Solve {c}{8x+15y=289}{x^2+y^2=17^2}

$$\left. \begin{array} { c } { 8 x + 15 y = 289 } \\ { x ^ { 2 } + y ^ { 2 } = 17 ^ { 2 } } \end{array} \right.$$

Solve for x, y

$$x=8$$

$$y=15$$

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.

$$8x+15y=289,y^{2}+x^{2}=289$$

Solve 8x+15y=289 for x by isolating x on the left hand side of the equal sign.

$$8x+15y=289$$

Subtract 15y from both sides of the equation.

$$8x=-15y+289$$

Divide both sides by 8.

$$x=-\frac{15}{8}y+\frac{289}{8}$$

Substitute -\frac{15}{8}y+\frac{289}{8} for x in the other equation, y^{2}+x^{2}=289.

$$y^{2}+\left(-\frac{15}{8}y+\frac{289}{8}\right)^{2}=289$$

Square -\frac{15}{8}y+\frac{289}{8}.

$$y^{2}+\frac{225}{64}y^{2}-\frac{4335}{32}y+\frac{83521}{64}=289$$

Add y^{2} to \frac{225}{64}y^{2}.

$$\frac{289}{64}y^{2}-\frac{4335}{32}y+\frac{83521}{64}=289$$

Subtract 289 from both sides of the equation.

$$\frac{289}{64}y^{2}-\frac{4335}{32}y+\frac{65025}{64}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-\frac{15}{8}\right)^{2} for a, 1\times \frac{289}{8}\left(-\frac{15}{8}\right)\times 2 for b, and \frac{65025}{64} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-\frac{4335}{32}\right)±\sqrt{\left(-\frac{4335}{32}\right)^{2}-4\times \frac{289}{64}\times \frac{65025}{64}}}{2\times \frac{289}{64}}$$

Square 1\times \frac{289}{8}\left(-\frac{15}{8}\right)\times 2.

$$y=\frac{-\left(-\frac{4335}{32}\right)±\sqrt{\frac{18792225}{1024}-4\times \frac{289}{64}\times \frac{65025}{64}}}{2\times \frac{289}{64}}$$

Multiply -4 times 1+1\left(-\frac{15}{8}\right)^{2}.

$$y=\frac{-\left(-\frac{4335}{32}\right)±\sqrt{\frac{18792225}{1024}-\frac{289}{16}\times \frac{65025}{64}}}{2\times \frac{289}{64}}$$

Multiply -\frac{289}{16} times \frac{65025}{64} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{4335}{32}\right)±\sqrt{\frac{18792225-18792225}{1024}}}{2\times \frac{289}{64}}$$

Add \frac{18792225}{1024} to -\frac{18792225}{1024} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{4335}{32}\right)±\sqrt{0}}{2\times \frac{289}{64}}$$

Take the square root of 0.

$$y=-\frac{-\frac{4335}{32}}{2\times \frac{289}{64}}$$

The opposite of 1\times \frac{289}{8}\left(-\frac{15}{8}\right)\times 2 is \frac{4335}{32}.

$$y=\frac{\frac{4335}{32}}{2\times \frac{289}{64}}$$

Multiply 2 times 1+1\left(-\frac{15}{8}\right)^{2}.

$$y=\frac{\frac{4335}{32}}{\frac{289}{32}}$$

Divide \frac{4335}{32} by \frac{289}{32} by multiplying \frac{4335}{32} by the reciprocal of \frac{289}{32}.

$$y=15$$

There are two solutions for y: 15 and 15. Substitute 15 for y in the equation x=-\frac{15}{8}y+\frac{289}{8} to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{15}{8}\times 15+\frac{289}{8}$$

Multiply -\frac{15}{8} times 15.

$$x=\frac{-225+289}{8}$$

Add -\frac{15}{8}\times 15 to \frac{289}{8}.

$$x=8$$

The system is now solved.

$$x=8,y=15\text{ or }x=8,y=15$$