Solve x ^ 2 + y ^ 2 - 6x + uy + u = c | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$x^{2}+y^{2}-6x+uy+u=c$$

Solve for u (complex solution)

$\left\{\begin{matrix}u=-\frac{x^{2}-6x+y^{2}-c}{y+1}\text{, }&y\neq -1\\u\in \mathrm{C}\text{, }&c=x^{2}-6x+1\text{ and }y=-1\end{matrix}\right.$

Steps for Solving Linear Equation

Subtract $x^{2}$ from both sides.

$$y^{2}-6x+uy+u=c-x^{2}$$

Subtract $y^{2}$ from both sides.

$$-6x+uy+u=c-x^{2}-y^{2}$$

Add $6x$ to both sides.

$$uy+u=c-x^{2}-y^{2}+6x$$

Combine all terms containing $u$.

$$\left(y+1\right)u=c-x^{2}-y^{2}+6x$$

The equation is in standard form.

$$\left(y+1\right)u=c-y^{2}+6x-x^{2}$$

Divide both sides by $y+1$.

$$\frac{\left(y+1\right)u}{y+1}=\frac{c-y^{2}+6x-x^{2}}{y+1}$$

Dividing by $y+1$ undoes the multiplication by $y+1$.

$$u=\frac{c-y^{2}+6x-x^{2}}{y+1}$$

Solve for u

$\left\{\begin{matrix}u=-\frac{x^{2}-6x+y^{2}-c}{y+1}\text{, }&y\neq -1\\u\in \mathrm{R}\text{, }&c=x^{2}-6x+1\text{ and }y=-1\end{matrix}\right.$

Steps for Solving Linear Equation

Subtract $x^{2}$ from both sides.

$$y^{2}-6x+uy+u=c-x^{2}$$

Subtract $y^{2}$ from both sides.

$$-6x+uy+u=c-x^{2}-y^{2}$$

Add $6x$ to both sides.

$$uy+u=c-x^{2}-y^{2}+6x$$

Combine all terms containing $u$.

$$\left(y+1\right)u=c-x^{2}-y^{2}+6x$$

The equation is in standard form.

$$\left(y+1\right)u=c-y^{2}+6x-x^{2}$$

Divide both sides by $y+1$.

$$\frac{\left(y+1\right)u}{y+1}=\frac{c-y^{2}+6x-x^{2}}{y+1}$$

Dividing by $y+1$ undoes the multiplication by $y+1$.

$$u=\frac{c-y^{2}+6x-x^{2}}{y+1}$$

Solve for c

$c=x^{2}-6x+y^{2}+uy+u$