Solve 2x+y=10quad,quadx^2+y^2=25

$$2 x + y = 10 \quad , \quad x ^ { 2 } + y ^ { 2 } = 25$$

Solve for x, y

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

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

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.

$$2x+y=10,y^{2}+x^{2}=25$$

Solve 2x+y=10 for x by isolating x on the left hand side of the equal sign.

$$2x+y=10$$

Subtract y from both sides of the equation.

$$2x=-y+10$$

Divide both sides by 2.

$$x=-\frac{1}{2}y+5$$

Substitute -\frac{1}{2}y+5 for x in the other equation, y^{2}+x^{2}=25.

$$y^{2}+\left(-\frac{1}{2}y+5\right)^{2}=25$$

Square -\frac{1}{2}y+5.

$$y^{2}+\frac{1}{4}y^{2}-5y+25=25$$

Add y^{2} to \frac{1}{4}y^{2}.

$$\frac{5}{4}y^{2}-5y+25=25$$

Subtract 25 from both sides of the equation.

$$\frac{5}{4}y^{2}-5y=0$$

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

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

Take the square root of \left(-5\right)^{2}.

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

The opposite of 1\times 5\left(-\frac{1}{2}\right)\times 2 is 5.

$$y=\frac{5±5}{2\times \frac{5}{4}}$$

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

$$y=\frac{5±5}{\frac{5}{2}}$$

Now solve the equation y=\frac{5±5}{\frac{5}{2}} when ± is plus. Add 5 to 5.

$$y=\frac{10}{\frac{5}{2}}$$

Divide 10 by \frac{5}{2} by multiplying 10 by the reciprocal of \frac{5}{2}.

$$y=4$$

Now solve the equation y=\frac{5±5}{\frac{5}{2}} when ± is minus. Subtract 5 from 5.

$$y=\frac{0}{\frac{5}{2}}$$

Divide 0 by \frac{5}{2} by multiplying 0 by the reciprocal of \frac{5}{2}.

$$y=0$$

There are two solutions for y: 4 and 0. Substitute 4 for y in the equation x=-\frac{1}{2}y+5 to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{1}{2}\times 4+5$$

Multiply -\frac{1}{2} times 4.

$$x=-2+5$$

Add -\frac{1}{2}\times 4 to 5.

$$x=3$$

Now substitute 0 for y in the equation x=-\frac{1}{2}y+5 and solve to find the corresponding solution for x that satisfies both equations.

$$x=5$$

The system is now solved.

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