Solve 5x+7/3xy-2x-5=1

$$\frac { 5 x + 7 } { 3 x y } - 2 x - 5 = 1$$

Solve for y

$$y=\frac{5x+7}{6x\left(x+3\right)}$$

$$x\neq -\frac{7}{5}\text{ and }x\neq -3\text{ and }x\neq 0$$

Steps for Solving Linear Equation

Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 3xy.

$$5x+7-2x\times 3xy+3xy\left(-5\right)=3xy$$

Multiply x and x to get x^{2}.

$$5x+7-2x^{2}\times 3y+3xy\left(-5\right)=3xy$$

Multiply -2 and 3 to get -6.

$$5x+7-6x^{2}y+3xy\left(-5\right)=3xy$$

Multiply 3 and -5 to get -15.

$$5x+7-6x^{2}y-15xy=3xy$$

Subtract 3xy from both sides.

$$5x+7-6x^{2}y-15xy-3xy=0$$

Combine -15xy and -3xy to get -18xy.

$$5x+7-6x^{2}y-18xy=0$$

Subtract 5x from both sides. Anything subtracted from zero gives its negation.

$$7-6x^{2}y-18xy=-5x$$

Subtract 7 from both sides.

$$-6x^{2}y-18xy=-5x-7$$

Combine all terms containing y.

$$\left(-6x^{2}-18x\right)y=-5x-7$$

Divide both sides by -18x-6x^{2}.

$$\frac{\left(-6x^{2}-18x\right)y}{-6x^{2}-18x}=\frac{-5x-7}{-6x^{2}-18x}$$

Dividing by -18x-6x^{2} undoes the multiplication by -18x-6x^{2}.

$$y=\frac{-5x-7}{-6x^{2}-18x}$$

Divide -5x-7 by -18x-6x^{2}.

$$y=\frac{5x+7}{6x\left(x+3\right)}$$

Variable y cannot be equal to 0.

$$y=\frac{5x+7}{6x\left(x+3\right)}\text{, }y\neq 0$$

Solve for x

$$x=-\frac{\sqrt{324y^{2}-12y+25}+18y-5}{12y}$$

$$x=-\frac{-\sqrt{324y^{2}-12y+25}+18y-5}{12y}\text{, }y\neq 0$$