Solving Equations with Algebraic Fractions

Hello! Today we are going to take a look at how to solve equations that have algebraic fractions. But first, let’s review what algebraic fractions are. An algebraic fraction is any fraction that contains an algebraic expression. In other words, it’s a fraction that has a variable in it anywhere. For example,

\(\frac{5}{x}\), \(\frac{2x-9}{7}\), and \(\frac{3x}{2}\)

 

are all algebraic fractions.

So, now that we know what they are, let’s jump into a problem where we have to solve these. So we’re going to take a look at:

\(\frac{x}{8}=7\)

 

This is a nice and simple one. We know that in order to solve an equation, we need to isolate our variable. Here we can see that our variable, \(x\), is being divided by 8. So, to undo this division, we simply need to multiply both sides by 8.

\(8\cdot \frac{x}{8}=7\cdot 8\)

 

When we do that, our division is canceled out and we’re left with \(x\) on the left side, and \(7\times 8=56\), so:

\(x=56\)

 

Not too hard, right? Let’s try one that’s a little bit more complicated.

\(\frac{3x}{4}-\frac{2x}{12}=7\)

 

To solve this, I’m first going to simplify my fractions on the left side. To do this, I’ll need to convert \(\frac{3x}{4}\) to a fraction with a denominator of 12. I can do this by multiplying both the numerator and denominator by 3.

\(\frac{3x\cdot 3}{4\cdot 3}=\frac{9x}{12}\)

 

So that gives me the fraction, \(\frac{9x}{12}\). And then the rest of our equation is going to stay the same, so:

\(\frac{9x}{12}-\frac{2x}{12}=7\)

 

Since our denominators are the same, we can now subtract our numerators, so \(9x-2x=7x\).

\(\frac{7x}{12}=7\)

 

Another way we can write this is:

\(\frac{7}{12}\cdot x=7\)

 

So if we look at it this way, the way that we can get rid of this \(\frac{7}{12}\) over here is by multiplying by the reciprocal. So the reciprocal of \(\frac{7}{12}\), is \(\frac{12}{7}\), so we’ll multiply by that on both sides.

\(\frac{12}{7}\cdot \frac{7}{12}\cdot x=7\cdot \frac{12}{7}\)

 

When we do that, it cancels our fractions out over here, and we’re left with \(x\), which is what we want. And then if we do that over here, we can simplify first by doing \(7\div 7\), which gives us 1, and then do \(1\times 12\), which is 12.

\(x=12\)

 

So \(x=12\), which is our answer.


Let’s take a look at another problem.

\(\frac{4x}{5}=\frac{3x}{10}+9\)

 

This one we’re going to solve a little bit differently. For this problem, I’m going to solve for \(x\) by eliminating the fractions in our first step. To do this, we simply multiply the entire equation by the least common multiple of the denominators, of the fractions in the problem.

In this case, our denominators are 5 and 10, and the least common multiple of 5 and 10 is 10. So multiply the entire equation by 10.

\(10(\frac{4x}{5}=\frac{3x}{10}+9)\)

 

So that means we’re going to multiply each part of our equation by 10. So, \(10\cdot \frac{4x}{5}\). We can do \(10\div 5\) first, because remember it doesn’t matter if you multiply or divide first, multiplication and division can happen simultaneously. So for this, I’m going to divide \(10\div 5\) and get 2, and then multiply by \(4x\) to get \(8x\). If I multiply 10 by \(\frac{3x}{10}\), our 10s will cancel out and we’ll be left with \(3x\). And \(10\cdot 9=90\).

\(8x=3x+90\)

 

Now we can solve it like a normal algebra problem . So I’m going to subtract \(3x\) from both sides.

\(8x-3x=3x-3x+90\)

 

That gives us:

\(5x=90\)

 

And then we divide both sides by 5.

\(\frac{5x}{5}=\frac{90}{5}\)

 

That gives us:

\(x=18\)

 

Either of these methods will help you get the right answer when solving equations with algebraic fractions, so feel free to use whichever one you feel most comfortable with.

Before we go, I want to show you one more problem you might come across.

\(\frac{x+17}{5}=21\)

 

So we have an expression in the numerator instead of just a variable, and maybe a number being multiplied by it, so it’s a little bit different than the ones we’ve looked at before. So, to solve problems like these, we first get rid of the fraction by multiplying by the denominator on both sides of the equation. So we’re going to multiply by 5 on both sides.

\(5\cdot \frac{x+17}{5}=21\cdot 5\)

 

That cancels out our denominator, so we no longer have a fraction, and we’re just left with the expression in the numerator: \(x+17\). And \(21\cdot 5=105\).

\(x+17=105\)

 

So now we’re going to solve it like normal. We’ll subtract 17 from both sides,

\(x+17-17=105-17\)
 
\(x=88\)

 

And that’s it! I hope this video helped you better understand how to solve equations with algebraic fractions. Thanks for watching, and happy studying!

Practice Questions

Question #1:

 
Solve the equation: \(\frac{2x}{3}-\frac{4x}{9}=6\).

\(x=18\)
\(x=27\)
\(x=12\)
\(x=2\)
Answer:

We can combine the fractions on the left side by converting \(\frac{2x}{3}\) to a fraction with a denominator of 9. To do so, multiply the first fraction by \(\frac{3}{3}\).

\(\frac{3}{3}\cdot\frac{2x}{3}-\frac{4x}{9}=6\)
 
\(\frac{6x}{9}-\frac{4x}{9}=6\)
 
\(\frac{2x}{9}=6\)

Now, multiply both sides by 9.

\(9\cdot\frac{2x}{9}=6\cdot9\)
 
\(2x=54\)

Finally, divide both sides by 2.

\(\frac{2x}{2}=\frac{54}{2}\)
 
\(x=27\)

Question #2:

 
Solve the equation: \(\frac{5x}{6}=\frac{11x}{15}+2\).

\(x=-\frac{22}{9}\)
\(x=\frac{2}{3}\)
\(x=20\)
\(x=11\)
Answer:

We can multiply both sides of the equation by the least common multiple of the denominators. The least common multiple of 6 and 15 is 30, so multiply the entire equation by 30

\(30\left(\frac{5}{6}x=\frac{11}{15}x+2\right)\)
 
\(\frac{30\cdot5}{6}x=\frac{30\cdot11}{15}x+30\cdot2\)
 
\(25x=22x+60\)

To get the variable terms on the left side, subtract 22x from both sides of the equation.

\(25x-22x=22x+60-22x\)
 
\(3x=60\)

Now, divide both sides by 3.

\(\frac{3x}{3}=\frac{60}{3}\)
 
\(x=20\)

Question #3:

Solve the equation: \(2x+79=5\).

\(x=-4\)
\(x=10\)
\(x=-1\)
\(x=19\)
Answer:

We can get rid of the fraction by multiplying by the denominator on both sides of the equation. That is, multiply both sides of the equation by 9.

\(9\cdot \frac{2x+7}{9}=5\cdot9\)
 
\(2x+7=45\)

Now, subtract 7 from both sides.

\(2x+7-7=45-7\)
 
\(2x=38\)

Then, divide both sides by 2.

\(\frac{2x}{2}=\frac{38}{2}\)
 
\(x=19\)

Question #4:

 
The quotient of thirteen less than a number and 7 equals twice the number. What is the number?

-1

5

-2

7

Answer:

Let x be the number you are trying to find. Since quotient means to divide, we can write the quotient of thirteen less than a number and 7 as \(\frac{x-13}{7}\). Twice the number can be written as \(2x\). Set these two expressions equal to each other to get the equation:

\(\frac{x-13}{7}=2x\)

To solve for x, first, get rid of the fraction by multiplying by the denominator on both sides of the equation. That is, multiply both sides of the equation by 7.

\(7\cdot\frac{x-13}{7}=2x\cdot7\)
 
\(x-13=14x\)

To get the variable terms on the left side, subtract \(14x\) from both sides of the equation.

\(x-13-14x=14x-14x\)
 
\(-13x-13=0\)

Now, add the constant of 13 to both sides.

\(-13x-13+13=0+13\)
 
\(-13x=13\)

Then, divide both sides by –13.

\(\frac{-13x}{-13}=\frac{13}{-13}\)
 
\(x=-1\)

So, the number is –1.

Question #5:

 
You and a friend play on the same soccer team. Both of you decide to save money to buy a soccer goal to practice shooting. One-third of the money, in dollars, you and your friend have saved is $30 less than twice of what you have saved. If your friend has saved $25, how much money have you saved?

$46

$23

$10

$17

Answer:

Let x be the amount of money you have saved. Since your friend has saved $25, we can write one-third of the money you and your friend have saved as \(\frac{x+25}{3}\). Thirty dollars less than twice of what you have saved can be written as \(2x-30\). Setting these two expressions equal to each other gives us the equation:

\(\frac{x+25}{3}=2x-30\)

To solve for x, first, get rid of the fraction by multiplying by the denominator on both sides of the equation. That is, multiply both sides of the equation by 3.

\(3\cdot\frac{x+25}{3}=3(2x-30)\)
 
\(x+25=6x-90\)

To get the variable terms on the left side, subtract \(6x\) from both sides of the equation.

\(x+25-6x=6x-90-6x\)
 
\(-5x+25=-90\)

Now, subtract 25 from both sides.

\(-5x+25-25=-90-25\)
 
\(-5x=-115\)

Then, divide both sides by –5.

\(\frac{-5x}{-5}=\frac{115}{-5}\)
 
\(x=23\)

So, you have saved $23.

 

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by Mometrix Test Preparation | This Page Last Updated: May 12, 2022