Question

what is the solution to the pair of simultaneous equations?
$2x + y = 5 \\ 3x - 2y = 4$
A.
$x = 7 \: and \: y = - 9$
B.
$x = 1 \: and \: y = 3$
C.
$x = 2 \: and \: y = 1$
D.
$x = 6 \: and \: y = - 7$
​

Answer 1

★ Solution :-

$\sf \leadsto 2x + y = 5 \: \: - - - (i)$

$\sf \leadsto 3x - 2y = 4 \: \: - - - (ii)$

Now, by first equation,

$\sf \leadsto 2x + y = 5$

$\sf \leadsto 2x = 5 - y$

$\sf \leadsto x = \dfrac{5 - y}{2}$

Now, finding the value of y by second equation,

$\sf \leadsto 3x - 2y = 4$

$\sf \leadsto 3 \bigg( \dfrac{5 - y}{2} \bigg) - 2y = 4$

$\sf \leadsto \dfrac{15 - 3y}{2} - 2y = 4$

$\sf \leadsto \dfrac{15 - 3y - 4y}{2} = 4$

$\sf \leadsto \dfrac{15 - 7y}{2} = 4$

$\sf \leadsto 15 - 7y = 4(2)$

$\sf \leadsto 15 - 7y = 8$

$\sf \leadsto - 7y = 8 - 15$

$\sf \leadsto - 7y = - 7$

$\sf \leadsto y = \dfrac{ - 7}{ - 7}$

$\sf \leadsto y = 1$

Now, finding the value of x by first equation,

$\sf \leadsto 2x + y = 5$

$\sf \leadsto 2x + 1 = 5$

$\sf \leadsto 2x = 5 - 1$

$\sf \leadsto 2x = 4$

$\sf \leadsto x = \dfrac{4}{2}$

$\sf \leadsto x = 2$

Therefore, the values of x and y are 2 and 1 respectively.