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.
[tex]x=\frac{3-3y}{5}[/tex].
Substitute this x value in equation x+y= -1.
[tex]\frac{3-3y}{5}+y=-1[/tex].
After taking LCM for the denominator,
[tex]\frac{3-3y+5y}{5}= -1[/tex].
3-3y+5y=-5.
2y=-8.
y=-4.
For [tex]x=\frac{3-3y}{5}[/tex] substitute the y value.
[tex]x=\frac{3-3(-4)}{5}[/tex].
[tex]x=\frac{3+12}{5}[/tex].
[tex]x=\frac{15}{5}[/tex].
x=3.
∴ The solution is (3,-4).
