Solve {l}{x^2+y^2=49}{x+2y=7}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 49 } \\ { x + 2 y = 7 } \end{array} \right.$$

Solve for x, y

$$x=7\text{, }y=0$$

$$x=-\frac{21}{5}=-4.2\text{, }y=\frac{28}{5}=5.6$$

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.

$$x+2y=7,y^{2}+x^{2}=49$$

Solve x+2y=7 for x by isolating x on the left hand side of the equal sign.

$$x+2y=7$$

Subtract 2y from both sides of the equation.

$$x=-2y+7$$

Substitute -2y+7 for x in the other equation, y^{2}+x^{2}=49.

$$y^{2}+\left(-2y+7\right)^{2}=49$$

Square -2y+7.

$$y^{2}+4y^{2}-28y+49=49$$

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

$$5y^{2}-28y+49=49$$

Subtract 49 from both sides of the equation.

$$5y^{2}-28y=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-2\right)^{2} for a, 1\times 7\left(-2\right)\times 2 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}}}{2\times 5}$$

Take the square root of \left(-28\right)^{2}.

$$y=\frac{-\left(-28\right)±28}{2\times 5}$$

The opposite of 1\times 7\left(-2\right)\times 2 is 28.

$$y=\frac{28±28}{2\times 5}$$

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

$$y=\frac{28±28}{10}$$

Now solve the equation y=\frac{28±28}{10} when ± is plus. Add 28 to 28.

$$y=\frac{56}{10}$$

Reduce the fraction \frac{56}{10} to lowest terms by extracting and canceling out 2.

$$y=\frac{28}{5}$$

Now solve the equation y=\frac{28±28}{10} when ± is minus. Subtract 28 from 28.

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

Divide 0 by 10.

$$y=0$$

There are two solutions for y: \frac{28}{5} and 0. Substitute \frac{28}{5} for y in the equation x=-2y+7 to find the corresponding solution for x that satisfies both equations.

$$x=-2\times \frac{28}{5}+7$$

Multiply -2 times \frac{28}{5}.

$$x=-\frac{56}{5}+7$$

Add -2\times \frac{28}{5} to 7.

$$x=-\frac{21}{5}$$

Now substitute 0 for y in the equation x=-2y+7 and solve to find the corresponding solution for x that satisfies both equations.

$$x=7$$

The system is now solved.

$$x=-\frac{21}{5},y=\frac{28}{5}\text{ or }x=7,y=0$$