Question

3x-2y=1
2x-3y=-6
by elimination method​

Answer 1

Answer:

x = 3 and y = 4

Step-by-step explanation:

Solution;

Given;

3x - 2y = 1................(i)

2x - 3y = -6.............(ii)

Multiplying (i) by 2 and (ii) by 3, we get,

6x - 4y = 2.................(iii)

6x - 9y = -18..............(iv)

Subtracting (iv) from (iii), we get,

6x - 4y = 2

6x - 9y = -18

(-) (+) (+)

_________

5y = 20

5y = 20

or, y = 20/5

or, y = 4

Putting y = 4 in (i)

3x - 2y = 1

or, 3x - 2x4 = 1

or, 3x = 1 + 8

or, 3x = 9

or, x = 9/3

or, x = 3

Hence, x = 3 and y = 4

Answer 2

$\huge\mathbf\green{Solution:-}$

Given :-

• 3x -2y = 1 ---------(1)
• 2x-3y = -6 ----------(2)

Find :-

• Value of x & y by, elimination method .

${\blue{\underline{\underline{\bold{Explanation:-}}}}}$

Elimination Method:-

Multiply equation(1) with 2 and equation (2) with 3

• 2(3x - 2y) = 2(1)
• 3(2x - 3y) = 3(-6)

We get,

• 6x - 4y = 2
• 6x - 9y = -18

Subtract equation (1) and (2). Eliminate x

We get,

==> 9y -4y = 18 + 2

==> 5y = 20

==> y = 20/5

==> y = 4

Keep Value of y in equ(1)

==> 3x - 2×4 = 1

==> 3x = 1+8

==> 3x = 9

==>x = 9/3

==>x = 3

Hence,

• Value of x = -2
• Value of y = 1

________________

${\green{\underline{\underline{\bold{Answer\: Verification:-}}}}}$

Case(1).

Keep value of x & y in equ(1)

==> 3×3 - 2×4 = 1

==> 9 - 8 = 1

==> 1 = 1

L.H.S. = R.H.S.

______________________

Case(2).

Keep value of x & y in equ(2)

==> 2(3) - 3(4) = -6

==> 6 - 12 = -6

==> -6 = -6

L.H.S. = R.H.S.

$\huge\mathcal\red{Hence\:Proved}$