Solve by Factoring 8x^2-19x=2x^2-x

$8 x^{2} - 19 x = 2 x^{2} - x$

Step 1

Move all the expressions to the left side of the equation.

Subtract $2 x^{2}$ from both sides of the equation.

$8 x^{2} - 19 x - 2 x^{2} = - x$

Add $x$ to both sides of the equation.

$8 x^{2} - 19 x - 2 x^{2} + x = 0$

Step 2

Simplify $8 x^{2} - 19 x - 2 x^{2} + x$ .

Subtract $2 x^{2}$ from $8 x^{2}$ .

$6 x^{2} - 19 x + x = 0$

Add $- 19 x$ and $x$ .

$6 x^{2} - 18 x = 0$

Step 3

Factor $6 x$ out of $6 x^{2} - 18 x$ .

Factor $6 x$ out of $6 x^{2}$ .

$6 x (x) - 18 x = 0$

Factor $6 x$ out of $- 18 x$ .

$6 x (x) + 6 x (- 3) = 0$

Factor $6 x$ out of $6 x (x) + 6 x (- 3)$ .

$6 x (x - 3) = 0$

Step 4

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x - 3 = 0$

Step 5

Set $x$ equal to $0$ .

$x = 0$

Step 6

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 7

The final solution is all the values that make $6 x (x - 3) = 0$ true.

$x = 0, 3$