Solve {l}{x^2+y^2-9=0}{2x-y+6=0}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } - 9 = 0 } \\ { 2 x - y + 6 = 0 } \end{array} \right.$$

Solve for x, y

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

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

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.

$$2x-y+6=0,y^{2}+x^{2}-9=0$$

Solve 2x-y+6=0 for x by isolating x on the left hand side of the equal sign.

$$2x-y+6=0$$

Subtract 6 from both sides of the equation.

$$2x-y=-6$$

Subtract -y from both sides of the equation.

$$2x=y-6$$

Divide both sides by 2.

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

Substitute \frac{1}{2}y-3 for x in the other equation, y^{2}+x^{2}-9=0.

$$y^{2}+\left(\frac{1}{2}y-3\right)^{2}-9=0$$

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

$$y^{2}+\frac{1}{4}y^{2}-3y+9-9=0$$

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

$$\frac{5}{4}y^{2}-3y+9-9=0$$

Add 1\left(-3\right)^{2} to -9.

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

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

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

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

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

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

$$y=\frac{3±3}{2\times \frac{5}{4}}$$

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

$$y=\frac{3±3}{\frac{5}{2}}$$

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

$$y=\frac{6}{\frac{5}{2}}$$

Divide 6 by \frac{5}{2} by multiplying 6 by the reciprocal of \frac{5}{2}.

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

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

$$y=\frac{0}{\frac{5}{2}}$$

Divide 0 by \frac{5}{2} by multiplying 0 by the reciprocal of \frac{5}{2}.

$$y=0$$

There are two solutions for y: \frac{12}{5} and 0. Substitute \frac{12}{5} for y in the equation x=\frac{1}{2}y-3 to find the corresponding solution for x that satisfies both equations.

$$x=\frac{1}{2}\times \frac{12}{5}-3$$

Multiply \frac{1}{2} times \frac{12}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{6}{5}-3$$

Add \frac{1}{2}\times \frac{12}{5} to -3.

$$x=-\frac{9}{5}$$

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

$$x=-3$$

The system is now solved.

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