Solve 3x^2+y^2=28,3x+y=8

$$3 x ^ { 2 } + y ^ { 2 } = 28,3 x + y = 8$$

Solve for x, y

$$x=3\text{, }y=-1$$

$$x=1\text{, }y=5$$

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.

$$3x+y=8,y^{2}+3x^{2}=28$$

Solve 3x+y=8 for x by isolating x on the left hand side of the equal sign.

$$3x+y=8$$

Subtract y from both sides of the equation.

$$3x=-y+8$$

Divide both sides by 3.

$$x=-\frac{1}{3}y+\frac{8}{3}$$

Substitute -\frac{1}{3}y+\frac{8}{3} for x in the other equation, y^{2}+3x^{2}=28.

$$y^{2}+3\left(-\frac{1}{3}y+\frac{8}{3}\right)^{2}=28$$

Square -\frac{1}{3}y+\frac{8}{3}.

$$y^{2}+3\left(\frac{1}{9}y^{2}-\frac{16}{9}y+\frac{64}{9}\right)=28$$

Multiply 3 times \frac{1}{9}y^{2}-\frac{16}{9}y+\frac{64}{9}.

$$y^{2}+\frac{1}{3}y^{2}-\frac{16}{3}y+\frac{64}{3}=28$$

Add y^{2} to \frac{1}{3}y^{2}.

$$\frac{4}{3}y^{2}-\frac{16}{3}y+\frac{64}{3}=28$$

Subtract 28 from both sides of the equation.

$$\frac{4}{3}y^{2}-\frac{16}{3}y-\frac{20}{3}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+3\left(-\frac{1}{3}\right)^{2} for a, 3\times \frac{8}{3}\left(-\frac{1}{3}\right)\times 2 for b, and -\frac{20}{3} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-\frac{16}{3}\right)±\sqrt{\left(-\frac{16}{3}\right)^{2}-4\times \frac{4}{3}\left(-\frac{20}{3}\right)}}{2\times \frac{4}{3}}$$

Square 3\times \frac{8}{3}\left(-\frac{1}{3}\right)\times 2.

$$y=\frac{-\left(-\frac{16}{3}\right)±\sqrt{\frac{256}{9}-4\times \frac{4}{3}\left(-\frac{20}{3}\right)}}{2\times \frac{4}{3}}$$

Multiply -4 times 1+3\left(-\frac{1}{3}\right)^{2}.

$$y=\frac{-\left(-\frac{16}{3}\right)±\sqrt{\frac{256}{9}-\frac{16}{3}\left(-\frac{20}{3}\right)}}{2\times \frac{4}{3}}$$

Multiply -\frac{16}{3} times -\frac{20}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{16}{3}\right)±\sqrt{\frac{256+320}{9}}}{2\times \frac{4}{3}}$$

Add \frac{256}{9} to \frac{320}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{16}{3}\right)±\sqrt{64}}{2\times \frac{4}{3}}$$

Take the square root of 64.

$$y=\frac{-\left(-\frac{16}{3}\right)±8}{2\times \frac{4}{3}}$$

The opposite of 3\times \frac{8}{3}\left(-\frac{1}{3}\right)\times 2 is \frac{16}{3}.

$$y=\frac{\frac{16}{3}±8}{2\times \frac{4}{3}}$$

Multiply 2 times 1+3\left(-\frac{1}{3}\right)^{2}.

$$y=\frac{\frac{16}{3}±8}{\frac{8}{3}}$$

Now solve the equation y=\frac{\frac{16}{3}±8}{\frac{8}{3}} when ± is plus. Add \frac{16}{3} to 8.

$$y=\frac{\frac{40}{3}}{\frac{8}{3}}$$

Divide \frac{40}{3} by \frac{8}{3} by multiplying \frac{40}{3} by the reciprocal of \frac{8}{3}.

$$y=5$$

Now solve the equation y=\frac{\frac{16}{3}±8}{\frac{8}{3}} when ± is minus. Subtract 8 from \frac{16}{3}.

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

Divide -\frac{8}{3} by \frac{8}{3} by multiplying -\frac{8}{3} by the reciprocal of \frac{8}{3}.

$$y=-1$$

There are two solutions for y: 5 and -1. Substitute 5 for y in the equation x=-\frac{1}{3}y+\frac{8}{3} to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{1}{3}\times 5+\frac{8}{3}$$

Multiply -\frac{1}{3} times 5.

$$x=\frac{-5+8}{3}$$

Add -\frac{1}{3}\times 5 to \frac{8}{3}.

$$x=1$$

Now substitute -1 for y in the equation x=-\frac{1}{3}y+\frac{8}{3} and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{1}{3}\left(-1\right)+\frac{8}{3}$$

Multiply -\frac{1}{3} times -1.

$$x=\frac{1+8}{3}$$

Add -\left(-\frac{1}{3}\right) to \frac{8}{3}.

$$x=3$$

The system is now solved.

$$x=1,y=5\text{ or }x=3,y=-1$$