Solve {l}{y=x-1}{x^2/2+y^2=1}

$$\left\{ \begin{array} { l } { y = x - 1 } \\ { \frac { x ^ { 2 } } { 2 } + y ^ { 2 } = 1 } \end{array} \right.$$

Solve for y, x

$$x=0\text{, }y=-1$$

$$x=\frac{4}{3}\approx 1.333333333\text{, }y=\frac{1}{3}\approx 0.333333333$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Subtract x from both sides.

$$y-x=-1$$

Consider the second equation. Multiply both sides of the equation by 2.

$$x^{2}+2y^{2}=2$$

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.

$$y-x=-1,x^{2}+2y^{2}=2$$

Solve y-x=-1 for y by isolating y on the left hand side of the equal sign.

$$y-x=-1$$

Subtract -x from both sides of the equation.

$$y=x-1$$

Substitute x-1 for y in the other equation, x^{2}+2y^{2}=2.

$$x^{2}+2\left(x-1\right)^{2}=2$$

Square x-1.

$$x^{2}+2\left(x^{2}-2x+1\right)=2$$

Multiply 2 times x^{2}-2x+1.

$$x^{2}+2x^{2}-4x+2=2$$

Add x^{2} to 2x^{2}.

$$3x^{2}-4x+2=2$$

Subtract 2 from both sides of the equation.

$$3x^{2}-4x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+2\times 1^{2} for a, 2\left(-1\right)\times 1\times 2 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}}}{2\times 3}$$

Take the square root of \left(-4\right)^{2}.

$$x=\frac{-\left(-4\right)±4}{2\times 3}$$

The opposite of 2\left(-1\right)\times 1\times 2 is 4.

$$x=\frac{4±4}{2\times 3}$$

Multiply 2 times 1+2\times 1^{2}.

$$x=\frac{4±4}{6}$$

Now solve the equation x=\frac{4±4}{6} when ± is plus. Add 4 to 4.

$$x=\frac{8}{6}$$

Reduce the fraction \frac{8}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{4}{3}$$

Now solve the equation x=\frac{4±4}{6} when ± is minus. Subtract 4 from 4.

$$x=\frac{0}{6}$$

Divide 0 by 6.

$$x=0$$

There are two solutions for x: \frac{4}{3} and 0. Substitute \frac{4}{3} for x in the equation y=x-1 to find the corresponding solution for y that satisfies both equations.

$$y=\frac{4}{3}-1$$

Add 1\times \frac{4}{3} to -1.

$$y=\frac{1}{3}$$

Now substitute 0 for x in the equation y=x-1 and solve to find the corresponding solution for y that satisfies both equations.

$$y=-1$$

The system is now solved.

$$y=\frac{1}{3},x=\frac{4}{3}\text{ or }y=-1,x=0$$