Question

SOLVE THE FOLLOWING LINEAR EQUATIONS BY SUBSTITUTION METHOD:
(a). X-2Y=1
2X+Y=7
(b). X+2Y=5
2X-3Y= - 4
( c). 2X-Y=1
X+2Y=8
(d) X+3Y=15
2X-Y=2
(e) X+Y=6
2X-Y =0​

Answer 1

Answer:-

(a)

$x - 2y = 1 - - - 1 \\ 2x + y = 7 - - - 2 \\ from \: - 1 \\ x = 1+ 2y \\ substitute \: in \: equation - 2 \\ 2(1 + 2y) + y = 7 \\ 2 + 4y + y = 7 \\ 5y = 5 \\ y = 1 \: and \: x = 3$

(b)

$x + 2y = 5 - - - - 1 \\ 2x - 3y = - 4 - - - 2 \\ from \: - 1 \\ x = 5 - 2y \\ 2(5 - 2y) - 3y = - 4 \\ 10 - 4y - 3y = - 4 \\ - 7y = - 14 \\ y = 2 \: \: and \: \: x = 1$

(c)

$2x - y = 1 - - 1 \\ x + 2y = 8 - - 2 \\ from \: - 2 \\ x = 8 - 2y \\ 2(8 - 2y) - y = 1 \\ 16 - 4y - y = 1 \\ - 5y = - 15 \\ y = 3 \: \: and \: \: x = 2$

(d)

$x + 3y = 15 - - - 1 \\ 2x - y = 2 - - - - 2 \\ from \: - 1 \\ x = 15 - 3y \\ 2(15 - 3y) - y = 2 \\ 30 - 6y - y = 2 \\ - 7y = - 28 \\ y = 4 \: \: and \: \: x = 3$

(e)

$x + y = 6 - - - - 1 \\ 2x - y = 0 - - - 2 \\ from \: - 1 \\ x = 6 - y \\ 2(6 - y) - y = 0 \\ 12 - 2y - y = 0 \\ - 3y = - 12 \\ y = 4 \: \: and \: \: x = 2$