Question

y=-2x+6,3y-x+3=0....Can somebody please explain how to solve this system using substitution???

Answer 1

Since one of your equations directly gives you y in terms of x, replace y in the second equation with its equivalent in terms of x. Now solve for x. When you've got a value for x, plug it into the first equation to get y. It's really quite easy :-)

Answer 2

Hint:

\[3(-2x+6)-x+3=0\]

Answer 3

Y=-2x+6

3y-x+3=0 Substitute in the other equation for y
3(-2x+6)-x+3=0 now multiply the parenthesis
-6x+18-x+3=0 Now combine like terms
-7x+21=0 add -7x to both sides
21=-7x divide both sides by -7x
3=x

now figure Y so..
Y=-2(3)+6
y=-6+6
y=0

Checking:
3y - x + 3 = 0
3(0) - 3 + 3 = 0
0-3+3=0

Answer 4

Thank you both :)