Solve {l}{y=2x-5}{x^2+y^2=25}

$$\left\{ \begin{array} { l } { y = 2 x - 5 } \\ { x ^ { 2 } + y ^ { 2 } = 25 } \end{array} \right.$$

Solve for y, x

$$x=0\text{, }y=-5$$

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

Steps Using Substitution and Quadratic Formula

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

$$y-2x=-5$$

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.

$$y-2x=-5,x^{2}+y^{2}=25$$

Solve y-2x=-5 for y by isolating y on the left hand side of the equal sign.

$$y-2x=-5$$

Subtract -2x from both sides of the equation.

$$y=2x-5$$

Substitute 2x-5 for y in the other equation, x^{2}+y^{2}=25.

$$x^{2}+\left(2x-5\right)^{2}=25$$

Square 2x-5.

$$x^{2}+4x^{2}-20x+25=25$$

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

$$5x^{2}-20x+25=25$$

Subtract 25 from both sides of the equation.

$$5x^{2}-20x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 2^{2} for a, 1\left(-5\right)\times 2\times 2 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Take the square root of \left(-20\right)^{2}.

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

The opposite of 1\left(-5\right)\times 2\times 2 is 20.

$$x=\frac{20±20}{2\times 5}$$

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

$$x=\frac{20±20}{10}$$

Now solve the equation x=\frac{20±20}{10} when ± is plus. Add 20 to 20.

$$x=\frac{40}{10}$$

Divide 40 by 10.

$$x=4$$

Now solve the equation x=\frac{20±20}{10} when ± is minus. Subtract 20 from 20.

$$x=\frac{0}{10}$$

Divide 0 by 10.

$$x=0$$

There are two solutions for x: 4 and 0. Substitute 4 for x in the equation y=2x-5 to find the corresponding solution for y that satisfies both equations.

$$y=2\times 4-5$$

Multiply 2 times 4.

$$y=8-5$$

Add 2\times 4 to -5.

$$y=3$$

Now substitute 0 for x in the equation y=2x-5 and solve to find the corresponding solution for y that satisfies both equations.

$$y=-5$$

The system is now solved.

$$y=3,x=4\text{ or }y=-5,x=0$$