Solve {l}{ax/b-by/a=a+b}{ax-bu=2ab}

$$\left. \begin{array} { l } { \frac { a x } { b } - \frac { b y } { a } = a + b } \\ { a x - b u = 2 a b } \end{array} \right.$$

Solve for x, y

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

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

$$b\neq 0\text{ and }a\neq 0$$

Steps Using Substitution

Consider the second equation. Add bu to both sides.

$$ax=2ab+bu$$

Consider the first equation. Multiply both sides of the equation by ab, the least common multiple of b,a.

$$aax-bby=aba+abb$$

Multiply a and a to get a^{2}.

$$a^{2}x-bby=aba+abb$$

Multiply b and b to get b^{2}.

$$a^{2}x-b^{2}y=aba+abb$$

Multiply a and a to get a^{2}.

$$a^{2}x-b^{2}y=a^{2}b+abb$$

Multiply b and b to get b^{2}.

$$a^{2}x-b^{2}y=a^{2}b+ab^{2}$$

Reorder the terms.

$$xa^{2}-yb^{2}=ab^{2}+ba^{2}$$

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.

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

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

$$ax=bu+2ab$$

Divide both sides by a.

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

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

$$a^{2}\times \frac{b\left(u+2a\right)}{a}+\left(-b^{2}\right)y=ab^{2}+ba^{2}$$

Multiply a^{2} times \frac{b\left(u+2a\right)}{a}.

$$ab\left(u+2a\right)+\left(-b^{2}\right)y=ab^{2}+ba^{2}$$

Subtract b\left(u+2a\right)a from both sides of the equation.

$$\left(-b^{2}\right)y=ab\left(-u+b-a\right)$$

Divide both sides by -b^{2}.

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

The system is now solved.

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