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

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

Solve for x, y

$$x=\frac{6\sqrt{2}+4}{7}\approx 1.783611625\text{, }y=\frac{3-6\sqrt{2}}{7}\approx -0.783611625$$

$$x=\frac{4-6\sqrt{2}}{7}\approx -0.640754482\text{, }y=\frac{6\sqrt{2}+3}{7}\approx 1.640754482$$

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.

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

Solve x+y-1=0 for x by isolating x on the left hand side of the equal sign.

$$x+y-1=0$$

Add 1 to both sides of the equation.

$$x+y=1$$

Subtract y from both sides of the equation.

$$x=-y+1$$

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

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

Square -y+1.

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

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

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

Add \frac{1}{3}y^{2} to \frac{1}{4}y^{2}.

$$\frac{7}{12}y^{2}-\frac{1}{2}y+\frac{1}{4}=1$$

Subtract 1 from both sides of the equation.

$$\frac{7}{12}y^{2}-\frac{1}{2}y-\frac{3}{4}=0$$

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

$$y=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\left(-\frac{1}{2}\right)^{2}-4\times \frac{7}{12}\left(-\frac{3}{4}\right)}}{2\times \frac{7}{12}}$$

Square \frac{1}{4}\times 1\left(-1\right)\times 2.

$$y=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\frac{1}{4}-4\times \frac{7}{12}\left(-\frac{3}{4}\right)}}{2\times \frac{7}{12}}$$

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

$$y=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\frac{1}{4}-\frac{7}{3}\left(-\frac{3}{4}\right)}}{2\times \frac{7}{12}}$$

Multiply -\frac{7}{3} times -\frac{3}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\frac{1+7}{4}}}{2\times \frac{7}{12}}$$

Add \frac{1}{4} to \frac{7}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{1}{2}\right)±\sqrt{2}}{2\times \frac{7}{12}}$$

The opposite of \frac{1}{4}\times 1\left(-1\right)\times 2 is \frac{1}{2}.

$$y=\frac{\frac{1}{2}±\sqrt{2}}{2\times \frac{7}{12}}$$

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

$$y=\frac{\frac{1}{2}±\sqrt{2}}{\frac{7}{6}}$$

Now solve the equation y=\frac{\frac{1}{2}±\sqrt{2}}{\frac{7}{6}} when ± is plus. Add \frac{1}{2} to \sqrt{2}.

$$y=\frac{\sqrt{2}+\frac{1}{2}}{\frac{7}{6}}$$

Divide \frac{1}{2}+\sqrt{2} by \frac{7}{6} by multiplying \frac{1}{2}+\sqrt{2} by the reciprocal of \frac{7}{6}.

$$y=\frac{6\sqrt{2}+3}{7}$$

Now solve the equation y=\frac{\frac{1}{2}±\sqrt{2}}{\frac{7}{6}} when ± is minus. Subtract \sqrt{2} from \frac{1}{2}.

$$y=\frac{\frac{1}{2}-\sqrt{2}}{\frac{7}{6}}$$

Divide \frac{1}{2}-\sqrt{2} by \frac{7}{6} by multiplying \frac{1}{2}-\sqrt{2} by the reciprocal of \frac{7}{6}.

$$y=\frac{3-6\sqrt{2}}{7}$$

There are two solutions for y: \frac{6\sqrt{2}+3}{7} and \frac{3-6\sqrt{2}}{7}. Substitute \frac{6\sqrt{2}+3}{7} for y in the equation x=-y+1 to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{6\sqrt{2}+3}{7}+1$$

Now substitute \frac{3-6\sqrt{2}}{7} for y in the equation x=-y+1 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{3-6\sqrt{2}}{7}+1$$

The system is now solved.

$$x=-\frac{6\sqrt{2}+3}{7}+1,y=\frac{6\sqrt{2}+3}{7}\text{ or }x=-\frac{3-6\sqrt{2}}{7}+1,y=\frac{3-6\sqrt{2}}{7}$$