Solve 3x+53^5/3x^2+2x^3=y

$$\frac { 3 x + 53 ^ { 5 } } { 3 x ^ { 2 } + 2 x ^ { 3 } } = y$$

Solve for y

$$y=\frac{3x+418195493}{\left(2x+3\right)x^{2}}$$

$$x\neq -\frac{3}{2}\text{ and }x\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by \left(2x+3\right)x^{2}.

$$3x+53^{5}=y\left(2x+3\right)x^{2}$$

Calculate 53 to the power of 5 and get 418195493.

$$3x+418195493=y\left(2x+3\right)x^{2}$$

Use the distributive property to multiply y by 2x+3.

$$3x+418195493=\left(2yx+3y\right)x^{2}$$

Use the distributive property to multiply 2yx+3y by x^{2}.

$$3x+418195493=2yx^{3}+3yx^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$2yx^{3}+3yx^{2}=3x+418195493$$

Combine all terms containing y.

$$\left(2x^{3}+3x^{2}\right)y=3x+418195493$$

Divide both sides by 3x^{2}+2x^{3}.

$$\frac{\left(2x^{3}+3x^{2}\right)y}{2x^{3}+3x^{2}}=\frac{3x+418195493}{2x^{3}+3x^{2}}$$

Dividing by 3x^{2}+2x^{3} undoes the multiplication by 3x^{2}+2x^{3}.

$$y=\frac{3x+418195493}{2x^{3}+3x^{2}}$$

Divide 3x+418195493 by 3x^{2}+2x^{3}.

$$y=\frac{3x+418195493}{\left(2x+3\right)x^{2}}$$