Solve {c}{4x-y=6}{4x^2+y^2=8}

$$\left\{ \begin{array} { c } { 4 x - y = 6 } \\ { 4 x ^ { 2 } + y ^ { 2 } = 8 } \end{array} \right.$$

Solve for x, y

$$x=\frac{7}{5}=1.4\text{, }y=-\frac{2}{5}=-0.4$$

$$x=1\text{, }y=-2$$

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.

$$4x-y=6,y^{2}+4x^{2}=8$$

Solve 4x-y=6 for x by isolating x on the left hand side of the equal sign.

$$4x-y=6$$

Subtract -y from both sides of the equation.

$$4x=y+6$$

Divide both sides by 4.

$$x=\frac{1}{4}y+\frac{3}{2}$$

Substitute \frac{1}{4}y+\frac{3}{2} for x in the other equation, y^{2}+4x^{2}=8.

$$y^{2}+4\left(\frac{1}{4}y+\frac{3}{2}\right)^{2}=8$$

Square \frac{1}{4}y+\frac{3}{2}.

$$y^{2}+4\left(\frac{1}{16}y^{2}+\frac{3}{4}y+\frac{9}{4}\right)=8$$

Multiply 4 times \frac{1}{16}y^{2}+\frac{3}{4}y+\frac{9}{4}.

$$y^{2}+\frac{1}{4}y^{2}+3y+9=8$$

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

$$\frac{5}{4}y^{2}+3y+9=8$$

Subtract 8 from both sides of the equation.

$$\frac{5}{4}y^{2}+3y+1=0$$

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

$$y=\frac{-3±\sqrt{3^{2}-4\times \frac{5}{4}}}{2\times \frac{5}{4}}$$

Square 4\times \frac{3}{2}\times \frac{1}{4}\times 2.

$$y=\frac{-3±\sqrt{9-4\times \frac{5}{4}}}{2\times \frac{5}{4}}$$

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

$$y=\frac{-3±\sqrt{9-5}}{2\times \frac{5}{4}}$$

Add 9 to -5.

$$y=\frac{-3±\sqrt{4}}{2\times \frac{5}{4}}$$

Take the square root of 4.

$$y=\frac{-3±2}{2\times \frac{5}{4}}$$

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

$$y=\frac{-3±2}{\frac{5}{2}}$$

Now solve the equation y=\frac{-3±2}{\frac{5}{2}} when ± is plus. Add -3 to 2.

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

Divide -1 by \frac{5}{2} by multiplying -1 by the reciprocal of \frac{5}{2}.

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

Now solve the equation y=\frac{-3±2}{\frac{5}{2}} when ± is minus. Subtract 2 from -3.

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

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

$$y=-2$$

There are two solutions for y: -\frac{2}{5} and -2. Substitute -\frac{2}{5} for y in the equation x=\frac{1}{4}y+\frac{3}{2} to find the corresponding solution for x that satisfies both equations.

$$x=\frac{1}{4}\left(-\frac{2}{5}\right)+\frac{3}{2}$$

Multiply \frac{1}{4} times -\frac{2}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{1}{10}+\frac{3}{2}$$

Add -\frac{2}{5}\times \frac{1}{4} to \frac{3}{2}.

$$x=\frac{7}{5}$$

Now substitute -2 for y in the equation x=\frac{1}{4}y+\frac{3}{2} and solve to find the corresponding solution for x that satisfies both equations.

$$x=\frac{1}{4}\left(-2\right)+\frac{3}{2}$$

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

$$x=\frac{-1+3}{2}$$

Add -2\times \frac{1}{4} to \frac{3}{2}.

$$x=1$$

The system is now solved.

$$x=\frac{7}{5},y=-\frac{2}{5}\text{ or }x=1,y=-2$$