I assume you want to solve for x. The idea is to get the "x" isolated on one side of the equation. To do this, first subtract 6 from both sides (left and right of the equal sign). This will get rid of the 6 on the left side. Next, in the result, divide both sides by 10 - this will get rid of the 10 on the left side.
6x2-2x+36 = 5x2+10x 6x2-5x2-2x-10x+36 = 0 x2-12x+36 = 0 (x-6)(x-6) = 0 x = 6 or x = 6 It has two equal roots.
x + 13x + 10x = 50 - 6
6-x+12 = 10x+7 6+12-7 = 10x+x 11 = 11x x = 1
X equals 6
-6x-6 -or- -6(x+1)
6x2-2x+36 = 5x2+10x 6x2-5x2-2x-10x+36 = 0 x2-12x+36 = 0 (x-6)(x-6) = 0 x = 6 or x = 6 It has two equal roots.
x + 13x + 10x = 50 - 6
6-x+12 = 10x+7 6+12-7 = 10x+x 11 = 11x x = 1
X equals 6
-6x-6 -or- -6(x+1)
y=-10x-4
6-3x = 5x-10x+2 4 = 8x-10x 4 = -2x x = -2
12x - 6 = 10x +2 12x - 10x = 2 + 6 2x = 8 x = 4
=>2x - 12x + 36 = 0 =>10x = 36=>x = 3.6Another contributor's answer:This is a quadratic equation question:x2-12x+36 = 0When factorised:(x-6)(x-6) = 0
-6
10*6+8*7 = 116
Substitute the values of x and y in 10x + 7y, thus: 10x + 7y = 10*9 + 7*6 = 90 + 42 = 132