2x+3y=12 how do I solve for x and y
2x+3y= 12
To solve for the x intercept, y=0
2x+3(0)= 12
2x= 12
Divide both sides by 2
2x/2x= 12/2x
X= 6
So:(6,0)
To solve for y intercept, x=0
2(0)+3y= 12
3y= 12
Divide both sides by 3
3y/3y= 12/3
Y= 4
So: (0,4)
2x + 3y = 122x - 2x + 3y = 12 - 2x3y = 12 - 2x 3 3 y = 4 - 2/3x2x + 3(4 - 2/3x) = 122x + 12 - 2x = 122x - 2x + 12 = 120x + 12 = 12 -12 -12 0x = 0 0 0 x = 02(0) + 3y = 120 + 3y = 12-0 -0 3y = 12 3 3 y = 4 (x, y) = (0, 4)