Solve {l}{x-2y-5=0}{x^2+y^2=50}

$$\left. \begin{array} { l } { x - 2 y - 5 = 0 } \\ { x ^ { 2 } + y ^ { 2 } = 50 } \end{array} \right.$$

Solve for x, y

$$x=7\text{, }y=1$$

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

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.

$$x-2y-5=0,y^{2}+x^{2}=50$$

Solve x-2y-5=0 for x by isolating x on the left hand side of the equal sign.

$$x-2y-5=0$$

Add 5 to both sides of the equation.

$$x-2y=5$$

Subtract -2y from both sides of the equation.

$$x=2y+5$$

Substitute 2y+5 for x in the other equation, y^{2}+x^{2}=50.

$$y^{2}+\left(2y+5\right)^{2}=50$$

Square 2y+5.

$$y^{2}+4y^{2}+20y+25=50$$

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

$$5y^{2}+20y+25=50$$

Subtract 50 from both sides of the equation.

$$5y^{2}+20y-25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 2^{2} for a, 1\times 5\times 2\times 2 for b, and -25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-20±\sqrt{20^{2}-4\times 5\left(-25\right)}}{2\times 5}$$

Square 1\times 5\times 2\times 2.

$$y=\frac{-20±\sqrt{400-4\times 5\left(-25\right)}}{2\times 5}$$

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

$$y=\frac{-20±\sqrt{400-20\left(-25\right)}}{2\times 5}$$

Multiply -20 times -25.

$$y=\frac{-20±\sqrt{400+500}}{2\times 5}$$

Add 400 to 500.

$$y=\frac{-20±\sqrt{900}}{2\times 5}$$

Take the square root of 900.

$$y=\frac{-20±30}{2\times 5}$$

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

$$y=\frac{-20±30}{10}$$

Now solve the equation y=\frac{-20±30}{10} when ± is plus. Add -20 to 30.

$$y=\frac{10}{10}$$

Divide 10 by 10.

$$y=1$$

Now solve the equation y=\frac{-20±30}{10} when ± is minus. Subtract 30 from -20.

$$y=-\frac{50}{10}$$

Divide -50 by 10.

$$y=-5$$

There are two solutions for y: 1 and -5. Substitute 1 for y in the equation x=2y+5 to find the corresponding solution for x that satisfies both equations.

$$x=2+5$$

Add 1\times 2 to 5.

$$x=7$$

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

$$x=2\left(-5\right)+5$$

Multiply 2 times -5.

$$x=-10+5$$

Add -5\times 2 to 5.

$$x=-5$$

The system is now solved.

$$x=7,y=1\text{ or }x=-5,y=-5$$