Problem: (3r+2)(r-1)= -(7r-7)
I need to factor this quadratic equation then check by substitution
(3r + 2)(r - 1) = -(7r - 7)
3r^2 - 3r + 2r - 2 = -7r + 7
3r^2 - r - 2 = -7r + 7
3r^2 - r - 2 + 7r = -7r + 7 + 7r
3r^2 + 6r - 2 = 7
3r^2 + 6r - 2 - 7 = 7 - 7
3r^2 + 6r - 9 = 0
(3r - 3)(r + 3) = 0
To check it, you need to solve for r. Set both factors equal to zero and solve.
3r - 3 = 0
3r = 3
r = 1
r + 3 = 0
r = -3
Substitute each of those values into your original equation to make sure they work.
(3r + 2)(r - 1) = -(7r - 7)
(3(1) + 2)(1 - 1) = -(7(1) - 7)
(3 + 2)(1 - 1) = -(7 - 7)
(5)(0) = -(0)
0 = 0
(3r + 2)(r - 1) = -(7r - 7)
(3(-3) + 2)(-3 - 1) = -(7(-3) - 7)
(-9 + 2)(-3 - 1) = -(-21 - 7)
(-7)(-4) = -(-28)
28 = 28
Both answers work.
Answer: r = 1 and r = -3