Solve {l}{x^2+(a-3)(a+3)=(x+a)^2}{7a-{2ax-[4,5ax+3a/4-}

$$\left. \begin{array} { l } { x ^ { 2 } + ( a - 3 ) ( a + 3 ) = ( x + a ) ^ { 2 } } \\ { 7 a - \{ 2 a x - [ 4,5 a x + 3 a / 4 - } \end{array} \right.$$

Solve for x

$$x=-\frac{9}{2a}$$

$$a\neq 0$$

Steps for Solving Linear Equation

Consider \left(a-3\right)\left(a+3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$x^{2}+a^{2}-9=\left(x+a\right)^{2}$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(x+a\right)^{2}.

$$x^{2}+a^{2}-9=x^{2}+2xa+a^{2}$$

Subtract x^{2} from both sides.

$$x^{2}+a^{2}-9-x^{2}=2xa+a^{2}$$

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

$$a^{2}-9=2xa+a^{2}$$

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

$$2xa+a^{2}=a^{2}-9$$

Subtract a^{2} from both sides.

$$2xa=a^{2}-9-a^{2}$$

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

$$2xa=-9$$

The equation is in standard form.

$$2ax=-9$$

Divide both sides by 2a.

$$\frac{2ax}{2a}=-\frac{9}{2a}$$

Dividing by 2a undoes the multiplication by 2a.

$$x=-\frac{9}{2a}$$