Solve {l}{x^2+y^2=25}{x=3y-5}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 25 } \\ { x = 3 y - 5 } \end{array} \right.$$

Solve for x, y

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

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

Steps Using Substitution and Quadratic Formula

Consider the second equation. Subtract 3y from both sides.

$$x-3y=-5$$

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.

$$x-3y=-5,y^{2}+x^{2}=25$$

Solve x-3y=-5 for x by isolating x on the left hand side of the equal sign.

$$x-3y=-5$$

Subtract -3y from both sides of the equation.

$$x=3y-5$$

Substitute 3y-5 for x in the other equation, y^{2}+x^{2}=25.

$$y^{2}+\left(3y-5\right)^{2}=25$$

Square 3y-5.

$$y^{2}+9y^{2}-30y+25=25$$

Add y^{2} to 9y^{2}.

$$10y^{2}-30y+25=25$$

Subtract 25 from both sides of the equation.

$$10y^{2}-30y=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 3^{2} for a, 1\left(-5\right)\times 2\times 3 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-30\right)±\sqrt{\left(-30\right)^{2}}}{2\times 10}$$

Take the square root of \left(-30\right)^{2}.

$$y=\frac{-\left(-30\right)±30}{2\times 10}$$

The opposite of 1\left(-5\right)\times 2\times 3 is 30.

$$y=\frac{30±30}{2\times 10}$$

Multiply 2 times 1+1\times 3^{2}.

$$y=\frac{30±30}{20}$$

Now solve the equation y=\frac{30±30}{20} when ± is plus. Add 30 to 30.

$$y=\frac{60}{20}$$

Divide 60 by 20.

$$y=3$$

Now solve the equation y=\frac{30±30}{20} when ± is minus. Subtract 30 from 30.

$$y=\frac{0}{20}$$

Divide 0 by 20.

$$y=0$$

There are two solutions for y: 3 and 0. Substitute 3 for y in the equation x=3y-5 to find the corresponding solution for x that satisfies both equations.

$$x=3\times 3-5$$

Multiply 3 times 3.

$$x=9-5$$

Add 3\times 3 to -5.

$$x=4$$

Now substitute 0 for y in the equation x=3y-5 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-5$$

The system is now solved.

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