Solve {l}{y=x+9}{14x^2+y^2=81}

$$\left\{ \begin{array}{l}{ y = x + 9 }\\{ 14 x ^ { 2 } + y ^ { 2 } = 81 }\end{array} \right.$$

Solve for y, x

$$x=0\text{, }y=9$$

$$x=-\frac{6}{5}=-1.2\text{, }y=\frac{39}{5}=7.8$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Subtract x from both sides.

$$y-x=9$$

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.

$$y-x=9,14x^{2}+y^{2}=81$$

Solve y-x=9 for y by isolating y on the left hand side of the equal sign.

$$y-x=9$$

Subtract -x from both sides of the equation.

$$y=x+9$$

Substitute x+9 for y in the other equation, 14x^{2}+y^{2}=81.

$$14x^{2}+\left(x+9\right)^{2}=81$$

Square x+9.

$$14x^{2}+x^{2}+18x+81=81$$

Add 14x^{2} to x^{2}.

$$15x^{2}+18x+81=81$$

Subtract 81 from both sides of the equation.

$$15x^{2}+18x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 14+1\times 1^{2} for a, 1\times 9\times 1\times 2 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-18±\sqrt{18^{2}}}{2\times 15}$$

Take the square root of 18^{2}.

$$x=\frac{-18±18}{2\times 15}$$

Multiply 2 times 14+1\times 1^{2}.

$$x=\frac{-18±18}{30}$$

Now solve the equation x=\frac{-18±18}{30} when ± is plus. Add -18 to 18.

$$x=\frac{0}{30}$$

Divide 0 by 30.

$$x=0$$

Now solve the equation x=\frac{-18±18}{30} when ± is minus. Subtract 18 from -18.

$$x=-\frac{36}{30}$$

Reduce the fraction \frac{-36}{30} to lowest terms by extracting and canceling out 6.

$$x=-\frac{6}{5}$$

There are two solutions for x: 0 and -\frac{6}{5}. Substitute 0 for x in the equation y=x+9 to find the corresponding solution for y that satisfies both equations.

$$y=9$$

Now substitute -\frac{6}{5} for x in the equation y=x+9 and solve to find the corresponding solution for y that satisfies both equations.

$$y=-\frac{6}{5}+9$$

Add -\frac{6}{5} to 9.

$$y=\frac{39}{5}$$

The system is now solved.

$$y=9,x=0\text{ or }y=\frac{39}{5},x=-\frac{6}{5}$$