Lesson 5: Equations with Fractions

New Concept

You will learn to solve equations with algebraic fractions by multiplying each term by the Least Common Denominator (LCD). This technique eliminates denominators, transforming the problem into a familiar linear or quadratic equation.

What’s next

Next, you'll master this technique through interactive examples and practice problems, learning to spot and avoid extraneous solutions along the way.

Property

To solve an equation with fractions, we multiply each side of the equation by the denominator of the fraction. This will clear the fraction and give us an equivalent equation without fractions. If the equation contains more than one fraction, we can clear all the denominators at once by multiplying both sides by the LCD of the fractions. We must multiply each term of an equation by the LCD, whether or not the term is a fraction.

Examples

• To solve $\frac{50}{10 - x} = 5$, we multiply both sides by $10 - x$ to get $50 = 5(10 - x)$. Distributing gives $50 = 50 - 5x$, which simplifies to $5x=0$, so $x=0$.

• Solve $\frac{5}{x} - 2 = \frac{7}{2x + 1}$. The LCD is $x(2x+1)$. Multiplying each term gives $x(2x+1)(\frac{5}{x}) - x(2x+1)(2) = x(2x+1)(\frac{7}{2x+1})$. This simplifies to $5(2x+1) - 2x(2x+1) = 7x$, which becomes a quadratic equation to solve.

Property

A proportion is a statement that two ratios are equal. For example, $\frac{a}{b} = \frac{c}{d}$.

If $\frac{a}{b} = \frac{c}{d}$, then $ad = bc$, as long as $b, d \neq 0$. This shortcut is known as cross-multiplying.

Examples

• To solve the proportion $\frac{8}{3} = \frac{x}{9}$, we can cross-multiply. This gives $8 \cdot 9 = 3 \cdot x$, so $72 = 3x$. Dividing by 3, we find $x=24$.

Property

An apparent solution that does not satisfy the original equation is called an extraneous solution. Whenever we multiply an equation by an expression containing the variable, we should check that the solution obtained does not cause any of the fractions to be undefined. An algebraic fraction is undefined for any values of $x$ that make its denominator equal to zero.

Examples

• Solve $\frac{x}{x-5} = \frac{5}{x-5} + 3$. Multiplying by the LCD, $x-5$, gives $x = 5 + 3(x-5)$, which simplifies to $x = 5 + 3x - 15$, or $-2x = -10$, so $x=5$. Since $x=5$ makes the original denominator zero, it is an extraneous solution.

• Solve $\frac{x^2}{x-4} = \frac{16}{x-4}$. Multiplying by $x-4$ gives $x^2=16$, so $x=4$ or $x=-4$. The value $x=4$ is an extraneous solution because it makes the denominator zero. The only valid solution is $x=-4$.

Property

To solve a formula for a specific variable that appears in a fraction, we may need to clear the fractions. First, multiply both sides of the equation by the denominator to eliminate it. Then, use the distributive law and collect all terms involving the desired variable on one side. If the variable appears in more than one term, factor it out, then isolate it by dividing.

Examples

• To solve the formula $S = \frac{a}{1 - r}$ for $r$, first multiply by $1-r$ to get $S(1-r) = a$. Distribute to get $S - Sr = a$. Isolate the term with $r$: $-Sr = a - S$. Finally, divide by $-S$ to get $r = \frac{a-S}{-S}$ or $r = \frac{S-a}{S}$.

• Solve the formula $m = \frac{y - k}{x - h}$ for $x$. Multiply by $x-h$ to get $m(x-h) = y-k$. Distribute: $mx - mh = y-k$. Isolate the $x$ term: $mx = y - k + mh$. Divide by $m$: $x = \frac{y-k+mh}{m}$.