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

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

Solve for x, y

$$x=\frac{a+\sqrt{2}+3}{a+4}$$

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

$$a\neq -4$$

Steps Using Substitution

Consider the second equation. Subtract y from both sides.

$$a-4x+\sqrt{2}-y=0$$

Subtract a from both sides. Anything subtracted from zero gives its negation.

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

Subtract \sqrt{2} from both sides.

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

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.

$$ax-y=3,-4x-y=-a-\sqrt{2}$$

Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.

$$ax-y=3$$

Add y to both sides of the equation.

$$ax=y+3$$

Divide both sides by a.

$$x=\frac{1}{a}\left(y+3\right)$$

Multiply \frac{1}{a} times y+3.

$$x=\frac{1}{a}y+\frac{3}{a}$$

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

$$-4\left(\frac{1}{a}y+\frac{3}{a}\right)-y=-a-\sqrt{2}$$

Multiply -4 times \frac{3+y}{a}.

$$\left(-\frac{4}{a}\right)y-\frac{12}{a}-y=-a-\sqrt{2}$$

Add -\frac{4y}{a} to -y.

$$\left(-1-\frac{4}{a}\right)y-\frac{12}{a}=-a-\sqrt{2}$$

Add \frac{12}{a} to both sides of the equation.

$$\left(-1-\frac{4}{a}\right)y=-a-\sqrt{2}+\frac{12}{a}$$

Divide both sides by -\frac{4}{a}-1.

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

Substitute -\frac{12-\sqrt{2}a-a^{2}}{4+a} for y in x=\frac{1}{a}y+\frac{3}{a}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{1}{a}\left(-\frac{-a^{2}-\sqrt{2}a+12}{a+4}\right)+\frac{3}{a}$$

Multiply \frac{1}{a} times -\frac{12-\sqrt{2}a-a^{2}}{4+a}.

$$x=-\frac{-a^{2}-\sqrt{2}a+12}{a\left(a+4\right)}+\frac{3}{a}$$

Add \frac{3}{a} to -\frac{12-\sqrt{2}a-a^{2}}{a\left(4+a\right)}.

$$x=\frac{a+\sqrt{2}+3}{a+4}$$

The system is now solved.

$$x=\frac{a+\sqrt{2}+3}{a+4},y=-\frac{-a^{2}-\sqrt{2}a+12}{a+4}$$

Steps Using Elimination

Consider the second equation. Subtract y from both sides.

$$a-4x+\sqrt{2}-y=0$$

Subtract a from both sides. Anything subtracted from zero gives its negation.

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

Subtract \sqrt{2} from both sides.

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

In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.

$$ax-y=3,-4x-y=-a-\sqrt{2}$$

Subtract -4x-y=-a-\sqrt{2} from ax-y=3 by subtracting like terms on each side of the equal sign.

$$ax+4x-y+y=3+a+\sqrt{2}$$

Add -y to y. Terms -y and y cancel out, leaving an equation with only one variable that can be solved.

$$ax+4x=3+a+\sqrt{2}$$

Add ax to 4x.

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

Add 3 to a+\sqrt{2}.

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

Divide both sides by a+4.

$$x=\frac{a+\sqrt{2}+3}{a+4}$$

Substitute \frac{3+a+\sqrt{2}}{a+4} for x in -4x-y=-a-\sqrt{2}. Because the resulting equation contains only one variable, you can solve for y directly.

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

Multiply -4 times \frac{3+a+\sqrt{2}}{a+4}.

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

Add \frac{4\left(3+a+\sqrt{2}\right)}{a+4} to both sides of the equation.

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

Divide both sides by -1.

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

The system is now solved.

$$x=\frac{a+\sqrt{2}+3}{a+4},y=-\frac{-a^{2}-\sqrt{2}a+12}{a+4}$$