5x + 2 = 22 5x = 22 - 2 5x = 20 5x/5 = 20/5 x=4
For 5x+y=1, you would subtract 5x from each side, so you would get y=1-5x For 3x+2y=2, you would subtract 3x from each side, and then divide by 2. 2y=2-3x y=1-(3/2)x
(3)(2)(2)5x=60 12(5x)=60 12(5x)/12=60/12 5x=5 x=1
(x-3)(x-2)
To solve for x: x2 = 5x + 2 x2 - 5x = 2 x2 - 5x + (5/2)2 = 2 + (5/2)2 (x - 5/2)2 = 8/4 + 25/4 x - 5/2 = ± √(33/4) x = (5 ± √33) / 2
This cannot be solved. It needs 2 equations to solve for 2 unknowns.
5x + 2 = 22 5x = 22 - 2 5x = 20 5x/5 = 20/5 x=4
3
For 5x+y=1, you would subtract 5x from each side, so you would get y=1-5x For 3x+2y=2, you would subtract 3x from each side, and then divide by 2. 2y=2-3x y=1-(3/2)x
5x+2 = 3x-6 5x-3x = -6-2 2x = -8 x = -4
(3)(2)(2)5x=60 12(5x)=60 12(5x)/12=60/12 5x=5 x=1
5x-3 = 11+12x 5x-12x = 11+3 -7x = 14 x = -2
35x*2=5x Well, I can only think of one way to solve this. Forgive if I'm wrong 'cause I'm only in 9th grade. I've tried it the other way where you solve the equation for zero, but it didn't work. So here's the only way I saw, I apologise if it is wrong. 35x*2=5x -5x......-5x 30x*2=0 .......-2-2 ..30x=-2 ...30x/30-2/30 x=-1/15
(x-3)(x-2)
To solve for x: x2 = 5x + 2 x2 - 5x = 2 x2 - 5x + (5/2)2 = 2 + (5/2)2 (x - 5/2)2 = 8/4 + 25/4 x - 5/2 = ± √(33/4) x = (5 ± √33) / 2
2x+3x-2=0=5x=0+2=5x=2=5x_5=2_5=x=0.4
-5x + 30 = 20 -5x + 30 - 30 = 20 - 30 -5x = -10 -5x / -5 = -10 / -5 x = 2