Solve {l}{y^2=13-x^2}{3x+2y=13}quad(48)^circ

$$\left.\begin{array} { l } { y ^ { 2 } = 13 - x ^ { 2 } } \\ { 3 x + 2 y = 13 } \end{array} \right\} \quad ( 48 ) ^ { \circ }$$

Solve for y, x

$$x=3$$

$$y=2$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Add x^{2} to both sides.

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

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.

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

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

$$3x+2y=13$$

Subtract 2y from both sides of the equation.

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

Divide both sides by 3.

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

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

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

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

$$y^{2}+\frac{4}{9}y^{2}-\frac{52}{9}y+\frac{169}{9}=13$$

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

$$\frac{13}{9}y^{2}-\frac{52}{9}y+\frac{169}{9}=13$$

Subtract 13 from both sides of the equation.

$$\frac{13}{9}y^{2}-\frac{52}{9}y+\frac{52}{9}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-\frac{2}{3}\right)^{2} for a, 1\times \frac{13}{3}\left(-\frac{2}{3}\right)\times 2 for b, and \frac{52}{9} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-\frac{52}{9}\right)±\sqrt{\left(-\frac{52}{9}\right)^{2}-4\times \frac{13}{9}\times \frac{52}{9}}}{2\times \frac{13}{9}}$$

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

$$y=\frac{-\left(-\frac{52}{9}\right)±\sqrt{\frac{2704}{81}-4\times \frac{13}{9}\times \frac{52}{9}}}{2\times \frac{13}{9}}$$

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

$$y=\frac{-\left(-\frac{52}{9}\right)±\sqrt{\frac{2704}{81}-\frac{52}{9}\times \frac{52}{9}}}{2\times \frac{13}{9}}$$

Multiply -\frac{52}{9} times \frac{52}{9} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{52}{9}\right)±\sqrt{\frac{2704-2704}{81}}}{2\times \frac{13}{9}}$$

Add \frac{2704}{81} to -\frac{2704}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{52}{9}\right)±\sqrt{0}}{2\times \frac{13}{9}}$$

Take the square root of 0.

$$y=-\frac{-\frac{52}{9}}{2\times \frac{13}{9}}$$

The opposite of 1\times \frac{13}{3}\left(-\frac{2}{3}\right)\times 2 is \frac{52}{9}.

$$y=\frac{\frac{52}{9}}{2\times \frac{13}{9}}$$

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

$$y=\frac{\frac{52}{9}}{\frac{26}{9}}$$

Divide \frac{52}{9} by \frac{26}{9} by multiplying \frac{52}{9} by the reciprocal of \frac{26}{9}.

$$y=2$$

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

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

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

$$x=\frac{-4+13}{3}$$

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

$$x=3$$

The system is now solved.

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