Solve 7:2*y=x^3/(1-x)^2

$$7 : 2 \cdot y = \frac { x ^ { 3 } } { ( 1 - x ) ^ { 2 } }$$

Solve for y

$$y=\frac{2x^{3}}{7\left(x-1\right)^{2}}$$

$$x\neq 1$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 2\left(x-1\right)^{2}, the least common multiple of 2,\left(1-x\right)^{2}.

$$\frac{7}{2}y\times 2\left(x-1\right)^{2}=2x^{3}$$

Multiply \frac{7}{2} and 2 to get 7.

$$7y\left(x-1\right)^{2}=2x^{3}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$7y\left(x^{2}-2x+1\right)=2x^{3}$$

Use the distributive property to multiply 7y by x^{2}-2x+1.

$$7yx^{2}-14xy+7y=2x^{3}$$

Combine all terms containing y.

$$\left(7x^{2}-14x+7\right)y=2x^{3}$$

Divide both sides by 7-14x+7x^{2}.

$$\frac{\left(7x^{2}-14x+7\right)y}{7x^{2}-14x+7}=\frac{2x^{3}}{7x^{2}-14x+7}$$

Dividing by 7-14x+7x^{2} undoes the multiplication by 7-14x+7x^{2}.

$$y=\frac{2x^{3}}{7x^{2}-14x+7}$$

Divide 2x^{3} by 7-14x+7x^{2}.

$$y=\frac{2x^{3}}{7\left(x-1\right)^{2}}$$