Solve {l}{x^2-3y^2=4}{x+y=6}

$$\left\{ \begin{array} { l } { x ^ { 2 } - 3 y ^ { 2 } = 4 } \\ { x + y = 6 } \end{array} \right.$$

Solve for x, y

$$x=14\text{, }y=-8$$

$$x=4\text{, }y=2$$

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.

$$x+y=6,-3y^{2}+x^{2}=4$$

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

$$x+y=6$$

Subtract y from both sides of the equation.

$$x=-y+6$$

Substitute -y+6 for x in the other equation, -3y^{2}+x^{2}=4.

$$-3y^{2}+\left(-y+6\right)^{2}=4$$

Square -y+6.

$$-3y^{2}+y^{2}-12y+36=4$$

Add -3y^{2} to y^{2}.

$$-2y^{2}-12y+36=4$$

Subtract 4 from both sides of the equation.

$$-2y^{2}-12y+32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3+1\left(-1\right)^{2} for a, 1\times 6\left(-1\right)\times 2 for b, and 32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\left(-2\right)\times 32}}{2\left(-2\right)}$$

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

$$y=\frac{-\left(-12\right)±\sqrt{144-4\left(-2\right)\times 32}}{2\left(-2\right)}$$

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

$$y=\frac{-\left(-12\right)±\sqrt{144+8\times 32}}{2\left(-2\right)}$$

Multiply 8 times 32.

$$y=\frac{-\left(-12\right)±\sqrt{144+256}}{2\left(-2\right)}$$

Add 144 to 256.

$$y=\frac{-\left(-12\right)±\sqrt{400}}{2\left(-2\right)}$$

Take the square root of 400.

$$y=\frac{-\left(-12\right)±20}{2\left(-2\right)}$$

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

$$y=\frac{12±20}{2\left(-2\right)}$$

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

$$y=\frac{12±20}{-4}$$

Now solve the equation y=\frac{12±20}{-4} when ± is plus. Add 12 to 20.

$$y=\frac{32}{-4}$$

Divide 32 by -4.

$$y=-8$$

Now solve the equation y=\frac{12±20}{-4} when ± is minus. Subtract 20 from 12.

$$y=-\frac{8}{-4}$$

Divide -8 by -4.

$$y=2$$

There are two solutions for y: -8 and 2. Substitute -8 for y in the equation x=-y+6 to find the corresponding solution for x that satisfies both equations.

$$x=-\left(-8\right)+6$$

Multiply -1 times -8.

$$x=8+6$$

Add -8\left(-1\right) to 6.

$$x=14$$

Now substitute 2 for y in the equation x=-y+6 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-2+6$$

Add -2 to 6.

$$x=4$$

The system is now solved.

$$x=14,y=-8\text{ or }x=4,y=2$$