Solve 1-a(x+1)+b(x+1)=acx

$$1 - a ( x + 1 ) + b ( x + 1 ) = a c x$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=\frac{bx+b+1}{cx+x+1}\text{, }&c=-1\text{ or }x\neq -\frac{1}{c+1}\\a\in \mathrm{C}\text{, }&x=\frac{c}{c+1}-1\text{ and }b=-1-\frac{1}{c}\text{ and }c\neq -1\text{ and }c\neq 0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply a by x+1.

$$1-\left(ax+a\right)+b\left(x+1\right)=acx$$

To find the opposite of ax+a, find the opposite of each term.

$$1-ax-a+b\left(x+1\right)=acx$$

Use the distributive property to multiply b by x+1.

$$1-ax-a+bx+b=acx$$

Subtract acx from both sides.

$$1-ax-a+bx+b-acx=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$-ax-a+bx+b-acx=-1$$

Subtract bx from both sides.

$$-ax-a+b-acx=-1-bx$$

Subtract b from both sides.

$$-ax-a-acx=-1-bx-b$$

Combine all terms containing a.

$$\left(-x-1-cx\right)a=-1-bx-b$$

The equation is in standard form.

$$\left(-cx-x-1\right)a=-bx-b-1$$

Divide both sides by -x-1-cx.

$$\frac{\left(-cx-x-1\right)a}{-cx-x-1}=\frac{-bx-b-1}{-cx-x-1}$$

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

$$a=\frac{-bx-b-1}{-cx-x-1}$$

Divide -1-bx-b by -x-1-cx.

$$a=\frac{bx+b+1}{cx+x+1}$$

Solve for b (complex solution)

$$\left\{\begin{matrix}b=-\frac{1-a-ax-acx}{x+1}\text{, }&x\neq -1\\b\in \mathrm{C}\text{, }&a=-\frac{1}{c}\text{ and }x=-1\text{ and }c\neq 0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply a by x+1.

$$1-\left(ax+a\right)+b\left(x+1\right)=acx$$

To find the opposite of ax+a, find the opposite of each term.

$$1-ax-a+b\left(x+1\right)=acx$$

Use the distributive property to multiply b by x+1.

$$1-ax-a+bx+b=acx$$

Subtract 1 from both sides.

$$-ax-a+bx+b=acx-1$$

Add ax to both sides.

$$-a+bx+b=acx-1+ax$$

Add a to both sides.

$$bx+b=acx-1+ax+a$$

Combine all terms containing b.

$$\left(x+1\right)b=acx-1+ax+a$$

The equation is in standard form.

$$\left(x+1\right)b=acx+ax+a-1$$

Divide both sides by x+1.

$$\frac{\left(x+1\right)b}{x+1}=\frac{acx+ax+a-1}{x+1}$$

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

$$b=\frac{acx+ax+a-1}{x+1}$$

Solve for a

$$\left\{\begin{matrix}a=\frac{bx+b+1}{cx+x+1}\text{, }&c=-1\text{ or }x\neq -\frac{1}{c+1}\\a\in \mathrm{R}\text{, }&x=\frac{c}{c+1}-1\text{ and }b=-1-\frac{1}{c}\text{ and }c\neq -1\text{ and }c\neq 0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply a by x+1.

$$1-\left(ax+a\right)+b\left(x+1\right)=acx$$

To find the opposite of ax+a, find the opposite of each term.

$$1-ax-a+b\left(x+1\right)=acx$$

Use the distributive property to multiply b by x+1.

$$1-ax-a+bx+b=acx$$

Subtract acx from both sides.

$$1-ax-a+bx+b-acx=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$-ax-a+bx+b-acx=-1$$

Subtract bx from both sides.

$$-ax-a+b-acx=-1-bx$$

Subtract b from both sides.

$$-ax-a-acx=-1-bx-b$$

Combine all terms containing a.

$$\left(-x-1-cx\right)a=-1-bx-b$$

The equation is in standard form.

$$\left(-cx-x-1\right)a=-bx-b-1$$

Divide both sides by -x-1-cx.

$$\frac{\left(-cx-x-1\right)a}{-cx-x-1}=\frac{-bx-b-1}{-cx-x-1}$$

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

$$a=\frac{-bx-b-1}{-cx-x-1}$$

Divide -1-bx-b by -x-1-cx.

$$a=\frac{bx+b+1}{cx+x+1}$$

Solve for b

$$\left\{\begin{matrix}b=-\frac{1-a-ax-acx}{x+1}\text{, }&x\neq -1\\b\in \mathrm{R}\text{, }&a=-\frac{1}{c}\text{ and }x=-1\text{ and }c\neq 0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply a by x+1.

$$1-\left(ax+a\right)+b\left(x+1\right)=acx$$

To find the opposite of ax+a, find the opposite of each term.

$$1-ax-a+b\left(x+1\right)=acx$$

Use the distributive property to multiply b by x+1.

$$1-ax-a+bx+b=acx$$

Subtract 1 from both sides.

$$-ax-a+bx+b=acx-1$$

Add ax to both sides.

$$-a+bx+b=acx-1+ax$$

Add a to both sides.

$$bx+b=acx-1+ax+a$$

Combine all terms containing b.

$$\left(x+1\right)b=acx-1+ax+a$$

The equation is in standard form.

$$\left(x+1\right)b=acx+ax+a-1$$

Divide both sides by x+1.

$$\frac{\left(x+1\right)b}{x+1}=\frac{acx+ax+a-1}{x+1}$$

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

$$b=\frac{acx+ax+a-1}{x+1}$$