Question

elimination method ​

Answer 1

Answer:

x=8

y=4

add both the eqn

I hope u get it .....

make me brainlist...

Answer 2

$\frak{\pmb{\underline\red{Elimination \: Method}}}:-$

$\sf1) 2x - 3y = 4 , - x + 3y = 4 \\ \\ \sf 2x - 3y = 4 .. ( 1 ) \\ \sf- x + 3y = 4 .. ( 2 ) \\ \\ \sf 1 × 2x - 3y = 4 .. ( 1 ) \\ \sf 2 × - x + 3y = 4 .. ( 2 ) \\ \\ \sf \cancel{ 2x} - 3y = 4 \\\sf \cancel{- 2x }+ 6y = 8 \\ \sf \underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \sf 3y = 12\\ \\ \sf \: y = \frac{ \cancel{12} \: ^{4} }{ \cancel{3}} \\ \\ {\underline{\boxed{\bf\red{ y = 4}}}}$

$\sf{Substitute \: \red{y = 4} \: in \: eq.. ( 1 )}\\ \\ \sf \: 2x - 3y = 4 \\ \\ \sf 2x - 3(4) = 4 \\ \\ \sf 2x - 12 = 4 \\ \\ \sf \: 2x = 4 + 12 \\ \\ \sf 2x = 16 \\ \\ \sf x = \frac{\cancel {16}\:^{8}}{\cancel{2}} \\ \\ {\underline{\boxed{\bf\red{x = 8}}}}$