Solve -(l+m)-lx^2+(2l+m)x=0

$$- ( l + m ) - l x ^ { 2 } + ( 2 l + m ) x = 0$$

Solve for l (complex solution)

$$\left\{\begin{matrix}l=-\frac{m}{1-x}\text{, }&x\neq 1\\l\in \mathrm{C}\text{, }&x=1\end{matrix}\right.$$

Steps for Solving Linear Equation

To find the opposite of l+m, find the opposite of each term.

$$-l-m-lx^{2}+\left(2l+m\right)x=0$$

Use the distributive property to multiply 2l+m by x.

$$-l-m-lx^{2}+2lx+mx=0$$

Subtract mx from both sides. Anything subtracted from zero gives its negation.

$$-l-m-lx^{2}+2lx=-mx$$

Add m to both sides.

$$-l-lx^{2}+2lx=-mx+m$$

Combine all terms containing l.

$$\left(-1-x^{2}+2x\right)l=-mx+m$$

The equation is in standard form.

$$\left(-x^{2}+2x-1\right)l=m-mx$$

Divide both sides by -1-x^{2}+2x.

$$\frac{\left(-x^{2}+2x-1\right)l}{-x^{2}+2x-1}=\frac{m-mx}{-x^{2}+2x-1}$$

Dividing by -1-x^{2}+2x undoes the multiplication by -1-x^{2}+2x.

$$l=\frac{m-mx}{-x^{2}+2x-1}$$

Divide -mx+m by -1-x^{2}+2x.

$$l=\frac{m}{x-1}$$

Solve for m (complex solution)

$$\left\{\begin{matrix}\\m=l\left(x-1\right)\text{, }&\text{unconditionally}\\m\in \mathrm{C}\text{, }&x=1\end{matrix}\right.$$

Steps for Solving Linear Equation

To find the opposite of l+m, find the opposite of each term.

$$-l-m-lx^{2}+\left(2l+m\right)x=0$$

Use the distributive property to multiply 2l+m by x.

$$-l-m-lx^{2}+2lx+mx=0$$

Subtract 2lx from both sides. Anything subtracted from zero gives its negation.

$$-l-m-lx^{2}+mx=-2lx$$

Add l to both sides.

$$-m-lx^{2}+mx=-2lx+l$$

Add lx^{2} to both sides.

$$-m+mx=-2lx+l+lx^{2}$$

Combine all terms containing m.

$$\left(-1+x\right)m=-2lx+l+lx^{2}$$

The equation is in standard form.

$$\left(x-1\right)m=lx^{2}-2lx+l$$

Divide both sides by x-1.

$$\frac{\left(x-1\right)m}{x-1}=\frac{l\left(x-1\right)^{2}}{x-1}$$

Dividing by x-1 undoes the multiplication by x-1.

$$m=\frac{l\left(x-1\right)^{2}}{x-1}$$

Divide l\left(-1+x\right)^{2} by x-1.

$$m=l\left(x-1\right)$$

Solve for l

$$\left\{\begin{matrix}l=-\frac{m}{1-x}\text{, }&x\neq 1\\l\in \mathrm{R}\text{, }&x=1\end{matrix}\right.$$

Steps for Solving Linear Equation

To find the opposite of l+m, find the opposite of each term.

$$-l-m-lx^{2}+\left(2l+m\right)x=0$$

Use the distributive property to multiply 2l+m by x.

$$-l-m-lx^{2}+2lx+mx=0$$

Subtract mx from both sides. Anything subtracted from zero gives its negation.

$$-l-m-lx^{2}+2lx=-mx$$

Add m to both sides.

$$-l-lx^{2}+2lx=-mx+m$$

Combine all terms containing l.

$$\left(-1-x^{2}+2x\right)l=-mx+m$$

The equation is in standard form.

$$\left(-x^{2}+2x-1\right)l=m-mx$$

Divide both sides by -1-x^{2}+2x.

$$\frac{\left(-x^{2}+2x-1\right)l}{-x^{2}+2x-1}=\frac{m-mx}{-x^{2}+2x-1}$$

Dividing by -1-x^{2}+2x undoes the multiplication by -1-x^{2}+2x.

$$l=\frac{m-mx}{-x^{2}+2x-1}$$

Divide -mx+m by -1-x^{2}+2x.

$$l=\frac{m}{x-1}$$

Solve for m

$$\left\{\begin{matrix}\\m=l\left(x-1\right)\text{, }&\text{unconditionally}\\m\in \mathrm{R}\text{, }&x=1\end{matrix}\right.$$

Steps for Solving Linear Equation

To find the opposite of l+m, find the opposite of each term.

$$-l-m-lx^{2}+\left(2l+m\right)x=0$$

Use the distributive property to multiply 2l+m by x.

$$-l-m-lx^{2}+2lx+mx=0$$

Subtract 2lx from both sides. Anything subtracted from zero gives its negation.

$$-l-m-lx^{2}+mx=-2lx$$

Add l to both sides.

$$-m-lx^{2}+mx=-2lx+l$$

Add lx^{2} to both sides.

$$-m+mx=-2lx+l+lx^{2}$$

Combine all terms containing m.

$$\left(-1+x\right)m=-2lx+l+lx^{2}$$

The equation is in standard form.

$$\left(x-1\right)m=lx^{2}-2lx+l$$

Divide both sides by -1+x.

$$\frac{\left(x-1\right)m}{x-1}=\frac{l\left(x-1\right)^{2}}{x-1}$$

Dividing by -1+x undoes the multiplication by -1+x.

$$m=\frac{l\left(x-1\right)^{2}}{x-1}$$

Divide l\left(-1+x\right)^{2} by -1+x.

$$m=l\left(x-1\right)$$