Solve {l}{x^2/3+y^2/1=1}{x+y=2}

$$\left\{ \begin{array} { l } { \frac { x ^ { 2 } } { 3 } + \frac { y ^ { 2 } } { 1 } = 1 } \\ { x + y = 2 } \end{array} \right.$$

Solve for x, y

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

$$y=\frac{1}{2}=0.5$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Multiply both sides of the equation by 3.

$$x^{2}+3y^{2}=3$$

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.

$$x+y=2,3y^{2}+x^{2}=3$$

Solve x+y=2 for x by isolating x on the left hand side of the equal sign.

$$x+y=2$$

Subtract y from both sides of the equation.

$$x=-y+2$$

Substitute -y+2 for x in the other equation, 3y^{2}+x^{2}=3.

$$3y^{2}+\left(-y+2\right)^{2}=3$$

Square -y+2.

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

Add 3y^{2} to y^{2}.

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

Subtract 3 from both sides of the equation.

$$4y^{2}-4y+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3+1\left(-1\right)^{2} for a, 1\times 2\left(-1\right)\times 2 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 1\times 2\left(-1\right)\times 2.

$$y=\frac{-\left(-4\right)±\sqrt{16-4\times 4}}{2\times 4}$$

Multiply -4 times 3+1\left(-1\right)^{2}.

$$y=\frac{-\left(-4\right)±\sqrt{16-16}}{2\times 4}$$

Add 16 to -16.

$$y=\frac{-\left(-4\right)±\sqrt{0}}{2\times 4}$$

Take the square root of 0.

$$y=-\frac{-4}{2\times 4}$$

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

$$y=\frac{4}{2\times 4}$$

Multiply 2 times 3+1\left(-1\right)^{2}.

$$y=\frac{4}{8}$$

Reduce the fraction \frac{4}{8} to lowest terms by extracting and canceling out 4.

$$y=\frac{1}{2}$$

There are two solutions for y: \frac{1}{2} and \frac{1}{2}. Substitute \frac{1}{2} for y in the equation x=-y+2 to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{1}{2}+2$$

Add -\frac{1}{2} to 2.

$$x=\frac{3}{2}$$

The system is now solved.

$$x=\frac{3}{2},y=\frac{1}{2}\text{ or }x=\frac{3}{2},y=\frac{1}{2}$$