Question

solve these by using substitution method​

Answer 1

Answer:

9x-10y=-12___eq1

2x+3y=13___eq2

3y=13-2x___eq3

y=13-2x/3

put eq 3 in eq1

9x-10(13-2x) /3=-12

9x-130+20x/3=-12

29x-130=-12×3

29x-130=-36

29x=130-36

29x=94

x=94/29

Answer 2

Question

$\tt \: 9x - 10y = - 12 \\ \tt \: 2x + 3y = 13 \: \: \: \: \: \:$

Solve by substitution method.

Answer

$\bf \: 9x - 10y = - 12 \\ \bf \: 2x + 3y = 13 \: \: \: \: \: \:$

Let's calculate

$\rm \: 9x - 10y = - 12 \\ \rm \underline{2x + 3y = 13 \: } \\ \rm-7x-13y=-26$

$\implies\rm7x + 13y=26 \\ \rm\implies\rm13y = 26 - 7x \\ \implies\rm \: y = \frac{26 - 7x}{13}$

Putting value of y in the above mentioned

$\bf \: 2x + 3y = 13$

$\leadsto \tt \: 2x + 3(\rm \frac{26 - 7x}{13}) = 13$

$\leadsto \tt \: 2x + (\rm \frac{72 - 21x}{13}) = 13$

$\leadsto \tt \: (\rm \frac{26x + 72 - 21x}{13}) = 13$

$\leadsto \tt \: (\rm \frac{5x + 72 }{13}) = 13$

$\leadsto \rm \: 5x + 72 = 169$

$\leadsto \rm \: 5x = 169 - 72$

$\leadsto \rm \: 5x = 97$

$\leadsto \rm \: x = \frac{97}{5}$

Again

$\bf \: 2x + 3y = 13$

$\leadsto \tt \: 2x + 3y = 13$

$\leadsto \tt \: 2( \frac{97}{5 })+ 3y = 13$

$\leadsto \frac{194}{5} + 3y = 13$

$\tt\leadsto 3y = 13 - \frac{194}{5}$

$\tt\leadsto 3y = \frac{65 - 194}{5}$

$\tt\leadsto y = \frac{65 - 194}{15}$

$\tt\leadsto y = \frac{ - 129}{ \: \: 15}$

Hence

$\color{brown}\boxed{ x = \frac{97}{5}}$

${\tt \color{cyan}\boxed{ y = \frac{ - 129}{15}}}$