Question

Find the solution for this system of equations.

Answer 1

Let,
12x + 15y =34--------(1)
-6x + 5y =3-------(2)

(2) x 2 = -12x + 10y =6-------(3)

(3) + (1)

20y =40
y= 2

substitute y=2 into (1)
12x + 15(2) =34
12x =4
x= 1/3

Answer 2

Answer:

$x=\frac{5}{6}$

$y=\frac{8}{5}$

Step-by-step explanation:

$12x + 15=34$

$-6x + 5y=3$

In elimination method we try to make the coefficient of one variable same

LEts multiply the second equation by 2

$-6x + 5y=3$ * 2

$-12x + 10y =6$

Now add it with first equation

$12x + 15y=34$

$-12x + 10y =6$

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       $25y=40$

Divide by 25 on both sides

$y=\frac{8}{5}$

Now plug it in first equation and find out x

$12x + 15y=34$

$12x + 15(\frac{8}{5})=34$

$12x + 24=34$, subtract 24

$12x =10$

Divide by 12 on both sides

$x=\frac{5}{6}$