Solve 4ax+2b-ax^2-bx+c=-x^2+5x

$$4 a x + 2 b - a x ^ { 2 } - b x + c = - x ^ { 2 } + 5 x$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=-\frac{x^{2}-bx-5x+c+2b}{x\left(4-x\right)}\text{, }&x\neq 4\text{ and }x\neq 0\\a\in \mathrm{C}\text{, }&\left(b=-\frac{c}{2}\text{ and }x=0\right)\text{ or }\left(b=\frac{c}{2}-2\text{ and }x=4\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Add bx to both sides.

$$4ax+2b-ax^{2}+c=-x^{2}+5x+bx$$

Subtract c from both sides.

$$4ax+2b-ax^{2}=-x^{2}+5x+bx-c$$

Subtract 2b from both sides.

$$4ax-ax^{2}=-x^{2}+5x+bx-c-2b$$

Combine all terms containing a.

$$\left(4x-x^{2}\right)a=-x^{2}+5x+bx-c-2b$$

The equation is in standard form.

$$\left(4x-x^{2}\right)a=-x^{2}+bx+5x-c-2b$$

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

$$\frac{\left(4x-x^{2}\right)a}{4x-x^{2}}=\frac{-x^{2}+bx+5x-c-2b}{4x-x^{2}}$$

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

$$a=\frac{-x^{2}+bx+5x-c-2b}{4x-x^{2}}$$

Divide -x^{2}+5x+bx-c-2b by -x^{2}+4x.

$$a=\frac{-x^{2}+bx+5x-c-2b}{x\left(4-x\right)}$$

Solve for b (complex solution)

$$\left\{\begin{matrix}b=-\frac{c-5x+4ax+x^{2}-ax^{2}}{2-x}\text{, }&x\neq 2\\b\in \mathrm{C}\text{, }&a=-\frac{c}{4}+\frac{3}{2}\text{ and }x=2\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract c from both sides.

$$4ax+2b-ax^{2}-bx=-x^{2}+5x-c$$

Subtract 4ax from both sides.

$$2b-ax^{2}-bx=-x^{2}+5x-c-4ax$$

Add ax^{2} to both sides.

$$2b-bx=-x^{2}+5x-c-4ax+ax^{2}$$

Combine all terms containing b.

$$\left(2-x\right)b=-x^{2}+5x-c-4ax+ax^{2}$$

The equation is in standard form.

$$\left(2-x\right)b=ax^{2}-x^{2}-4ax+5x-c$$

Divide both sides by -x+2.

$$\frac{\left(2-x\right)b}{2-x}=\frac{ax^{2}-x^{2}-4ax+5x-c}{2-x}$$

Dividing by -x+2 undoes the multiplication by -x+2.

$$b=\frac{ax^{2}-x^{2}-4ax+5x-c}{2-x}$$

Solve for a

$$\left\{\begin{matrix}a=-\frac{x^{2}-bx-5x+c+2b}{x\left(4-x\right)}\text{, }&x\neq 4\text{ and }x\neq 0\\a\in \mathrm{R}\text{, }&\left(b=-\frac{c}{2}\text{ and }x=0\right)\text{ or }\left(b=\frac{c}{2}-2\text{ and }x=4\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Add bx to both sides.

$$4ax+2b-ax^{2}+c=-x^{2}+5x+bx$$

Subtract c from both sides.

$$4ax+2b-ax^{2}=-x^{2}+5x+bx-c$$

Subtract 2b from both sides.

$$4ax-ax^{2}=-x^{2}+5x+bx-c-2b$$

Combine all terms containing a.

$$\left(4x-x^{2}\right)a=-x^{2}+5x+bx-c-2b$$

The equation is in standard form.

$$\left(4x-x^{2}\right)a=-x^{2}+bx+5x-c-2b$$

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

$$\frac{\left(4x-x^{2}\right)a}{4x-x^{2}}=\frac{-x^{2}+bx+5x-c-2b}{4x-x^{2}}$$

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

$$a=\frac{-x^{2}+bx+5x-c-2b}{4x-x^{2}}$$

Divide -x^{2}+5x+bx-c-2b by 4x-x^{2}.

$$a=\frac{-x^{2}+bx+5x-c-2b}{x\left(4-x\right)}$$

Solve for b

$$\left\{\begin{matrix}b=-\frac{c-5x+4ax+x^{2}-ax^{2}}{2-x}\text{, }&x\neq 2\\b\in \mathrm{R}\text{, }&a=-\frac{c}{4}+\frac{3}{2}\text{ and }x=2\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract c from both sides.

$$4ax+2b-ax^{2}-bx=-x^{2}+5x-c$$

Subtract 4ax from both sides.

$$2b-ax^{2}-bx=-x^{2}+5x-c-4ax$$

Add ax^{2} to both sides.

$$2b-bx=-x^{2}+5x-c-4ax+ax^{2}$$

Combine all terms containing b.

$$\left(2-x\right)b=-x^{2}+5x-c-4ax+ax^{2}$$

The equation is in standard form.

$$\left(2-x\right)b=ax^{2}-x^{2}-4ax+5x-c$$

Divide both sides by -x+2.

$$\frac{\left(2-x\right)b}{2-x}=\frac{ax^{2}-x^{2}-4ax+5x-c}{2-x}$$

Dividing by -x+2 undoes the multiplication by -x+2.

$$b=\frac{ax^{2}-x^{2}-4ax+5x-c}{2-x}$$