Solve 1x^2+3x=10x^2+bx+c

$$1 x ^ { 2 } + 3 x = 10 x ^ { 2 } + b x + c$$

Solve for b (complex solution)

$$\left\{\begin{matrix}b=-9x-\frac{c}{x}+3\text{, }&x\neq 0\\b\in \mathrm{C}\text{, }&c=0\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$10x^{2}+bx+c=1x^{2}+3x$$

Subtract 10x^{2} from both sides.

$$bx+c=1x^{2}+3x-10x^{2}$$

Combine 1x^{2} and -10x^{2} to get -9x^{2}.

$$bx+c=-9x^{2}+3x$$

Subtract c from both sides.

$$bx=-9x^{2}+3x-c$$

The equation is in standard form.

$$xb=-9x^{2}+3x-c$$

Divide both sides by x.

$$\frac{xb}{x}=\frac{-9x^{2}+3x-c}{x}$$

Dividing by x undoes the multiplication by x.

$$b=\frac{-9x^{2}+3x-c}{x}$$

Divide -9x^{2}+3x-c by x.

$$b=-9x-\frac{c}{x}+3$$

Solve for b

$$\left\{\begin{matrix}b=-9x-\frac{c}{x}+3\text{, }&x\neq 0\\b\in \mathrm{R}\text{, }&c=0\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$10x^{2}+bx+c=1x^{2}+3x$$

Subtract 10x^{2} from both sides.

$$bx+c=1x^{2}+3x-10x^{2}$$

Combine 1x^{2} and -10x^{2} to get -9x^{2}.

$$bx+c=-9x^{2}+3x$$

Subtract c from both sides.

$$bx=-9x^{2}+3x-c$$

The equation is in standard form.

$$xb=-9x^{2}+3x-c$$

Divide both sides by x.

$$\frac{xb}{x}=\frac{-9x^{2}+3x-c}{x}$$

Dividing by x undoes the multiplication by x.

$$b=\frac{-9x^{2}+3x-c}{x}$$

Divide -9x^{2}+3x-c by x.

$$b=-9x-\frac{c}{x}+3$$

Solve for c

$$c=-x\left(9x+b-3\right)$$

Solution Steps

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

$$10x^{2}+bx+c=1x^{2}+3x$$

Subtract 10x^{2} from both sides.

$$bx+c=1x^{2}+3x-10x^{2}$$

Combine 1x^{2} and -10x^{2} to get -9x^{2}.

$$bx+c=-9x^{2}+3x$$

Subtract bx from both sides.

$$c=-9x^{2}+3x-bx$$