Solve AB/-10=5/12text{và}AC-AB=

$$\frac { A B } { - 10 } = \frac { 5 } { 12 } \text { và } A C - A B =$$

Solve for A (complex solution)

$$\left\{\begin{matrix}\\A=0\text{, }&\text{unconditionally}\\A\in \mathrm{C}\text{, }&B=\frac{25Cvà}{54}\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 60, the least common multiple of -10,12.

$$-6AB=25vàAC-60AB$$

Subtract 25vàAC from both sides.

$$-6AB-25vàAC=-60AB$$

Add 60AB to both sides.

$$-6AB-25vàAC+60AB=0$$

Combine -6AB and 60AB to get 54AB.

$$54AB-25vàAC=0$$

Combine all terms containing A.

$$\left(54B-25vàC\right)A=0$$

The equation is in standard form.

$$\left(54B-25Cvà\right)A=0$$

Divide 0 by -25Cvà+54B.

$$A=0$$

Solve for B (complex solution)

$$\left\{\begin{matrix}\\B=\frac{25Cvà}{54}\text{, }&\text{unconditionally}\\B\in \mathrm{C}\text{, }&A=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 60, the least common multiple of -10,12.

$$-6AB=25vàAC-60AB$$

Add 60AB to both sides.

$$-6AB+60AB=25vàAC$$

Combine -6AB and 60AB to get 54AB.

$$54AB=25vàAC$$

The equation is in standard form.

$$54AB=25ACvà$$

Divide both sides by 54A.

$$\frac{54AB}{54A}=\frac{25ACvà}{54A}$$

Dividing by 54A undoes the multiplication by 54A.

$$B=\frac{25ACvà}{54A}$$

Divide 25vàAC by 54A.

$$B=\frac{25Cvà}{54}$$

Solve for A

$$\left\{\begin{matrix}\\A=0\text{, }&\text{unconditionally}\\A\in \mathrm{R}\text{, }&B=\frac{25Cvà}{54}\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 60, the least common multiple of -10,12.

$$-6AB=25vàAC-60AB$$

Subtract 25vàAC from both sides.

$$-6AB-25vàAC=-60AB$$

Add 60AB to both sides.

$$-6AB-25vàAC+60AB=0$$

Combine -6AB and 60AB to get 54AB.

$$54AB-25vàAC=0$$

Combine all terms containing A.

$$\left(54B-25vàC\right)A=0$$

The equation is in standard form.

$$\left(54B-25Cvà\right)A=0$$

Divide 0 by -25Cvà+54B.

$$A=0$$

Solve for B

$$\left\{\begin{matrix}\\B=\frac{25Cvà}{54}\text{, }&\text{unconditionally}\\B\in \mathrm{R}\text{, }&A=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 60, the least common multiple of -10,12.

$$-6AB=25vàAC-60AB$$

Add 60AB to both sides.

$$-6AB+60AB=25vàAC$$

Combine -6AB and 60AB to get 54AB.

$$54AB=25vàAC$$

The equation is in standard form.

$$54AB=25ACvà$$

Divide both sides by 54A.

$$\frac{54AB}{54A}=\frac{25ACvà}{54A}$$

Dividing by 54A undoes the multiplication by 54A.

$$B=\frac{25ACvà}{54A}$$

Divide 25vàAC by 54A.

$$B=\frac{25Cvà}{54}$$