`\Rightarrow 4\times \frac{7y+1}{2}+3y=15`

`\Rightarrow \frac{28y+4}{2}+3y=15`

⇒ 28y + 4 + 6y = 30

`⇒ 34y = 26 ⇒ y = \frac{13}{17}`

Now, substituting the value of y in equation (1), we get

`2x – 7 × \frac{13}{17} = 1`

`⇒ 2x = 1 + \frac{91}{17} = \frac{108}{17} ⇒ x =\frac{108}{34} = \frac{54}{17}`

Hence,

`x = \frac{54}{17} , y = \frac{13}{17}`

(iv) We have,

3x – 5y = 1 …. (1)

5x + 2y = 19 …. (2)

From equation (1), we get;

`3x – 5y = 1 ⇒ x = \frac{5y+1}{3}`

Substituing the value of x in equation (2), we get

`⇒ 5 × \frac{5y+1}{3} + 2y = 19`

`⇒ 25y + 5 + 6y = 57 ⇒ 31y = 52`

Thus, `y = \frac{52}{31}`

Now, substituting the value of y in equation (1), we get

`3x – 5 × \frac{52}{31} = 1`

`⇒ 3x – \frac{260}{31} = 1 ⇒ 3x = \frac{291}{31}`

`⇒ x = \frac{97}{31}`

Hence, ` x = \frac{97}{31} , y = \frac{52}{31}`

(v) We have,

5x + 8y = 9 ….(1)

2x + 3y = 4 ….(2)

From equation (1), we get

`5x + 8y = 9 ⇒ x = \frac{9-8y}{5}`

Substituting the value of x in equation (2), we get

`⇒ 2 × \frac{9-8y}{5} + 3y = 4`

⇒ 18 – 16y + 15y = 20

⇒ –y = 2 or y = – 2

Now, substituting the value of y in equation (1), we get

5x + 8 (–2) = 9

5x = 25 ⇒ x = 5

Hence, x = 5, y = – 2.