Question

7.3 Students can solve a system of linear equations using elimination.

2x - y = 13
3x + y = 12

Check Solution:

Answer 1

Required solution :

Here we have been given with two equations,

• 2x - y = 13 (Equation No 1)
• 3x + y = 12 (Equation No 2)

From the first equation,

⇒ 2x - y = 13

⇒ -y = 13 - 2x

⇒ y = 2x - 13

Here we got a temporary value of y as 2x - 13 .

In second equation,

Substitute here the value of y as 2x - 13 inorder to get the value of x.

⇒ 3x + (2x - 13) = 12

⇒ 3x + 2x - 13 = 12

⇒ 5x - 13 = 12

⇒ 5x = 12 + 13

⇒ 5x = 25

⇒ x = 25 / 5

⇒ x = 5

★ Therefore, value of x is 5.

Finding out value of y :

Substitute the value of x in this equation.

⇒ y = 2x - 13

⇒ y = 2 (5) - 13

⇒ y = 2 × 5 - 13

⇒ y = 10 - 13

⇒ y = -3

★ Therefore, value of y is -3.

Answer 2

Topic : Linear equations in two variables.

Given :

• 2x - y = 13 ––– (i)
• 3x + y = 12 ––– (ii)

Solution :

Now,

(i) + (ii)

`\qquad \sf \: { \dashrightarrow 2x - y +(3x + y) = 13 + 12}`

`\qquad \sf \: { \dashrightarrow 2x \: \cancel{- \: y} +3x \: \cancel{+ \: y}= 25}`

`\qquad \sf \: { \dashrightarrow 2x \: +3x= 25}`

`\qquad \sf \: { \dashrightarrow 5x= 25}`

`\qquad \sf \: { \dashrightarrow x \: = (25)/(5) }`

`\qquad {\pmb{ \bf { \dashrightarrow x \: = 5 }}}`

Now, substituting the value of x in Eq (ii) :

`\qquad \sf \: { \dashrightarrow 3x + y = 12}`

`\qquad \sf \: { \dashrightarrow 3(5) + y = 12}`

`\qquad \sf \: { \dashrightarrow 15 + y = 12}`

`\qquad \sf \: { \dashrightarrow y = 12 - 15}`

`\qquad \pmb {\bf { \dashrightarrow y = - 3}}`

Therefore, The value of x = 5 and y = -3 .