Solve {c}{x^2+y^2=100}{3x-y=10}

$$\left. \begin{array} { c } { x ^ { 2 } + y ^ { 2 } = 100 } \\ { 3 x - y = 10 } \end{array} \right.$$

Solve for x, y

$$x=0\text{, }y=-10$$

$$x=6\text{, }y=8$$

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.

$$3x-y=10,y^{2}+x^{2}=100$$

Solve 3x-y=10 for x by isolating x on the left hand side of the equal sign.

$$3x-y=10$$

Subtract -y from both sides of the equation.

$$3x=y+10$$

Divide both sides by 3.

$$x=\frac{1}{3}y+\frac{10}{3}$$

Substitute \frac{1}{3}y+\frac{10}{3} for x in the other equation, y^{2}+x^{2}=100.

$$y^{2}+\left(\frac{1}{3}y+\frac{10}{3}\right)^{2}=100$$

Square \frac{1}{3}y+\frac{10}{3}.

$$y^{2}+\frac{1}{9}y^{2}+\frac{20}{9}y+\frac{100}{9}=100$$

Add y^{2} to \frac{1}{9}y^{2}.

$$\frac{10}{9}y^{2}+\frac{20}{9}y+\frac{100}{9}=100$$

Subtract 100 from both sides of the equation.

$$\frac{10}{9}y^{2}+\frac{20}{9}y-\frac{800}{9}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times \left(\frac{1}{3}\right)^{2} for a, 1\times \frac{10}{3}\times \frac{1}{3}\times 2 for b, and -\frac{800}{9} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\frac{20}{9}±\sqrt{\left(\frac{20}{9}\right)^{2}-4\times \frac{10}{9}\left(-\frac{800}{9}\right)}}{2\times \frac{10}{9}}$$

Square 1\times \frac{10}{3}\times \frac{1}{3}\times 2.

$$y=\frac{-\frac{20}{9}±\sqrt{\frac{400}{81}-4\times \frac{10}{9}\left(-\frac{800}{9}\right)}}{2\times \frac{10}{9}}$$

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

$$y=\frac{-\frac{20}{9}±\sqrt{\frac{400}{81}-\frac{40}{9}\left(-\frac{800}{9}\right)}}{2\times \frac{10}{9}}$$

Multiply -\frac{40}{9} times -\frac{800}{9} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\frac{20}{9}±\sqrt{\frac{400+32000}{81}}}{2\times \frac{10}{9}}$$

Add \frac{400}{81} to \frac{32000}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\frac{20}{9}±\sqrt{400}}{2\times \frac{10}{9}}$$

Take the square root of 400.

$$y=\frac{-\frac{20}{9}±20}{2\times \frac{10}{9}}$$

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

$$y=\frac{-\frac{20}{9}±20}{\frac{20}{9}}$$

Now solve the equation y=\frac{-\frac{20}{9}±20}{\frac{20}{9}} when ± is plus. Add -\frac{20}{9} to 20.

$$y=\frac{\frac{160}{9}}{\frac{20}{9}}$$

Divide \frac{160}{9} by \frac{20}{9} by multiplying \frac{160}{9} by the reciprocal of \frac{20}{9}.

$$y=8$$

Now solve the equation y=\frac{-\frac{20}{9}±20}{\frac{20}{9}} when ± is minus. Subtract 20 from -\frac{20}{9}.

$$y=-\frac{\frac{200}{9}}{\frac{20}{9}}$$

Divide -\frac{200}{9} by \frac{20}{9} by multiplying -\frac{200}{9} by the reciprocal of \frac{20}{9}.

$$y=-10$$

There are two solutions for y: 8 and -10. Substitute 8 for y in the equation x=\frac{1}{3}y+\frac{10}{3} to find the corresponding solution for x that satisfies both equations.

$$x=\frac{1}{3}\times 8+\frac{10}{3}$$

Multiply \frac{1}{3} times 8.

$$x=\frac{8+10}{3}$$

Add \frac{1}{3}\times 8 to \frac{10}{3}.

$$x=6$$

Now substitute -10 for y in the equation x=\frac{1}{3}y+\frac{10}{3} and solve to find the corresponding solution for x that satisfies both equations.

$$x=\frac{1}{3}\left(-10\right)+\frac{10}{3}$$

Multiply \frac{1}{3} times -10.

$$x=\frac{-10+10}{3}$$

Add -10\times \frac{1}{3} to \frac{10}{3}.

$$x=0$$

The system is now solved.

$$x=6,y=8\text{ or }x=0,y=-10$$