Respuesta :
To solve the system of equations we need to use elimination. But to do that we are required to rewrite the equations in implicit form.
[tex]ax+by+c=0[/tex]
So we have,
[tex]3x-y+3=0 \\ -2x-y+3=0[/tex]
Then multiply the first equation by 2 on both sides and second equation by 3 on both sides. Resulting with,
[tex]6x-2y+6=0 \\ -6x-3y+9=0[/tex]
Adding these two equations eliminates the first term in both since 6x - 6x = 0. Hence,
[tex]-5y+15=0\Longrightarrow y=3[/tex]
Now that we know the value of y we can insert it in either one of the equations in the system. I'll pick first one to get x.
[tex]3=3x+3\Longrightarrow 3x=0\Longrightarrow x=0[/tex]
So the solution to this system of equation are [tex]\boxed{x=0},\boxed{y=3}[/tex]
Hope this helps.
r3t40
Answer:
one solution (0, 3)
Step-by-step explanation:
The left sides of both equations are equal, so you can equate the right sides
3х + 3 = -2х + 3
3х + 2х = 3 - 3
5х = 0
х = 0
Put value of x in equation 1 to find value of y
y = 3x + 3 = 3*0 + 3 = 3
Тhe system of equations has one solution (0, 3)