Solve 3(2a-7)/5ac=b

$$\frac { 3 ( 2 a - 7 ) } { 5 a c } = b$$

Solve for a

$$a=\frac{21}{6-5bc}$$

$$\left(b=0\text{ or }c\neq \frac{6}{5b}\right)\text{ and }c\neq 0$$

Steps for Solving Linear Equation

Variable a cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 5ac.

$$3\left(2a-7\right)=b\times 5ac$$

Use the distributive property to multiply 3 by 2a-7.

$$6a-21=b\times 5ac$$

Subtract b\times 5ac from both sides.

$$6a-21-b\times 5ac=0$$

Multiply -1 and 5 to get -5.

$$6a-21-5bac=0$$

Add 21 to both sides. Anything plus zero gives itself.

$$6a-5bac=21$$

Combine all terms containing a.

$$\left(6-5bc\right)a=21$$

Divide both sides by 6-5cb.

$$\frac{\left(6-5bc\right)a}{6-5bc}=\frac{21}{6-5bc}$$

Dividing by 6-5cb undoes the multiplication by 6-5cb.

$$a=\frac{21}{6-5bc}$$

Variable a cannot be equal to 0.

$$a=\frac{21}{6-5bc}\text{, }a\neq 0$$

Solve for b

$$b=\frac{3\left(2a-7\right)}{5ac}$$

$$a\neq 0\text{ and }c\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 5ac.

$$3\left(2a-7\right)=b\times 5ac$$

Use the distributive property to multiply 3 by 2a-7.

$$6a-21=b\times 5ac$$

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

$$b\times 5ac=6a-21$$

The equation is in standard form.

$$5acb=6a-21$$

Divide both sides by 5ac.

$$\frac{5acb}{5ac}=\frac{6a-21}{5ac}$$

Dividing by 5ac undoes the multiplication by 5ac.

$$b=\frac{6a-21}{5ac}$$

Divide 6a-21 by 5ac.

$$b=\frac{3\left(2a-7\right)}{5ac}$$