Solve {l}{(a+b)a+(a-b)y=0a}{(a-b)x+(a+b)y=2b}

$$\left. \begin{array} { l } { ( a + b ) a + ( a - b ) y = 0 a } \\ { ( a - b ) x + ( a + b ) y = 2 b } \end{array} \right.$$

Solve for x, y

$$\left\{\begin{matrix}x=-\frac{2b^{2}-2ab-2ba^{2}-ab^{2}-a^{3}}{\left(a-b\right)^{2}}\text{, }y=-\frac{a\left(a+b\right)}{a-b}\text{, }&a\neq b\\x\in \mathrm{R}\text{, }y\in \mathrm{R}\text{, }&a=0\text{ and }b=0\end{matrix}\right.$$

Steps Using Substitution

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$$\left(a-b\right)y+a\left(a+b\right)=0,\left(a+b\right)y+\left(a-b\right)x=2b$$

Pick one of the two equations which is more simple to solve for y by isolating y on the left hand side of the equal sign.

$$\left(a-b\right)y+a\left(a+b\right)=0$$

Subtract \left(a+b\right)a from both sides of the equation.

$$\left(a-b\right)y=-a\left(a+b\right)$$

Divide both sides by a-b.

$$y=-\frac{a\left(a+b\right)}{a-b}$$

Substitute -\frac{\left(a+b\right)a}{a-b} for y in the other equation, \left(a+b\right)y+\left(a-b\right)x=2b.

$$\left(a+b\right)\left(-\frac{a\left(a+b\right)}{a-b}\right)+\left(a-b\right)x=2b$$

Multiply a+b times -\frac{\left(a+b\right)a}{a-b}.

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

Add \frac{a\left(a+b\right)^{2}}{a-b} to both sides of the equation.

$$\left(a-b\right)x=\frac{a^{3}+ab^{2}+2ab-2b^{2}+2ba^{2}}{a-b}$$

Divide both sides by a-b.

$$x=\frac{a^{3}+ab^{2}+2ab-2b^{2}+2ba^{2}}{\left(a-b\right)^{2}}$$

The system is now solved.

$$y=-\frac{a\left(a+b\right)}{a-b},x=\frac{a^{3}+ab^{2}+2ab-2b^{2}+2ba^{2}}{\left(a-b\right)^{2}}$$