Solve f(x)/g(x)=y^2-2x+3/2x^2+x-4

$$\frac { f ( x ) } { g ( x ) } = \frac { y ^ { 2 } - 2 x + 3 } { 2 x ^ { 2 } + x - 4 }$$

Solve for f

$$f=-\frac{g\left(2x-y^{2}-3\right)}{2x^{2}+x-4}$$

$$g\neq 0\text{ and }x\neq 0\text{ and }x\neq \frac{\sqrt{33}-1}{4}\text{ and }x\neq \frac{-\sqrt{33}-1}{4}$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 2gx\left(x-\left(-\frac{1}{4}\sqrt{33}-\frac{1}{4}\right)\right)\left(x-\left(\frac{1}{4}\sqrt{33}-\frac{1}{4}\right)\right), the least common multiple of gx,2x^{2}+x-4.

$$\left(2x^{2}-4+x\right)fx=gx\left(y^{2}-2x+3\right)$$

Use the distributive property to multiply 2x^{2}-4+x by f.

$$\left(2x^{2}f-4f+xf\right)x=gx\left(y^{2}-2x+3\right)$$

Use the distributive property to multiply 2x^{2}f-4f+xf by x.

$$2fx^{3}-4fx+fx^{2}=gx\left(y^{2}-2x+3\right)$$

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

$$2fx^{3}-4fx+fx^{2}=gxy^{2}-2gx^{2}+3gx$$

Combine all terms containing f.

$$\left(2x^{3}-4x+x^{2}\right)f=gxy^{2}-2gx^{2}+3gx$$

The equation is in standard form.

$$\left(2x^{3}+x^{2}-4x\right)f=3gx+gxy^{2}-2gx^{2}$$

Divide both sides by 2x^{3}-4x+x^{2}.

$$\frac{\left(2x^{3}+x^{2}-4x\right)f}{2x^{3}+x^{2}-4x}=\frac{gx\left(3+y^{2}-2x\right)}{2x^{3}+x^{2}-4x}$$

Dividing by 2x^{3}-4x+x^{2} undoes the multiplication by 2x^{3}-4x+x^{2}.

$$f=\frac{gx\left(3+y^{2}-2x\right)}{2x^{3}+x^{2}-4x}$$

Divide xg\left(y^{2}-2x+3\right) by 2x^{3}-4x+x^{2}.

$$f=\frac{g\left(3+y^{2}-2x\right)}{2x^{2}+x-4}$$

Solve for g

$$\left\{\begin{matrix}g=-\frac{f\left(2x^{2}+x-4\right)}{2x-y^{2}-3}\text{, }&x\neq \frac{\sqrt{33}-1}{4}\text{ and }x\neq \frac{-\sqrt{33}-1}{4}\text{ and }f\neq 0\text{ and }x\neq \frac{y^{2}+3}{2}\text{ and }x\neq 0\\g\neq 0\text{, }&f=0\text{ and }x=\frac{y^{2}+3}{2}\end{matrix}\right.$$

Steps for Solving Linear Equation

Variable g cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2gx\left(x-\left(-\frac{1}{4}\sqrt{33}-\frac{1}{4}\right)\right)\left(x-\left(\frac{1}{4}\sqrt{33}-\frac{1}{4}\right)\right), the least common multiple of gx,2x^{2}+x-4.

$$\left(2x^{2}-4+x\right)fx=xg\left(y^{2}-2x+3\right)$$

Use the distributive property to multiply 2x^{2}-4+x by f.

$$\left(2x^{2}f-4f+xf\right)x=xg\left(y^{2}-2x+3\right)$$

Use the distributive property to multiply 2x^{2}f-4f+xf by x.

$$2fx^{3}-4fx+fx^{2}=xg\left(y^{2}-2x+3\right)$$

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

$$2fx^{3}-4fx+fx^{2}=xgy^{2}-2gx^{2}+3xg$$

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

$$xgy^{2}-2gx^{2}+3xg=2fx^{3}-4fx+fx^{2}$$

Combine all terms containing g.

$$\left(xy^{2}-2x^{2}+3x\right)g=2fx^{3}-4fx+fx^{2}$$

The equation is in standard form.

$$\left(3x+xy^{2}-2x^{2}\right)g=2fx^{3}+fx^{2}-4fx$$

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

$$\frac{\left(3x+xy^{2}-2x^{2}\right)g}{3x+xy^{2}-2x^{2}}=\frac{fx\left(2x^{2}+x-4\right)}{3x+xy^{2}-2x^{2}}$$

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

$$g=\frac{fx\left(2x^{2}+x-4\right)}{3x+xy^{2}-2x^{2}}$$

Divide fx\left(2x^{2}-4+x\right) by xy^{2}-2x^{2}+3x.

$$g=\frac{f\left(2x^{2}+x-4\right)}{3+y^{2}-2x}$$

Variable g cannot be equal to 0.

$$g=\frac{f\left(2x^{2}+x-4\right)}{3+y^{2}-2x}\text{, }g\neq 0$$