Solve {l}{x^2+y^2=13}{2x+3y=13}

$$\left\{ \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 13 } \\ { 2 x + 3 y = 13 } \end{array} \right.$$

Solve for x, y

$$x=2$$

$$y=3$$

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.

$$2x+3y=13,y^{2}+x^{2}=13$$

Solve 2x+3y=13 for x by isolating x on the left hand side of the equal sign.

$$2x+3y=13$$

Subtract 3y from both sides of the equation.

$$2x=-3y+13$$

Divide both sides by 2.

$$x=-\frac{3}{2}y+\frac{13}{2}$$

Substitute -\frac{3}{2}y+\frac{13}{2} for x in the other equation, y^{2}+x^{2}=13.

$$y^{2}+\left(-\frac{3}{2}y+\frac{13}{2}\right)^{2}=13$$

Square -\frac{3}{2}y+\frac{13}{2}.

$$y^{2}+\frac{9}{4}y^{2}-\frac{39}{2}y+\frac{169}{4}=13$$

Add y^{2} to \frac{9}{4}y^{2}.

$$\frac{13}{4}y^{2}-\frac{39}{2}y+\frac{169}{4}=13$$

Subtract 13 from both sides of the equation.

$$\frac{13}{4}y^{2}-\frac{39}{2}y+\frac{117}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-\frac{3}{2}\right)^{2} for a, 1\times \frac{13}{2}\left(-\frac{3}{2}\right)\times 2 for b, and \frac{117}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-\frac{39}{2}\right)±\sqrt{\left(-\frac{39}{2}\right)^{2}-4\times \frac{13}{4}\times \frac{117}{4}}}{2\times \frac{13}{4}}$$

Square 1\times \frac{13}{2}\left(-\frac{3}{2}\right)\times 2.

$$y=\frac{-\left(-\frac{39}{2}\right)±\sqrt{\frac{1521}{4}-4\times \frac{13}{4}\times \frac{117}{4}}}{2\times \frac{13}{4}}$$

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

$$y=\frac{-\left(-\frac{39}{2}\right)±\sqrt{\frac{1521}{4}-13\times \frac{117}{4}}}{2\times \frac{13}{4}}$$

Multiply -13 times \frac{117}{4}.

$$y=\frac{-\left(-\frac{39}{2}\right)±\sqrt{\frac{1521-1521}{4}}}{2\times \frac{13}{4}}$$

Add \frac{1521}{4} to -\frac{1521}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{39}{2}\right)±\sqrt{0}}{2\times \frac{13}{4}}$$

Take the square root of 0.

$$y=-\frac{-\frac{39}{2}}{2\times \frac{13}{4}}$$

The opposite of 1\times \frac{13}{2}\left(-\frac{3}{2}\right)\times 2 is \frac{39}{2}.

$$y=\frac{\frac{39}{2}}{2\times \frac{13}{4}}$$

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

$$y=\frac{\frac{39}{2}}{\frac{13}{2}}$$

Divide \frac{39}{2} by \frac{13}{2} by multiplying \frac{39}{2} by the reciprocal of \frac{13}{2}.

$$y=3$$

There are two solutions for y: 3 and 3. Substitute 3 for y in the equation x=-\frac{3}{2}y+\frac{13}{2} to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{3}{2}\times 3+\frac{13}{2}$$

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

$$x=\frac{-9+13}{2}$$

Add -\frac{3}{2}\times 3 to \frac{13}{2}.

$$x=2$$

The system is now solved.

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