I think so.
r2 + 7r + 6 = [ (r+6) (r+1) ]
If that's the equation 6r + 7 = 13 + 7r, r = -6
12 + 8r - 6 = 7r - 5 - 3 8r +6 = 7r -8 r = -14
if the x in the formula means multiply (next time please use * to denote multiplication) then: 7r-38
how do i solve -3r+3r=-6(7+7r)-7(4-7r)
You are searching for the value of r.-7r + 6 = 9 - 6r + 1Notice the 9 and the 1 on the right side. Since they are common factors (terms not attached to x), they can be added.-7r + 6 = -6r + 10Subtract 6 from both sides to eliminate the 6 on the left.-7r = -6r + 4Add 7r to both sides, leaving r-terms on only one side.0 = r + 4Subtract 4 from both sides, so one side is constant and the other is the variable.-4 = rThe answer is -4 is equal to r.
x2+x-6 = (x-2)(x+3) when factored
6x2 + 5x - 6 is factored out to (2x + 3)(3x - 2).
6 + 3u can be factored into 2(3 + u)
x^2 +7x+6 = (x+1)(x+6) when factored
x2 + 5x + 6 = (x + 2)(x + 3)
It is: (x+1)(x+6) when factored
3x2-39x-90 = (3x+6)(x-15) when factored