Question

(і) x+2y +1 = 0, 2x-3y-12 = 0 cross multiplication method ​

Answer 1

${ \large \orange { \underline{answer : x = \frac{9}{7}and \: y = - 2 }}}$

$\mathfrak{ \large \red { \underline{ \underline{question}}}} \\ \implies{ \large{solve \: the \: equation \: by \: using \: crossmultiplication \: method }} \\ \implies{ \large{equation : x + 2y + 1 = 0 \: and \: 2x - 3y - 12 = 0}} \\ \\ \mathfrak{ \large \green { \underline{formula : }}} \\ \implies{ \large{x = \frac{(b1c2 - b2c1)}{(a1b2 - a2b1)} \: and \: y = \frac{(c1a2 - c2a1)}{(a1b2 - a2b1)} }} \\ \\ \mathfrak{ \large \blue { \underline{solution : }}} \\ \implies{ \large{a1 = 1 : b1 = 2 :c1 = 1 }} \\ \implies{ \large{a2 = 2 :b2 = - 3 :c2 = - 12 }} \\ \implies{ \large{ x = \frac{(2 \times - 12) - ( - 3 \times 1)}{(1 \times - 3) - (2 \times 2)} }} \\ \implies{ \large{y = \frac{(1 \times 2) - ( - 12 \times 1)}{(1 \times - 3) - (2 \times 2)} }} \\ \implies{ \large{x = \frac{ - 12 + 3}{ - 3 - 4} }} \\ \implies{ \large{y = \frac{2 + 12}{ - 3 - 4} }} \\ \implies{ \large{x = \frac{ - 9}{ - 7} }} \\ \implies{ \large{y = \frac{14}{ - 7} }} \\ { \large \orange { \underline{answer : x = \frac{9}{7}and \: y = - 2 }}}$