Solve A=(x+2)^2+(1-x)(2+x)-3

$$A = ( x + 2 ) ^ { 2 } + ( 1 - x ) ( 2 + x ) - 3$$

Solve for A

$$A=3\left(x+1\right)$$

Solution Steps

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

$$A=x^{2}+4x+4+\left(1-x\right)\left(2+x\right)-3$$

Use the distributive property to multiply 1-x by 2+x and combine like terms.

$$A=x^{2}+4x+4+2-x-x^{2}-3$$

Add 4 and 2 to get 6.

$$A=x^{2}+4x+6-x-x^{2}-3$$

Combine 4x and -x to get 3x.

$$A=x^{2}+3x+6-x^{2}-3$$

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

$$A=3x+6-3$$

Subtract 3 from 6 to get 3.

$$A=3x+3$$

Solve for x

$$x=\frac{A-3}{3}$$

Steps for Solving Linear Equation

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

$$A=x^{2}+4x+4+\left(1-x\right)\left(2+x\right)-3$$

Use the distributive property to multiply 1-x by 2+x and combine like terms.

$$A=x^{2}+4x+4+2-x-x^{2}-3$$

Add 4 and 2 to get 6.

$$A=x^{2}+4x+6-x-x^{2}-3$$

Combine 4x and -x to get 3x.

$$A=x^{2}+3x+6-x^{2}-3$$

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

$$A=3x+6-3$$

Subtract 3 from 6 to get 3.

$$A=3x+3$$

Swap sides so that all variable terms are on the left hand side.

$$3x+3=A$$

Subtract 3 from both sides.

$$3x=A-3$$

Divide both sides by 3.

$$\frac{3x}{3}=\frac{A-3}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x=\frac{A-3}{3}$$

Divide A-3 by 3.

$$x=\frac{A}{3}-1$$