Solve {l}{6x^2+y^2=5}{x+y=1}

$$\left. \begin{array} { l } { 6 x ^ { 2 } + y ^ { 2 } = 5 } \\ { x + y = 1 } \end{array} \right.$$

Solve for x, y

$$x=\frac{\sqrt{29}+1}{7}\approx 0.912166401\text{, }y=\frac{6-\sqrt{29}}{7}\approx 0.087833599$$

$$x=\frac{1-\sqrt{29}}{7}\approx -0.626452115\text{, }y=\frac{\sqrt{29}+6}{7}\approx 1.626452115$$

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.

$$x+y=1,y^{2}+6x^{2}=5$$

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

$$x+y=1$$

Subtract y from both sides of the equation.

$$x=-y+1$$

Substitute -y+1 for x in the other equation, y^{2}+6x^{2}=5.

$$y^{2}+6\left(-y+1\right)^{2}=5$$

Square -y+1.

$$y^{2}+6\left(y^{2}-2y+1\right)=5$$

Multiply 6 times y^{2}-2y+1.

$$y^{2}+6y^{2}-12y+6=5$$

Add y^{2} to 6y^{2}.

$$7y^{2}-12y+6=5$$

Subtract 5 from both sides of the equation.

$$7y^{2}-12y+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+6\left(-1\right)^{2} for a, 6\times 1\left(-1\right)\times 2 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 6\times 1\left(-1\right)\times 2.

$$y=\frac{-\left(-12\right)±\sqrt{144-4\times 7}}{2\times 7}$$

Multiply -4 times 1+6\left(-1\right)^{2}.

$$y=\frac{-\left(-12\right)±\sqrt{144-28}}{2\times 7}$$

Add 144 to -28.

$$y=\frac{-\left(-12\right)±\sqrt{116}}{2\times 7}$$

Take the square root of 116.

$$y=\frac{-\left(-12\right)±2\sqrt{29}}{2\times 7}$$

The opposite of 6\times 1\left(-1\right)\times 2 is 12.

$$y=\frac{12±2\sqrt{29}}{2\times 7}$$

Multiply 2 times 1+6\left(-1\right)^{2}.

$$y=\frac{12±2\sqrt{29}}{14}$$

Now solve the equation y=\frac{12±2\sqrt{29}}{14} when ± is plus. Add 12 to 2\sqrt{29}.

$$y=\frac{2\sqrt{29}+12}{14}$$

Divide 12+2\sqrt{29} by 14.

$$y=\frac{\sqrt{29}+6}{7}$$

Now solve the equation y=\frac{12±2\sqrt{29}}{14} when ± is minus. Subtract 2\sqrt{29} from 12.

$$y=\frac{12-2\sqrt{29}}{14}$$

Divide 12-2\sqrt{29} by 14.

$$y=\frac{6-\sqrt{29}}{7}$$

There are two solutions for y: \frac{6+\sqrt{29}}{7} and \frac{6-\sqrt{29}}{7}. Substitute \frac{6+\sqrt{29}}{7} for y in the equation x=-y+1 to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{\sqrt{29}+6}{7}+1$$

Now substitute \frac{6-\sqrt{29}}{7} for y in the equation x=-y+1 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{6-\sqrt{29}}{7}+1$$

The system is now solved.

$$x=-\frac{\sqrt{29}+6}{7}+1,y=\frac{\sqrt{29}+6}{7}\text{ or }x=-\frac{6-\sqrt{29}}{7}+1,y=\frac{6-\sqrt{29}}{7}$$