Solve 3x-y/2x+4y=191/2

$$\frac { 3 x - y } { 2 } x + 4 y = 19 \frac { 1 } { 2 }$$

Solve for y

$$y=-\frac{3\left(x^{2}-13\right)}{8-x}$$

$$x\neq 8$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 2.

$$\left(3x-y\right)x+8y=19\times 2+1$$

Use the distributive property to multiply 3x-y by x.

$$3x^{2}-yx+8y=19\times 2+1$$

Multiply 19 and 2 to get 38.

$$3x^{2}-yx+8y=38+1$$

Add 38 and 1 to get 39.

$$3x^{2}-yx+8y=39$$

Subtract 3x^{2} from both sides.

$$-yx+8y=39-3x^{2}$$

Combine all terms containing y.

$$\left(-x+8\right)y=39-3x^{2}$$

The equation is in standard form.

$$\left(8-x\right)y=39-3x^{2}$$

Divide both sides by -x+8.

$$\frac{\left(8-x\right)y}{8-x}=\frac{39-3x^{2}}{8-x}$$

Dividing by -x+8 undoes the multiplication by -x+8.

$$y=\frac{39-3x^{2}}{8-x}$$

Divide 39-3x^{2} by -x+8.

$$y=\frac{3\left(13-x^{2}\right)}{8-x}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{y^{2}-96y+468}+y}{6}$$

$$x=\frac{-\sqrt{y^{2}-96y+468}+y}{6}$$

Solve for x

$$x=\frac{\sqrt{y^{2}-96y+468}+y}{6}$$

$$x=\frac{-\sqrt{y^{2}-96y+468}+y}{6}\text{, }y\geq 6\sqrt{51}+48\text{ or }y\leq 48-6\sqrt{51}$$