Solve {l}{x^2+y^2=4}{3x+y=6}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 4 } \\ { 3 x + y = 6 } \end{array} \right.$$

Solve for x, y

$$x=\frac{8}{5}=1.6\text{, }y=\frac{6}{5}=1.2$$

$$x=2\text{, }y=0$$

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.

$$3x+y=6,y^{2}+x^{2}=4$$

Solve 3x+y=6 for x by isolating x on the left hand side of the equal sign.

$$3x+y=6$$

Subtract y from both sides of the equation.

$$3x=-y+6$$

Divide both sides by 3.

$$x=-\frac{1}{3}y+2$$

Substitute -\frac{1}{3}y+2 for x in the other equation, y^{2}+x^{2}=4.

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

Square -\frac{1}{3}y+2.

$$y^{2}+\frac{1}{9}y^{2}-\frac{4}{3}y+4=4$$

Add y^{2} to \frac{1}{9}y^{2}.

$$\frac{10}{9}y^{2}-\frac{4}{3}y+4=4$$

Subtract 4 from both sides of the equation.

$$\frac{10}{9}y^{2}-\frac{4}{3}y=0$$

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

$$y=\frac{-\left(-\frac{4}{3}\right)±\sqrt{\left(-\frac{4}{3}\right)^{2}}}{2\times \frac{10}{9}}$$

Take the square root of \left(-\frac{4}{3}\right)^{2}.

$$y=\frac{-\left(-\frac{4}{3}\right)±\frac{4}{3}}{2\times \frac{10}{9}}$$

The opposite of 1\times 2\left(-\frac{1}{3}\right)\times 2 is \frac{4}{3}.

$$y=\frac{\frac{4}{3}±\frac{4}{3}}{2\times \frac{10}{9}}$$

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

$$y=\frac{\frac{4}{3}±\frac{4}{3}}{\frac{20}{9}}$$

Now solve the equation y=\frac{\frac{4}{3}±\frac{4}{3}}{\frac{20}{9}} when ± is plus. Add \frac{4}{3} to \frac{4}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{\frac{8}{3}}{\frac{20}{9}}$$

Divide \frac{8}{3} by \frac{20}{9} by multiplying \frac{8}{3} by the reciprocal of \frac{20}{9}.

$$y=\frac{6}{5}$$

Now solve the equation y=\frac{\frac{4}{3}±\frac{4}{3}}{\frac{20}{9}} when ± is minus. Subtract \frac{4}{3} from \frac{4}{3} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{0}{\frac{20}{9}}$$

Divide 0 by \frac{20}{9} by multiplying 0 by the reciprocal of \frac{20}{9}.

$$y=0$$

There are two solutions for y: \frac{6}{5} and 0. Substitute \frac{6}{5} for y in the equation x=-\frac{1}{3}y+2 to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{1}{3}\times \frac{6}{5}+2$$

Multiply -\frac{1}{3} times \frac{6}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{2}{5}+2$$

Add -\frac{1}{3}\times \frac{6}{5} to 2.

$$x=\frac{8}{5}$$

Now substitute 0 for y in the equation x=-\frac{1}{3}y+2 and solve to find the corresponding solution for x that satisfies both equations.

$$x=2$$

The system is now solved.

$$x=\frac{8}{5},y=\frac{6}{5}\text{ or }x=2,y=0$$