Solve {l}{4x-15=3y}{8x^2-27y^2=45}

$$\left. \begin{array} { l } { 4 x - 15 = 3 y } \\ { 8 x ^ { 2 } - 27 y ^ { 2 } = 45 } \end{array} \right.$$

Solve for x, y

$$x=6\text{, }y=3$$

$$x=3\text{, }y=-1$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Subtract 3y from both sides.

$$4x-15-3y=0$$

Add 15 to both sides. Anything plus zero gives itself.

$$4x-3y=15$$

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.

$$4x-3y=15,-27y^{2}+8x^{2}=45$$

Solve 4x-3y=15 for x by isolating x on the left hand side of the equal sign.

$$4x-3y=15$$

Subtract -3y from both sides of the equation.

$$4x=3y+15$$

Divide both sides by 4.

$$x=\frac{3}{4}y+\frac{15}{4}$$

Substitute \frac{3}{4}y+\frac{15}{4} for x in the other equation, -27y^{2}+8x^{2}=45.

$$-27y^{2}+8\left(\frac{3}{4}y+\frac{15}{4}\right)^{2}=45$$

Square \frac{3}{4}y+\frac{15}{4}.

$$-27y^{2}+8\left(\frac{9}{16}y^{2}+\frac{45}{8}y+\frac{225}{16}\right)=45$$

Multiply 8 times \frac{9}{16}y^{2}+\frac{45}{8}y+\frac{225}{16}.

$$-27y^{2}+\frac{9}{2}y^{2}+45y+\frac{225}{2}=45$$

Add -27y^{2} to \frac{9}{2}y^{2}.

$$-\frac{45}{2}y^{2}+45y+\frac{225}{2}=45$$

Subtract 45 from both sides of the equation.

$$-\frac{45}{2}y^{2}+45y+\frac{135}{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -27+8\times \left(\frac{3}{4}\right)^{2} for a, 8\times \frac{15}{4}\times \frac{3}{4}\times 2 for b, and \frac{135}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-45±\sqrt{45^{2}-4\left(-\frac{45}{2}\right)\times \frac{135}{2}}}{2\left(-\frac{45}{2}\right)}$$

Square 8\times \frac{15}{4}\times \frac{3}{4}\times 2.

$$y=\frac{-45±\sqrt{2025-4\left(-\frac{45}{2}\right)\times \frac{135}{2}}}{2\left(-\frac{45}{2}\right)}$$

Multiply -4 times -27+8\times \left(\frac{3}{4}\right)^{2}.

$$y=\frac{-45±\sqrt{2025+90\times \frac{135}{2}}}{2\left(-\frac{45}{2}\right)}$$

Multiply 90 times \frac{135}{2}.

$$y=\frac{-45±\sqrt{2025+6075}}{2\left(-\frac{45}{2}\right)}$$

Add 2025 to 6075.

$$y=\frac{-45±\sqrt{8100}}{2\left(-\frac{45}{2}\right)}$$

Take the square root of 8100.

$$y=\frac{-45±90}{2\left(-\frac{45}{2}\right)}$$

Multiply 2 times -27+8\times \left(\frac{3}{4}\right)^{2}.

$$y=\frac{-45±90}{-45}$$

Now solve the equation y=\frac{-45±90}{-45} when ± is plus. Add -45 to 90.

$$y=\frac{45}{-45}$$

Divide 45 by -45.

$$y=-1$$

Now solve the equation y=\frac{-45±90}{-45} when ± is minus. Subtract 90 from -45.

$$y=-\frac{135}{-45}$$

Divide -135 by -45.

$$y=3$$

There are two solutions for y: -1 and 3. Substitute -1 for y in the equation x=\frac{3}{4}y+\frac{15}{4} to find the corresponding solution for x that satisfies both equations.

$$x=\frac{3}{4}\left(-1\right)+\frac{15}{4}$$

Multiply \frac{3}{4} times -1.

$$x=\frac{-3+15}{4}$$

Add -\frac{3}{4} to \frac{15}{4}.

$$x=3$$

Now substitute 3 for y in the equation x=\frac{3}{4}y+\frac{15}{4} and solve to find the corresponding solution for x that satisfies both equations.

$$x=\frac{3}{4}\times 3+\frac{15}{4}$$

Multiply \frac{3}{4} times 3.

$$x=\frac{9+15}{4}$$

Add \frac{3}{4}\times 3 to \frac{15}{4}.

$$x=6$$

The system is now solved.

$$x=3,y=-1\text{ or }x=6,y=3$$