Solve {l}{x^2+y^2=9}{2x-y=3}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 9 } \\ { 2 x - y = 3 } \end{array} \right.$$

Solve for x, y

$$x=0\text{, }y=-3$$

$$x=\frac{12}{5}=2.4\text{, }y=\frac{9}{5}=1.8$$

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.

$$2x-y=3,y^{2}+x^{2}=9$$

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

$$2x-y=3$$

Subtract -y from both sides of the equation.

$$2x=y+3$$

Divide both sides by 2.

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

Substitute \frac{1}{2}y+\frac{3}{2} for x in the other equation, y^{2}+x^{2}=9.

$$y^{2}+\left(\frac{1}{2}y+\frac{3}{2}\right)^{2}=9$$

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

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

Add y^{2} to \frac{1}{4}y^{2}.

$$\frac{5}{4}y^{2}+\frac{3}{2}y+\frac{9}{4}=9$$

Subtract 9 from both sides of the equation.

$$\frac{5}{4}y^{2}+\frac{3}{2}y-\frac{27}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times \left(\frac{1}{2}\right)^{2} for a, 1\times \frac{3}{2}\times \frac{1}{2}\times 2 for b, and -\frac{27}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\frac{3}{2}±\sqrt{\left(\frac{3}{2}\right)^{2}-4\times \frac{5}{4}\left(-\frac{27}{4}\right)}}{2\times \frac{5}{4}}$$

Square 1\times \frac{3}{2}\times \frac{1}{2}\times 2.

$$y=\frac{-\frac{3}{2}±\sqrt{\frac{9}{4}-4\times \frac{5}{4}\left(-\frac{27}{4}\right)}}{2\times \frac{5}{4}}$$

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

$$y=\frac{-\frac{3}{2}±\sqrt{\frac{9}{4}-5\left(-\frac{27}{4}\right)}}{2\times \frac{5}{4}}$$

Multiply -5 times -\frac{27}{4}.

$$y=\frac{-\frac{3}{2}±\sqrt{\frac{9+135}{4}}}{2\times \frac{5}{4}}$$

Add \frac{9}{4} to \frac{135}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\frac{3}{2}±\sqrt{36}}{2\times \frac{5}{4}}$$

Take the square root of 36.

$$y=\frac{-\frac{3}{2}±6}{2\times \frac{5}{4}}$$

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

$$y=\frac{-\frac{3}{2}±6}{\frac{5}{2}}$$

Now solve the equation y=\frac{-\frac{3}{2}±6}{\frac{5}{2}} when ± is plus. Add -\frac{3}{2} to 6.

$$y=\frac{\frac{9}{2}}{\frac{5}{2}}$$

Divide \frac{9}{2} by \frac{5}{2} by multiplying \frac{9}{2} by the reciprocal of \frac{5}{2}.

$$y=\frac{9}{5}$$

Now solve the equation y=\frac{-\frac{3}{2}±6}{\frac{5}{2}} when ± is minus. Subtract 6 from -\frac{3}{2}.

$$y=-\frac{\frac{15}{2}}{\frac{5}{2}}$$

Divide -\frac{15}{2} by \frac{5}{2} by multiplying -\frac{15}{2} by the reciprocal of \frac{5}{2}.

$$y=-3$$

There are two solutions for y: \frac{9}{5} and -3. Substitute \frac{9}{5} for y in the equation x=\frac{1}{2}y+\frac{3}{2} to find the corresponding solution for x that satisfies both equations.

$$x=\frac{1}{2}\times \frac{9}{5}+\frac{3}{2}$$

Multiply \frac{1}{2} times \frac{9}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{9}{10}+\frac{3}{2}$$

Add \frac{1}{2}\times \frac{9}{5} to \frac{3}{2}.

$$x=\frac{12}{5}$$

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

$$x=\frac{1}{2}\left(-3\right)+\frac{3}{2}$$

Multiply \frac{1}{2} times -3.

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

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

$$x=0$$

The system is now solved.

$$x=\frac{12}{5},y=\frac{9}{5}\text{ or }x=0,y=-3$$