 Lucille Davidson

2021-12-27

To solve: ${x}^{2}-6x=x\left(8+x\right)$ Jimmy Macias

Calculation:
By distributive property,
$x\left(8+x\right)=x\cdot 8+x\cdot x$
$x\left(8+x\right)=8x+{x}^{2}$
Replace
${x}^{2}-6x=8x+{x}^{2}$
Add $-{x}^{2}-8x$ on both sides,
${x}^{2}-6x-{x}^{2}-8x=8x+{x}^{2}-{x}^{2}-8x$
$0{x}^{2}-14x=0{x}^{2}+0x$
$-14x=0$
$x=0$
The value of x is 0. Dabanka4v

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation:
${x}^{2}-6\cdot x-\left(x\cdot \left(8+x\right)\right)=0$
$\left(\left({x}^{2}\right)-6x\right)-x\cdot \left(x+8\right)=0$
$-14x=0$
Multiply both sides of the equation by (-1): $14x=0$
Divide both sides of the equation by 14:
$x=0$ karton

Use the distributive property to multiply x by 8+x
${x}^{2}-6x=8x+{x}^{2}$
Subtract 8x from both sides.
${x}^{2}-6x-8x={x}^{2}$
Combine to get -14x.
${x}^{2}-14x={x}^{2}$
Subtract ${x}^{2}$ from both sides.
${x}^{2}-14x-{x}^{2}=0$
Combine to get 0.
-14x=0
Product of two numbers is equal to 0 if at least one of them is 0. Since -14 is not equal to 0, x must be equal to 0.
x=0

Do you have a similar question?