Solve {l}{3x-y+2=0}{x^2/16+y^2/4=1}

$$\left\{ \begin{array} { l } { 3 x - y + 2 = 0 } \\ { \frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 4 } = 1 } \end{array} \right.$$

Solve for x, y

$$x=0\text{, }y=2$$

$$x=-\frac{48}{37}\approx -1.297297297\text{, }y=-\frac{70}{37}\approx -1.891891892$$

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.

$$3x-y+2=0,\frac{1}{4}y^{2}+\frac{1}{16}x^{2}=1$$

Solve 3x-y+2=0 for x by isolating x on the left hand side of the equal sign.

$$3x-y+2=0$$

Subtract 2 from both sides of the equation.

$$3x-y=-2$$

Subtract -y from both sides of the equation.

$$3x=y-2$$

Divide both sides by 3.

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

Substitute \frac{1}{3}y-\frac{2}{3} for x in the other equation, \frac{1}{4}y^{2}+\frac{1}{16}x^{2}=1.

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

Square \frac{1}{3}y-\frac{2}{3}.

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

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

$$\frac{1}{4}y^{2}+\frac{1}{144}y^{2}-\frac{1}{36}y+\frac{1}{36}=1$$

Add \frac{1}{4}y^{2} to \frac{1}{144}y^{2}.

$$\frac{37}{144}y^{2}-\frac{1}{36}y+\frac{1}{36}=1$$

Subtract 1 from both sides of the equation.

$$\frac{37}{144}y^{2}-\frac{1}{36}y-\frac{35}{36}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{4}+\frac{1}{16}\times \left(\frac{1}{3}\right)^{2} for a, \frac{1}{16}\left(-\frac{2}{3}\right)\times \frac{1}{3}\times 2 for b, and -\frac{35}{36} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-\frac{1}{36}\right)±\sqrt{\left(-\frac{1}{36}\right)^{2}-4\times \frac{37}{144}\left(-\frac{35}{36}\right)}}{2\times \frac{37}{144}}$$

Square \frac{1}{16}\left(-\frac{2}{3}\right)\times \frac{1}{3}\times 2.

$$y=\frac{-\left(-\frac{1}{36}\right)±\sqrt{\frac{1}{1296}-4\times \frac{37}{144}\left(-\frac{35}{36}\right)}}{2\times \frac{37}{144}}$$

Multiply -4 times \frac{1}{4}+\frac{1}{16}\times \left(\frac{1}{3}\right)^{2}.

$$y=\frac{-\left(-\frac{1}{36}\right)±\sqrt{\frac{1}{1296}-\frac{37}{36}\left(-\frac{35}{36}\right)}}{2\times \frac{37}{144}}$$

Multiply -\frac{37}{36} times -\frac{35}{36} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{1}{36}\right)±\sqrt{\frac{1+1295}{1296}}}{2\times \frac{37}{144}}$$

Add \frac{1}{1296} to \frac{1295}{1296} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{1}{36}\right)±\sqrt{1}}{2\times \frac{37}{144}}$$

Take the square root of 1.

$$y=\frac{-\left(-\frac{1}{36}\right)±1}{2\times \frac{37}{144}}$$

The opposite of \frac{1}{16}\left(-\frac{2}{3}\right)\times \frac{1}{3}\times 2 is \frac{1}{36}.

$$y=\frac{\frac{1}{36}±1}{2\times \frac{37}{144}}$$

Multiply 2 times \frac{1}{4}+\frac{1}{16}\times \left(\frac{1}{3}\right)^{2}.

$$y=\frac{\frac{1}{36}±1}{\frac{37}{72}}$$

Now solve the equation y=\frac{\frac{1}{36}±1}{\frac{37}{72}} when ± is plus. Add \frac{1}{36} to 1.

$$y=\frac{\frac{37}{36}}{\frac{37}{72}}$$

Divide \frac{37}{36} by \frac{37}{72} by multiplying \frac{37}{36} by the reciprocal of \frac{37}{72}.

$$y=2$$

Now solve the equation y=\frac{\frac{1}{36}±1}{\frac{37}{72}} when ± is minus. Subtract 1 from \frac{1}{36}.

$$y=-\frac{\frac{35}{36}}{\frac{37}{72}}$$

Divide -\frac{35}{36} by \frac{37}{72} by multiplying -\frac{35}{36} by the reciprocal of \frac{37}{72}.

$$y=-\frac{70}{37}$$

There are two solutions for y: 2 and -\frac{70}{37}. Substitute 2 for y in the equation x=\frac{1}{3}y-\frac{2}{3} to find the corresponding solution for x that satisfies both equations.

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

Multiply \frac{1}{3} times 2.

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

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

$$x=0$$

Now substitute -\frac{70}{37} for y in the equation x=\frac{1}{3}y-\frac{2}{3} and solve to find the corresponding solution for x that satisfies both equations.

$$x=\frac{1}{3}\left(-\frac{70}{37}\right)-\frac{2}{3}$$

Multiply \frac{1}{3} times -\frac{70}{37} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{70}{111}-\frac{2}{3}$$

Add -\frac{70}{37}\times \frac{1}{3} to -\frac{2}{3}.

$$x=-\frac{48}{37}$$

The system is now solved.

$$x=0,y=2\text{ or }x=-\frac{48}{37},y=-\frac{70}{37}$$