Question

I need help plz

Answer 1

The solution for given linear equation systems is (3,-4).

Option: C

Step-by-step explanation:

The given equations are 5x+3y=3 and x+y= -1.

Let us isolate x value from equation 5x+3y=3.

5x = 3-3y.

$x=\frac{3-3y}{5}$.

Substitute this x value in equation x+y= -1.

$\frac{3-3y}{5}+y=-1$.

After taking LCM for the denominator,

$\frac{3-3y+5y}{5}= -1$.

3-3y+5y=-5.

2y=-8.

y=-4.

For $x=\frac{3-3y}{5}$ substitute the y value.

$x=\frac{3-3(-4)}{5}$.

$x=\frac{3+12}{5}$.

$x=\frac{15}{5}$.

x=3.

∴ The solution is (3,-4).