Solve A:(b+4):5=(A-1):(b-1):3

$$A : ( b + 4 ) : 5 = ( A - 1 ) : ( b - 1 ) : 3$$

Solve for A

$$A=\frac{5\left(b+4\right)}{2b+23}$$

$$b\neq -\frac{23}{2}\text{ and }b\neq -4\text{ and }b\neq 1$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 15, the least common multiple of 5,3.

$$3\times \frac{A}{b+4}=5\times \frac{A-1}{b-1}$$

Express 3\times \frac{A}{b+4} as a single fraction.

$$\frac{3A}{b+4}=5\times \frac{A-1}{b-1}$$

Express 5\times \frac{A-1}{b-1} as a single fraction.

$$\frac{3A}{b+4}=\frac{5\left(A-1\right)}{b-1}$$

Use the distributive property to multiply 5 by A-1.

$$\frac{3A}{b+4}=\frac{5A-5}{b-1}$$

Subtract \frac{5A-5}{b-1} from both sides.

$$\frac{3A}{b+4}-\frac{5A-5}{b-1}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of b+4 and b-1 is \left(b-1\right)\left(b+4\right). Multiply \frac{3A}{b+4} times \frac{b-1}{b-1}. Multiply \frac{5A-5}{b-1} times \frac{b+4}{b+4}.

$$\frac{3A\left(b-1\right)}{\left(b-1\right)\left(b+4\right)}-\frac{\left(5A-5\right)\left(b+4\right)}{\left(b-1\right)\left(b+4\right)}=0$$

Since \frac{3A\left(b-1\right)}{\left(b-1\right)\left(b+4\right)} and \frac{\left(5A-5\right)\left(b+4\right)}{\left(b-1\right)\left(b+4\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3A\left(b-1\right)-\left(5A-5\right)\left(b+4\right)}{\left(b-1\right)\left(b+4\right)}=0$$

Do the multiplications in 3A\left(b-1\right)-\left(5A-5\right)\left(b+4\right).

$$\frac{3Ab-3A-5Ab-20A+5b+20}{\left(b-1\right)\left(b+4\right)}=0$$

Combine like terms in 3Ab-3A-5Ab-20A+5b+20.

$$\frac{-2Ab-23A+5b+20}{\left(b-1\right)\left(b+4\right)}=0$$

Multiply both sides of the equation by \left(b-1\right)\left(b+4\right).

$$-2Ab-23A+5b+20=0$$

Subtract 5b from both sides. Anything subtracted from zero gives its negation.

$$-2Ab-23A+20=-5b$$

Subtract 20 from both sides.

$$-2Ab-23A=-5b-20$$

Combine all terms containing A.

$$\left(-2b-23\right)A=-5b-20$$

Divide both sides by -2b-23.

$$\frac{\left(-2b-23\right)A}{-2b-23}=\frac{-5b-20}{-2b-23}$$

Dividing by -2b-23 undoes the multiplication by -2b-23.

$$A=\frac{-5b-20}{-2b-23}$$

Divide -5b-20 by -2b-23.

$$A=\frac{5\left(b+4\right)}{2b+23}$$

Solve for b

$$b=-\frac{20-23A}{5-2A}$$

$$A\neq 0\text{ and }A\neq \frac{5}{2}\text{ and }A\neq 1$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 15, the least common multiple of 5,3.

$$3\times \frac{A}{b+4}=5\times \frac{A-1}{b-1}$$

Express 3\times \frac{A}{b+4} as a single fraction.

$$\frac{3A}{b+4}=5\times \frac{A-1}{b-1}$$

Express 5\times \frac{A-1}{b-1} as a single fraction.

$$\frac{3A}{b+4}=\frac{5\left(A-1\right)}{b-1}$$

Use the distributive property to multiply 5 by A-1.

$$\frac{3A}{b+4}=\frac{5A-5}{b-1}$$

Subtract \frac{5A-5}{b-1} from both sides.

$$\frac{3A}{b+4}-\frac{5A-5}{b-1}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of b+4 and b-1 is \left(b-1\right)\left(b+4\right). Multiply \frac{3A}{b+4} times \frac{b-1}{b-1}. Multiply \frac{5A-5}{b-1} times \frac{b+4}{b+4}.

$$\frac{3A\left(b-1\right)}{\left(b-1\right)\left(b+4\right)}-\frac{\left(5A-5\right)\left(b+4\right)}{\left(b-1\right)\left(b+4\right)}=0$$

Since \frac{3A\left(b-1\right)}{\left(b-1\right)\left(b+4\right)} and \frac{\left(5A-5\right)\left(b+4\right)}{\left(b-1\right)\left(b+4\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3A\left(b-1\right)-\left(5A-5\right)\left(b+4\right)}{\left(b-1\right)\left(b+4\right)}=0$$

Do the multiplications in 3A\left(b-1\right)-\left(5A-5\right)\left(b+4\right).

$$\frac{3Ab-3A-5Ab-20A+5b+20}{\left(b-1\right)\left(b+4\right)}=0$$

Combine like terms in 3Ab-3A-5Ab-20A+5b+20.

$$\frac{-2Ab-23A+5b+20}{\left(b-1\right)\left(b+4\right)}=0$$

Variable b cannot be equal to any of the values -4,1 since division by zero is not defined. Multiply both sides of the equation by \left(b-1\right)\left(b+4\right).

$$-2Ab-23A+5b+20=0$$

Add 23A to both sides. Anything plus zero gives itself.

$$-2Ab+5b+20=23A$$

Subtract 20 from both sides.

$$-2Ab+5b=23A-20$$

Combine all terms containing b.

$$\left(-2A+5\right)b=23A-20$$

The equation is in standard form.

$$\left(5-2A\right)b=23A-20$$

Divide both sides by -2A+5.

$$\frac{\left(5-2A\right)b}{5-2A}=\frac{23A-20}{5-2A}$$

Dividing by -2A+5 undoes the multiplication by -2A+5.

$$b=\frac{23A-20}{5-2A}$$

Variable b cannot be equal to any of the values -4,1.

$$b=\frac{23A-20}{5-2A}\text{, }b\neq -4\text{ and }b\neq 1$$