Solve {c}{y=x+4}{5x^2+y^2=16}

$$\left\{ \begin{array} { c } { y = x + 4 } \\ { 5 x ^ { 2 } + y ^ { 2 } = 16 } \end{array} \right.$$

Solve for y, x

$$x=0\text{, }y=4$$

$$x=-\frac{4}{3}\approx -1.333333333\text{, }y=\frac{8}{3}\approx 2.666666667$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Subtract x from both sides.

$$y-x=4$$

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.

$$y-x=4,5x^{2}+y^{2}=16$$

Solve y-x=4 for y by isolating y on the left hand side of the equal sign.

$$y-x=4$$

Subtract -x from both sides of the equation.

$$y=x+4$$

Substitute x+4 for y in the other equation, 5x^{2}+y^{2}=16.

$$5x^{2}+\left(x+4\right)^{2}=16$$

Square x+4.

$$5x^{2}+x^{2}+8x+16=16$$

Add 5x^{2} to x^{2}.

$$6x^{2}+8x+16=16$$

Subtract 16 from both sides of the equation.

$$6x^{2}+8x=0$$

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

$$x=\frac{-8±\sqrt{8^{2}}}{2\times 6}$$

Take the square root of 8^{2}.

$$x=\frac{-8±8}{2\times 6}$$

Multiply 2 times 5+1\times 1^{2}.

$$x=\frac{-8±8}{12}$$

Now solve the equation x=\frac{-8±8}{12} when ± is plus. Add -8 to 8.

$$x=\frac{0}{12}$$

Divide 0 by 12.

$$x=0$$

Now solve the equation x=\frac{-8±8}{12} when ± is minus. Subtract 8 from -8.

$$x=-\frac{16}{12}$$

Reduce the fraction \frac{-16}{12} to lowest terms by extracting and canceling out 4.

$$x=-\frac{4}{3}$$

There are two solutions for x: 0 and -\frac{4}{3}. Substitute 0 for x in the equation y=x+4 to find the corresponding solution for y that satisfies both equations.

$$y=4$$

Now substitute -\frac{4}{3} for x in the equation y=x+4 and solve to find the corresponding solution for y that satisfies both equations.

$$y=-\frac{4}{3}+4$$

Add -\frac{4}{3} to 4.

$$y=\frac{8}{3}$$

The system is now solved.

$$y=4,x=0\text{ or }y=\frac{8}{3},x=-\frac{4}{3}$$