Solve x^prime:y=1/2x^2-x+3/2

$$x ^ { \prime } : y = \frac { 1 } { 2 } x ^ { 2 } - x + \frac { 3 } { 2 }$$

Solve for y

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

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 2y, the least common multiple of y,2.

$$2\frac{\mathrm{d}}{\mathrm{d}x}(x)=\frac{1}{2}x^{2}\times 2y-x\times 2y+2y\times \frac{3}{2}$$

Multiply \frac{1}{2} and 2 to get 1.

$$2\frac{\mathrm{d}}{\mathrm{d}x}(x)=x^{2}y-x\times 2y+2y\times \frac{3}{2}$$

Multiply -1 and 2 to get -2.

$$2\frac{\mathrm{d}}{\mathrm{d}x}(x)=x^{2}y-2xy+2y\times \frac{3}{2}$$

Multiply 2 and \frac{3}{2} to get 3.

$$2\frac{\mathrm{d}}{\mathrm{d}x}(x)=x^{2}y-2xy+3y$$

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

$$x^{2}y-2xy+3y=2\frac{\mathrm{d}}{\mathrm{d}x}(x)$$

Combine all terms containing y.

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

The equation is in standard form.

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

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

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

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

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

Variable y cannot be equal to 0.

$$y=\frac{2}{x^{2}-2x+3}\text{, }y\neq 0$$

Solve for x

$$x=-\sqrt{-2+\frac{2}{y}}+1$$

$$x=\sqrt{-2+\frac{2}{y}}+1\text{, }y>0\text{ and }y\leq 1$$