Solve (2x+1)^2+(x-4)quad(x-1)=5(x+1)quad(x-1)

$$( 2 x + 1 ) ^ { 2 } + ( x - 4 ) \quad ( x - 1 ) = 5 ( x + 1 ) \quad ( x - 1 )$$

Solve for x

$$x=10$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

$$4x^{2}+4x+1+\left(x-4\right)\left(x-1\right)=5\left(x+1\right)\left(x-1\right)$$

Use the distributive property to multiply x-4 by x-1 and combine like terms.

$$4x^{2}+4x+1+x^{2}-5x+4=5\left(x+1\right)\left(x-1\right)$$

Combine 4x^{2} and x^{2} to get 5x^{2}.

$$5x^{2}+4x+1-5x+4=5\left(x+1\right)\left(x-1\right)$$

Combine 4x and -5x to get -x.

$$5x^{2}-x+1+4=5\left(x+1\right)\left(x-1\right)$$

Add 1 and 4 to get 5.

$$5x^{2}-x+5=5\left(x+1\right)\left(x-1\right)$$

Use the distributive property to multiply 5 by x+1.

$$5x^{2}-x+5=\left(5x+5\right)\left(x-1\right)$$

Use the distributive property to multiply 5x+5 by x-1 and combine like terms.

$$5x^{2}-x+5=5x^{2}-5$$

Subtract 5x^{2} from both sides.

$$5x^{2}-x+5-5x^{2}=-5$$

Combine 5x^{2} and -5x^{2} to get 0.

$$-x+5=-5$$

Subtract 5 from both sides.

$$-x=-5-5$$

Subtract 5 from -5 to get -10.

$$-x=-10$$

Multiply both sides by -1.

$$x=10$$