Solve {l}{ax-y=3}{(a-4)x+sqrt{2}=4}

$$\left\{ \begin{array} { l } { a x - y = 3 } \\ { ( a - 4 ) x + \sqrt { 2 } = 4 } \end{array} \right.$$

Solve for x, y

$$x=-\frac{\sqrt{2}-4}{a-4}$$

$$y=\frac{-\sqrt{2}a+a+12}{a-4}$$

$$a\neq 4$$

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.

$$\left(a-4\right)x+\sqrt{2}=4,ax-y=3$$

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.

$$\left(a-4\right)x+\sqrt{2}=4$$

Subtract \sqrt{2} from both sides of the equation.

$$\left(a-4\right)x=4-\sqrt{2}$$

Divide both sides by a-4.

$$x=\frac{4-\sqrt{2}}{a-4}$$

Substitute \frac{4-\sqrt{2}}{a-4} for x in the other equation, ax-y=3.

$$a\times \frac{4-\sqrt{2}}{a-4}-y=3$$

Multiply a times \frac{4-\sqrt{2}}{a-4}.

$$\frac{\left(4-\sqrt{2}\right)a}{a-4}-y=3$$

Subtract \frac{a\left(4-\sqrt{2}\right)}{a-4} from both sides of the equation.

$$-y=\frac{\sqrt{2}a-a-12}{a-4}$$

Divide both sides by -1.

$$y=-\frac{\sqrt{2}a-a-12}{a-4}$$

The system is now solved.

$$x=\frac{4-\sqrt{2}}{a-4},y=-\frac{\sqrt{2}a-a-12}{a-4}$$