x2 + 7x - 8 = (x + 8)(x - 1)
7x-5x=8 2x=8 x=4
9x-4 = 7x+8 9x-7x = 8+4 2x = 12 x = 6
Algebraic constants are numbers with no variable (7x+7x+8=14x+8. The constant would be 8)
Presumably this is an equation in which the value of x is to be found. 7x+8 = 4x-10 7x-4x = -10-8 3x = -18 x = -6
7x + 8 = 36 - 8 = - 8 -------------- 7x = 28 7x/7 = 28/7 x = 4
(7x + 8)(7x - 8)
x2+7x-8 = (x-1(x+8)
3/7x + 5 = 87x(3/7x + 5) = 8(7x)3(7x/7x) + (7x)(5) = 8(7x)3 + 35x = 56x3 + 35x - 35x = 56x - 35x3 = 21x3/21 = 21x/211/7 = x
x2 + 7x - 8 = (x + 8)(x - 1)
x2+7x-8 = (x-1(x+8)
7x + y = 8 Subtract 7x from each side of the equation: y = -7x + 8
Subtract 7x from each side: 8 = -9 Not possible, game over.
7x-5x=8 2x=8 x=4
(7x + 8)(7x - 9)
-8 + 7(6x) + 3 = -7x - 5 - 1-8 + 42x + 3 = -7x - 5 - 142x - 8 + 3 = -7x - 5 + (-1)42x - 5 = - 7x - 642x + 7x - 5 + 5 = - 7x + 7x - 6 + 549x = -149x/49 = -1/49x = -1/49
9x-4 = 7x+8 9x-7x = 8+4 2x = 12 x = 6