Solve {l}{8x^2-3y^2=6}{2x+y-5=0}

$$\left. \begin{array} { l } { 8 x ^ { 2 } - 3 y ^ { 2 } = 6 } \\ { 2 x + y - 5 = 0 } \end{array} \right.$$

Solve for x, y

$$x=\frac{27}{2}=13.5\text{, }y=-22$$

$$x=\frac{3}{2}=1.5\text{, }y=2$$

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.

$$2x+y-5=0,-3y^{2}+8x^{2}=6$$

Solve 2x+y-5=0 for x by isolating x on the left hand side of the equal sign.

$$2x+y-5=0$$

Add 5 to both sides of the equation.

$$2x+y=5$$

Subtract y from both sides of the equation.

$$2x=-y+5$$

Divide both sides by 2.

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

Substitute -\frac{1}{2}y+\frac{5}{2} for x in the other equation, -3y^{2}+8x^{2}=6.

$$-3y^{2}+8\left(-\frac{1}{2}y+\frac{5}{2}\right)^{2}=6$$

Square -\frac{1}{2}y+\frac{5}{2}.

$$-3y^{2}+8\left(\frac{1}{4}y^{2}-\frac{5}{2}y+\frac{25}{4}\right)=6$$

Multiply 8 times \frac{1}{4}y^{2}-\frac{5}{2}y+\frac{25}{4}.

$$-3y^{2}+2y^{2}-20y+50=6$$

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

$$-y^{2}-20y+50=6$$

Subtract 6 from both sides of the equation.

$$-y^{2}-20y+44=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3+8\left(-\frac{1}{2}\right)^{2} for a, 8\times \frac{5}{2}\left(-\frac{1}{2}\right)\times 2 for b, and 44 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\left(-1\right)\times 44}}{2\left(-1\right)}$$

Square 8\times \frac{5}{2}\left(-\frac{1}{2}\right)\times 2.

$$y=\frac{-\left(-20\right)±\sqrt{400-4\left(-1\right)\times 44}}{2\left(-1\right)}$$

Multiply -4 times -3+8\left(-\frac{1}{2}\right)^{2}.

$$y=\frac{-\left(-20\right)±\sqrt{400+4\times 44}}{2\left(-1\right)}$$

Multiply 4 times 44.

$$y=\frac{-\left(-20\right)±\sqrt{400+176}}{2\left(-1\right)}$$

Add 400 to 176.

$$y=\frac{-\left(-20\right)±\sqrt{576}}{2\left(-1\right)}$$

Take the square root of 576.

$$y=\frac{-\left(-20\right)±24}{2\left(-1\right)}$$

The opposite of 8\times \frac{5}{2}\left(-\frac{1}{2}\right)\times 2 is 20.

$$y=\frac{20±24}{2\left(-1\right)}$$

Multiply 2 times -3+8\left(-\frac{1}{2}\right)^{2}.

$$y=\frac{20±24}{-2}$$

Now solve the equation y=\frac{20±24}{-2} when ± is plus. Add 20 to 24.

$$y=\frac{44}{-2}$$

Divide 44 by -2.

$$y=-22$$

Now solve the equation y=\frac{20±24}{-2} when ± is minus. Subtract 24 from 20.

$$y=-\frac{4}{-2}$$

Divide -4 by -2.

$$y=2$$

There are two solutions for y: -22 and 2. Substitute -22 for y in the equation x=-\frac{1}{2}y+\frac{5}{2} to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{1}{2}\left(-22\right)+\frac{5}{2}$$

Multiply -\frac{1}{2} times -22.

$$x=11+\frac{5}{2}$$

Add -22\left(-\frac{1}{2}\right) to \frac{5}{2}.

$$x=\frac{27}{2}$$

Now substitute 2 for y in the equation x=-\frac{1}{2}y+\frac{5}{2} and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{1}{2}\times 2+\frac{5}{2}$$

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

$$x=-1+\frac{5}{2}$$

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

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

The system is now solved.

$$x=\frac{27}{2},y=-22\text{ or }x=\frac{3}{2},y=2$$