Lesson 4: Equations with Fractions

New Concept

Learn to solve equations with fractions by "clearing" the denominators. Multiplying every term by the Least Common Denominator (LCD) simplifies the equation, but be sure to check for extraneous solutions when variables are in the denominator.

What’s next

Get ready to apply this concept. You'll work through interactive examples, tackle practice problems, and see how to solve real-world motion and work problems.

Property

To solve an equation that involves an algebraic fraction, multiply both sides of the equation by the denominator of the fraction to 'clear' the fraction. For an equation like

Examples

• To solve $\frac{2x}{7} = 4$, multiply both sides by 7:
  $$7\left(\frac{2x}{7}\right) = (4)7$$
  , so
  $$2x = 28$$
  , and
  $$x = 14$$
  .
• To solve $\frac{40 + 8y}{y} = 12$, multiply both sides by $y$:
  $$y\left(\frac{40 + 8y}{y}\right) = (12)y$$
  , so
  $$40 + 8y = 12y$$
  . This simplifies to
  $$40 = 4y$$
  , and
  $$y = 10$$
  .
• To solve $\frac{z - 9}{2} = -5$, multiply both sides by 2:
  $$2\left(\frac{z - 9}{2}\right) = (-5)2$$
  , so
  $$z - 9 = -10$$
  , and
  $$z = -1$$
  .

Explanation

Think of this as undoing division. Since a fraction is a division problem, multiplying by the denominator cancels it out. This transforms a fractional equation into a simpler, linear one that is much easier to solve.

Property

If an equation contains more than one fraction, multiply both sides of the equation by the LCD (Least Common Denominator) of all the fractions. This will clear all the denominators at once. Be sure to multiply each term of the equation by the LCD, including terms that are not fractions.

Examples

• Solve $\frac{x}{2} + 5 = \frac{3x}{4}$. The LCD is 4. Multiply every term by 4:
  $$4\left(\frac{x}{2}\right) + 4(5) = 4\left(\frac{3x}{4}\right)$$
  , which simplifies to
  $$2x + 20 = 3x$$
  , so
  $$x = 20$$
  .
• Solve $\frac{y}{2} - \frac{y}{5} = 6$. The LCD is 10. Multiply every term by 10:
  $$10\left(\frac{y}{2}\right) - 10\left(\frac{y}{5}\right) = 10(6)$$
  , which simplifies to
  $$5y - 2y = 60$$
  , so
  $$3y = 60$$
  and
  $$y=20$$
  .
• Solve $\frac{a}{3} - 2 = \frac{a}{4} + \frac{1}{2}$. The LCD is 12. This gives
  $$12\left(\frac{a}{3}\right) - 12(2) = 12\left(\frac{a}{4}\right) + 12\left(\frac{1}{2}\right)$$
  , so
  $$4a - 24 = 3a + 6$$
  , and
  $$a = 30$$
  .

Explanation

Using the LCD is an efficient way to eliminate all fractions in a single step. By multiplying every term by the common denominator, each fraction simplifies to an integer or polynomial, leaving a straightforward equation to solve.

Property

Equations that involve algebraic fractions with variables in the denominator can also be solved using an LCD. The process remains the same: find the LCD for all fractions, which may include variable expressions, and multiply both sides of the equation by this LCD to clear the fractions.

Examples

• Solve $\frac{10}{x-3} = 5$. The LCD is $x-3$. Multiply both sides by $x-3$ to get
  $$10 = 5(x-3)$$
  , which is
  $$10 = 5x - 15$$
  . The solution is
  $$x=5$$
  .
• Solve $\frac{3}{2} + 4 = \frac{2x+7}{x-1}$. The LCD is $2(x-1)$. Multiplying gives
  $$3(x-1) + 8(x-1) = 2(2x+7)$$
  . This simplifies to
  $$11x - 11 = 4x + 14$$
  , so
  $$7x = 25$$
  and
  $$x = \frac{25}{7}$$
  .
• Solve $\frac{y}{y+1} = \frac{2}{3}$. The LCD is $3(y+1)$. Multiplying gives
  $$3y = 2(y+1)$$
  , so
  $$3y = 2y + 2$$
  , and
  $$y=2$$
  .

Explanation

Even if variables are in the denominator, the core strategy doesn't change. Find the LCD, multiply every term to clear fractions, and solve the resulting polynomial equation. Just be careful for extraneous solutions!

Property

When multiplying an equation by an expression containing a variable, a false solution may be introduced. Such a solution is called an extraneous solution. To check for extraneous solutions, substitute the possible solution into the original equation. If it causes any denominator in the original equation to equal zero, that solution is extraneous and must be discarded.

Examples

• Solve $5 + \frac{2}{x-4} = \frac{x-2}{x-4}$. Multiplying by the LCD $x-4$ gives
  $$5(x-4) + 2 = x-2$$
  , so
  $$5x - 20 + 2 = x-2$$
  , which simplifies to
  $$4x = 16$$
  and
  $$x=4$$
  . Since $x=4$ makes the original denominator zero, it is an extraneous solution. There is no solution.
• Solve $\frac{y}{y+2} - 1 = \frac{-2}{y+2}$. The LCD is $y+2$. This gives
  $$y - 1(y+2) = -2$$
  , so
  $$y - y - 2 = -2$$
  , which is
  $$-2=-2$$
  . The proposed solution is $y=-2$, which is extraneous. No solution.
• Solve $\frac{3a}{a-6} - 4 = \frac{18}{a-6}$. The LCD is $a-6$. This gives
  $$3a - 4(a-6) = 18$$
  , so
  $$3a - 4a + 24 = 18$$
  , which simplifies to
  $$-a = -6$$
  and
  $$a=6$$
  . This solution is extraneous.

Explanation

An extraneous solution is a tricky fake-out! It appears to be a valid answer, but it's invalid in the original context because it makes you divide by zero. Always plug your solution back in to check the original denominators.

Property

To solve a formula for a specific variable, first clear any fractions by multiplying the entire equation by the LCD. Next, gather all terms containing the desired variable on one side of the equation. If there are multiple terms with the variable, factor the variable out, and then divide both sides by the remaining factor to isolate it.

Examples

• Solve the formula $S = \frac{a}{1-r}$ for $r$. Multiply by $1-r$ to get
  $$S(1-r) = a$$
  . Distribute $S$ to get
  $$S - Sr = a$$
  . Then,
  $$-Sr = a - S$$
  , so
  $$r = \frac{a-S}{-S}$$
  or
  $$r = \frac{S-a}{S}$$
  .
• Solve $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ for $u$. The LCD is $fuv$. Multiply by the LCD to get
  $$uv = fv + fu$$
  . Move terms with $u$ to one side:
  $$uv - fu = fv$$
  . Factor out $u$:
  $$u(v-f) = fv$$
  . The solution is
  $$u = \frac{fv}{v-f}$$
  .
• Solve $h = \frac{2A}{b_1 + b_2}$ for $b_1$. Multiply by $b_1+b_2$ to get
  $$h(b_1+b_2) = 2A$$
  . Distribute $h$:
  $$hb_1 + hb_2 = 2A$$
  . Then
  $$hb_1 = 2A - hb_2$$
  , and
  $$b_1 = \frac{2A - hb_2}{h}$$
  .

Explanation

Rearranging formulas is like solving a puzzle. First, get rid of fractions. Then, herd all the pieces with your target variable to one side. If it appears in multiple terms, factor it out to isolate it.

Property

Motion problems use the formula distance = rate × time (

Examples

• A kayak travels 24 miles upstream and back in 9 hours. The river's current is 2 mph. Find the kayak's speed ($k$) in still water. The equation is
  $$\frac{24}{k-2} + \frac{24}{k+2} = 9$$
  . The solution is
  $$k \approx 5.5$$
  mph.
• A train travels 300 miles with a tailwind of 25 mph and 200 miles back against the same wind. The total travel time is 5 hours. Find the train's speed ($s$). The equation is
  $$\frac{300}{s+25} + \frac{200}{s-25} = 5$$
  . The solution is
  $$s=75$$
  mph.
• Maria jogs 6 miles to the park and then walks 4 miles back home. Her jogging speed is twice her walking speed ($w$). The total trip took 2 hours. Find her walking speed. The equation is
  $$\frac{6}{2w} + \frac{4}{w} = 2$$
  . This simplifies to
  $$\frac{7}{w}=2$$
  , so
  $$w=3.5$$
  mph.

Explanation

For tricky motion problems, the key is often time. By expressing time as distance divided by rate (

Property

The formula for work is: work rate × time = work completed (

Examples

• Tom can paint a room in 6 hours, and Jerry can do it in 12 hours. How long will it take them together? Let $t$ be the time together. The equation is
  $$\frac{t}{6} + \frac{t}{12} = 1$$
  . The LCD is 12, so
  $$2t + t = 12$$
  , and
  $$t=4$$
  hours.
• An inlet pipe fills a pool in 8 hours. A drain pipe empties it in 10 hours. If both are open, how long to fill the pool? Let $t$ be the time. The equation is
  $$\frac{t}{8} - \frac{t}{10} = 1$$
  . The LCD is 40, so
  $$5t - 4t = 40$$
  , and
  $$t=40$$
  hours.
• Printer A can print a manuscript in 3 hours. After working for 1 hour, Printer B joins, and they finish in 1 more hour. How long would it take Printer B alone? Let $b$ be B's time. The equation is
  $$1 = \frac{2}{3} + \frac{1}{b}$$
  . This gives
  $$\frac{1}{3} = \frac{1}{b}$$
  , so
  $$b=3$$
  hours.

Explanation

Work problems are about rates. Figure out how much of the job each person (or pipe) can do in one hour. Add those rates together to find their combined rate, then use that to find the time it takes to complete one whole job.