Solve {l}{x^2+y^2=25}{y=2x-2}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 25 } \\ { y = 2 x - 2 } \end{array} \right.$$

Solve for x, y

$$x=3\text{, }y=4$$

$$x=-\frac{7}{5}=-1.4\text{, }y=-\frac{24}{5}=-4.8$$

Steps Using Substitution and Quadratic Formula

Consider the second equation. Subtract 2x from both sides.

$$y-2x=-2$$

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.

$$y-2x=-2,x^{2}+y^{2}=25$$

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

$$y-2x=-2$$

Subtract -2x from both sides of the equation.

$$y=2x-2$$

Substitute 2x-2 for y in the other equation, x^{2}+y^{2}=25.

$$x^{2}+\left(2x-2\right)^{2}=25$$

Square 2x-2.

$$x^{2}+4x^{2}-8x+4=25$$

Add x^{2} to 4x^{2}.

$$5x^{2}-8x+4=25$$

Subtract 25 from both sides of the equation.

$$5x^{2}-8x-21=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 2^{2} for a, 1\left(-2\right)\times 2\times 2 for b, and -21 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 5\left(-21\right)}}{2\times 5}$$

Square 1\left(-2\right)\times 2\times 2.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 5\left(-21\right)}}{2\times 5}$$

Multiply -4 times 1+1\times 2^{2}.

$$x=\frac{-\left(-8\right)±\sqrt{64-20\left(-21\right)}}{2\times 5}$$

Multiply -20 times -21.

$$x=\frac{-\left(-8\right)±\sqrt{64+420}}{2\times 5}$$

Add 64 to 420.

$$x=\frac{-\left(-8\right)±\sqrt{484}}{2\times 5}$$

Take the square root of 484.

$$x=\frac{-\left(-8\right)±22}{2\times 5}$$

The opposite of 1\left(-2\right)\times 2\times 2 is 8.

$$x=\frac{8±22}{2\times 5}$$

Multiply 2 times 1+1\times 2^{2}.

$$x=\frac{8±22}{10}$$

Now solve the equation x=\frac{8±22}{10} when ± is plus. Add 8 to 22.

$$x=\frac{30}{10}$$

Divide 30 by 10.

$$x=3$$

Now solve the equation x=\frac{8±22}{10} when ± is minus. Subtract 22 from 8.

$$x=-\frac{14}{10}$$

Reduce the fraction \frac{-14}{10} to lowest terms by extracting and canceling out 2.

$$x=-\frac{7}{5}$$

There are two solutions for x: 3 and -\frac{7}{5}. Substitute 3 for x in the equation y=2x-2 to find the corresponding solution for y that satisfies both equations.

$$y=2\times 3-2$$

Multiply 2 times 3.

$$y=6-2$$

Add 2\times 3 to -2.

$$y=4$$

Now substitute -\frac{7}{5} for x in the equation y=2x-2 and solve to find the corresponding solution for y that satisfies both equations.

$$y=2\left(-\frac{7}{5}\right)-2$$

Multiply 2 times -\frac{7}{5}.

$$y=-\frac{14}{5}-2$$

Add -\frac{7}{5}\times 2 to -2.

$$y=-\frac{24}{5}$$

The system is now solved.

$$y=4,x=3\text{ or }y=-\frac{24}{5},x=-\frac{7}{5}$$