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

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

Solve for y, x

$$x=\frac{7}{5}=1.4\text{, }y=\frac{1}{5}=0.2$$

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

Steps Using Substitution and Quadratic Formula

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.

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

Solve y+2x=3 for y by isolating y on the left hand side of the equal sign.

$$y+2x=3$$

Subtract 2x from both sides of the equation.

$$y=-2x+3$$

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

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

Square -2x+3.

$$x^{2}+4x^{2}-12x+9=2$$

Add x^{2} to 4x^{2}.

$$5x^{2}-12x+9=2$$

Subtract 2 from both sides of the equation.

$$5x^{2}-12x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-2\right)^{2} for a, 1\times 3\left(-2\right)\times 2 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 5\times 7}}{2\times 5}$$

Square 1\times 3\left(-2\right)\times 2.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 5\times 7}}{2\times 5}$$

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

$$x=\frac{-\left(-12\right)±\sqrt{144-20\times 7}}{2\times 5}$$

Multiply -20 times 7.

$$x=\frac{-\left(-12\right)±\sqrt{144-140}}{2\times 5}$$

Add 144 to -140.

$$x=\frac{-\left(-12\right)±\sqrt{4}}{2\times 5}$$

Take the square root of 4.

$$x=\frac{-\left(-12\right)±2}{2\times 5}$$

The opposite of 1\times 3\left(-2\right)\times 2 is 12.

$$x=\frac{12±2}{2\times 5}$$

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

$$x=\frac{12±2}{10}$$

Now solve the equation x=\frac{12±2}{10} when ± is plus. Add 12 to 2.

$$x=\frac{14}{10}$$

Reduce the fraction \frac{14}{10} to lowest terms by extracting and canceling out 2.

$$x=\frac{7}{5}$$

Now solve the equation x=\frac{12±2}{10} when ± is minus. Subtract 2 from 12.

$$x=\frac{10}{10}$$

Divide 10 by 10.

$$x=1$$

There are two solutions for x: \frac{7}{5} and 1. Substitute \frac{7}{5} for x in the equation y=-2x+3 to find the corresponding solution for y that satisfies both equations.

$$y=-2\times \frac{7}{5}+3$$

Multiply -2 times \frac{7}{5}.

$$y=-\frac{14}{5}+3$$

Add -2\times \frac{7}{5} to 3.

$$y=\frac{1}{5}$$

Now substitute 1 for x in the equation y=-2x+3 and solve to find the corresponding solution for y that satisfies both equations.

$$y=-2+3$$

Add -2 to 3.

$$y=1$$

The system is now solved.

$$y=\frac{1}{5},x=\frac{7}{5}\text{ or }y=1,x=1$$