Question

Solve the following pair of equations: 1x+3y=1 6x−12y=2 , where x ≠ 0, y ≠ 0.

Answer 1

Answer:

Let

2x+3y

1

=u and

3x−2y

1

=v. Then, the given system of equations becomes

2

1

u+

7

12

v=

2

1

⇒7u+24v=7 ..(i)

and, 7u+4v=2 .(ii)

Substracting equation (ii) from equation (i), we get

20v=5⇒v=

4

1

Putting v=

4

1

in equation (i), we get

7u+6=7⇒u=

7

1

Now, u=

7

1

⇒

2x+3y

1

=

7

1

⇒2x+3y=7 (iii)

and, u=

4

1

⇒

3x−2y

1

=

4

1

⇒3x−2y=4 .(iv)

Multiplying equation (iii) by 2 and equation (iv) by 3, we get

4x+6y=14 .(v)

9x−6y=12 .(vi)

Adding equations (v) and (vi), we get

13x=26⇒x=2

Putting x=2 in equation(v), we get

8+6y=14⇒y=1

Hence, x=2,y=1 is the solution of the given system of equations.