带分数方程式

Solving Equations with Fractions

Variable in the Numerator

Sometimes, you are given an equation with a fraction that has a variable in the numerator. How do you solve for this variable?

Variable in the Denominator

Other times, you are given an equation with a fraction that has a variable in the denominator. Here's an example of how to solve:

Let's do an problem that encompasses both scenarios!

Step One: Isolate the Fraction with the Variable

Whether solving for a variable in the numerator or in the denominator, our first step is always to isolate the fraction with the variable. To do this, we subtract two thirds from both sides of the equation. $$\begin{align} \dfrac{3}{2y+5}+\dfrac{2}{3} &= 1\\ \dfrac{3}{2y+5}+\dfrac{2}{3}\,{\color{#8925AE}{-\,\dfrac{2}{3}}} &= 1\,{\color{#8925AE}{-\,\dfrac{2}{3}}}\\ \dfrac{3}{2y+5}+\cancel{\dfrac{2}{3}}-\cancel{\dfrac{2}{3}} &= \dfrac{3}{3}-\dfrac{2}{3}\\ \dfrac{3}{2y+5}&= \dfrac{3-2}{3}\\ \dfrac{3}{2y+5}&= \dfrac{1}{3} \end{align}$$

Step Two: Multiply Both Sides by the Denominator

Since the variable is in the denominator, our next step is to multiply both sides of the equation by the denominator of the fraction containing the variable. $$\begin{align} \dfrac{3}{2y+5}&= \dfrac{1}{3}\\ {\color{#8925AE}{(2y+5) \,\times }}\,\dfrac{3}{2y+5}&= \dfrac{1}{3} \,\color{#8925AE}{\times\, (2y+5)}\\ \cancel{(2y+5)} \times \dfrac{3}{\cancel{2y+5}}&= \dfrac{1 \times (2y+5)}{3}\\ 3&= \dfrac{2y+5}{3} \\ \end{align}$$

Step Three: Multiply Both Sides by the Reciprocal

Now we have an equation with a fraction in the numerator. We want to cancel out the denominator of this fraction to make everything easier to simplify later.

Let's take a closer look at our equation: $3= \frac{2y+5}{3}.$ Notice that on the right side, we are essentially multiplying $2y+5$ by $\frac{1}{3},$ since $(2y+5) \times \frac{1}{3} = \frac{2y+5}{3}.$

We are trying to get rid of the $3$ in the denominator. To do this, we first use a similar technique to using reciprocals to solve one-step equations. All we do is multiply both sides of the equation by the reciprocal of $\frac{1}{3},$ which is $\frac{3}{1}.$ Here's how it works: $$\begin{align} 3&= \dfrac{2y+5}{3}\\ {\color{#8925AE}{\dfrac{3}{1} \,\times}} \,3&= \dfrac{2y+5}{3}\,\color{#8925AE}{\times\, \dfrac{3}{1}} \end{align}$$

Step Four: Simplify and Solve for the Variable!

Now all we have to do is simplify to isolate our expression containing $y.$ Then we can solve for $y.$ $$\begin{align} \dfrac{3}{1} \times 3&= \dfrac{2y+5}{3}\times \dfrac{3}{1}\\ 9&= \dfrac{2y+5}{\cancel{3}}\times \dfrac{\cancel{3}}{1}\\ 9&=2y+5\\ 9\,{\color{#8925AE}{-\,5}}&=2y+5\,\color{#8925AE}{-\,5}\\ 4&=2y +\cancel{5}-\cancel{5}\\ 4&=2y\\ \dfrac{4}{\color{#8925AE}{2}}&=\dfrac{2y}{\color{#8925AE}{2}}\\ 2&=\dfrac{\cancel{2}y}{\cancel{2}}\\ 2&=y \end{align}$$ So $y=2.$

Alternate Way of Solving

If the previous methods took too long for your taste, there's a quicker way to solve equations with fractions: using cross-multiplication!

Let's use cross-multiplication to solve for $y$ in the example we just did. $$\begin{align} \dfrac{3}{2y+5}+\dfrac{2}{3} &= 1\\ \dfrac{3}{2y+5}&= \dfrac{1}{3}\\ (2y+5)(1)&=(3)(3) \text{ by cross-multiplication}\\ 2y+5&=9\\ 2y&=4\\ y&=2 \end{align}$$ Using this method, we get the same answer: $y=2.$

(Videos) Solving Linear Equations by Clearing Fractions

by mathman1024

Here's a video by mathman1024 showing you how to solve equations that have a lot of fractions. The simplest way to do so is by clearing fractions.

$\large{\text{Summary}}$

The steps he gives are:

1. Identify the LCD of all terms in the equation.
2. Multiply each terms by the LCD and simplify.
3. Solve the resulting equation.

The problem he looks at is: $\frac34x-\frac23=\frac56$. So we have denominators with $4$, $3$, and $6$. The least common denominator would be $12$ since all those numbers can go into that. So we multiply each term by $12$.

$$\begin{align} \frac34x-\frac23&=\frac56\\ {\color{#8925AE}{12\,\times}}\,\frac34x-{\color{#8925AE}{12\,\times}}\,\frac23&={\color{#8925AE}{12\,\times}}\,\frac56\\ 9x-8&=10\\ 9x-8\,{\color{#8925AE}{+\,8}}&=10\,\color{#8925AE}{+\,8}\\ 9x&=18\\ \frac{9x}{\color{#8925AE}{9}}&=\frac{18}{\color{#8925AE}{9}}\\ x&=2 \end{align}$$

(Video) Solving Linear Equations

by mathman1024

Try this next problem by yourself first.

1. $\frac25x+\frac1{15}=\frac34x-\frac{17}{20}$

(Video) Solving Linear Equations

by mathman1024

Finally, there's one more video with one more problem. Try this one on your own too!

1. $\frac78+\frac32x-2=\frac54x-\frac{11}{16}$

There are three key approaches when working with equations with fractions:

1. Isolate $x$ on one side of the equation (whether $x$ is part of the fraction or not).
2. If the only terms in your equation are fractions, you can cross-multiply them.
3. If you need to combine more than one fraction, find a common denominator.

The goal here (other than solving for $x$) is to keep your heart beating at its normal pace!

Equations with Fractions on One Side (Isolating)

Let's start by looking at an example: $$-7 = \frac{x}{3} + 2$$

The numerator in $\frac{x}{3}$ is the thing on top, so $x$!

To solve, we need to isolate any terms with $x$ on one side of the equation.

We can move the $2$ over to the other side of the equation using subtraction: $$\begin{align} -7 &= \frac{x}{3} + 2\\ -7 - 2 &= \frac{x}{3}\\ -9 &= \frac{x}{3} \end{align}$$

And we're almost there! All we need to do now is to get $x$ totally by itself on one side of the equation and solve!

Now we can get $3$ over to the other side of the equation using multiplication, leaving $x$ on its own. $$\begin{align} -9 &= \frac{x}{3}\\ -9 (3) &= x \\ -27 &= x \end{align}$$

Done! By isolating and then eliminating the fraction by multiplying both sides of the equation by the denominator, we have solved for $x$!

Equations with Fractions on Both Sides (Cross-Multiplying)

Now let's look an equation with more than one fraction involving $x$. The fractions in this equation also have different denominators, so that's something we'll need to be careful handling. $$\frac{x}{4} = 6 \left(\frac{x-2}{2}\right)$$

The first thing we want to do here is to simplify the equation as much as we can. To do this, carry out the multiplication on the right side of the equation. $$\frac{x}{4} = \frac{6x-12}{2}$$

(It might help to think about the $6$ on the outside of the parentheses as $\frac{6}{1}$. This helps show that the top of the fraction is multiplied by $6$ and the bottom stays the same because it's multiplied by $1$.)

When you've got fractions on both sides of an equation without any other non-fraction terms, you can cross-multiply to solve for $x$.

We've got that here. Recall that cross-multiplying means that we multiply terms on the diagonal, so that the top and bottom of the first fraction are multiplied by the bottom of the second, and the top and bottom of the second are multiplied by the bottom of the first.

$$\begin{align} \frac{x}{4} &= \frac{6x-12}{2}\\ \frac{2}{2} \left(\frac{x}{4}\right) &= \frac{4}{4} \left(\frac{6x-12}{2}\right)\\ \frac{2x}{8} &= \frac{24x - 48}{8}\\ \frac{2x}{8} &= \frac{24x - 48}{8} \end{align}$$

And now, since the denominators are the same on both sides, we can just get rid of them and get: $$2x = 24x - 48$$

Almost done! All that's left is to isolate $x$ (just like you would when solving other equations).

$$\begin{align} 2x &= 24x - 48\\ 2x - 24x &= - 48\\ -22x &= - 48\\ x &= \frac{- 48}{-22}\\ x &= \frac{24}{11}\\ \end{align}$$

(Video) Solving a Linear Equation

by PatrickJMT

How do you handle fractions in two step equations? Get rid of them! In this video, Patrick JMT uses a couple strategies to solve a linear equation with fractions.

$\large{\text{Summary}}$

In this video, he looks at the problem: $$5-\frac x3=\frac{4(x+1)}{6}$$

First, he makes sure that all his fractions are in the simplest form, so that all the denominators are as small as they can be. On the right hand side, the $4$ and $6$ can be simplified. $$\begin{align} 5-\frac x3&=\frac{4(x+1)}{6}\\ 5-\frac x3&=\frac{2(x+1)}{3} \end{align}$$

Then, he finds the lowest common denominator, in this example it's $3$, and multiplies the whole equation by it to get rid of all of the fractions. $$\begin{align} 5-\frac x3&=\frac{2(x+1)}{3}\\ {\color{#8925AE}{3\,\times}}\,\left(5-\frac x3\right)&={\color{#8925AE}{3\,\times}}\,\frac{2(x+1)}{3}\\ 15-x&=2(x+1) \end{align}$$

Then, all you have to do is solve the two-step equation how you would normally! $$\begin{align} 15-x&=2(x+1)\\ 15-x&=2x+2\\ 15-x\,{\color{#8925AE}{+\,x}}&=2x+2\,{\color{#8925AE}{+\,x}}\\ 15&=3x+2\\ 15\,{\color{#8925AE}{-\,2}}&=3x+2\,{\color{#8925AE}{-\,2}}\\ 13&=3x\\ \frac{13}{{\color{#8925AE}{3}}}&=\frac{3x}{{\color{#8925AE}{3}}}\\ \boxed{\frac{13}{3}} &= x \end{align}$$

Fractions and Equations

Solving an equation with a fraction isn't much different than solving an equation full of whole numbers. You group the variable terms on one side, the constants on the other, and then simplify.

Example $1$:

Find $x$ in $\frac{3}{2}x + \frac{1}{5} = \frac{3}{4}$.

First, group the variable terms on one side and the constants on the other: $$\begin{align} \frac{3}{2}x +\frac{1}{5} &=\frac{3}{4}\\ \frac{3}{2}x+\frac{1}{5}\,\color{#8925AE}{ - \,\frac{1}{5}} &=\frac{3}{4}\,\color{#8925AE}{ -\, \frac{1}{5}}\\ \end{align}$$

Next, simplify: $$\begin{align} \frac{3}{2}x &= \frac{15}{20} - \frac{4}{20}\\ \frac{3}{2}x &= \frac{11}{20}\\ {\color{#8925AE}{\frac{2}{3}\, \times}}\, \frac{3}{2}x &= \frac{11}{20} \,\color{#8925AE}{\times\,\frac{2}{3}}\\ x &= \frac{11}{30} \end{align}$$

Example $2$:

Find $x$ in $5-\frac{x}{3}= \frac{4(x+1)}{6}$.

First, reduce the $\frac{4}{6}$ on the right side of the equation and put $5$ into fraction form on the left side of the equation. $$\begin{align} 5-\frac{x}{3}&= \frac{4(x+1)}{6}\\ \frac{5}{1}-\frac{x}{3}&= \frac{2(x+1)}{3} \end{align}$$

Now, find the LCD for $\frac{5}{1}$ and $\frac{x}{3}$. $$\begin{align} \frac{5}{1}-\frac{x}{3}&= \frac{2(x+1)}{3}\\ {\color{#8925AE}{\frac{3}{3}} \times} \left(\frac{5}{1}\right)-\frac{x}{3}&= \frac{2(x+1)}{3}\\ \frac{15}{3}-\frac{x}{3}&= \frac{2(x+1)}{3}\\ \frac{15-x}{3}&= \frac{2(x+1)}{3}\\ \frac{15-x}{3}&= \frac{2(x+1)}{3} \end{align}$$

Now, we can multiply both sides by $3$. $$\begin{align} \frac{15-x}{3}&= \frac{2(x+1)}{3}\\ {\color{#8925AE}{3}\,\times}\,\frac{15-x}{3}&= {\color{#8925AE}{3}\,\times}\,\frac{2(x+1)}{3}\\ 15-x&= 2(x+1) \end{align}$$

Distribute the $2$ into $(x+1)$. $$\begin{align} 15-x&= 2(x+1)\\ 15-x&= {\color{#8925AE}2}x+{\color{#8925AE}2} \end{align}$$

Add $x$ to both sides. $$\begin{align} 15-x&= 2x+2\\ 15-x\,{\color{#8925AE}{+\,x}}&=2x+2\,{\color{#8925AE}{+\,x}}\\ 15&=3x+2 \end{align}$$

Subtract $2$ and divide by $3$. $$\begin{align} 15\,{\color{#8925AE}{-\,2}}&=3x+2\,{\color{#8925AE}{-\,2}}\\ 13&=3x\\ \frac{13}{\color{#8925AE}{3}}&=\frac{3x}{\color{#8925AE}{3}}\\ \frac{13}{3}&=x \end{align}$$

Now, try one on your own!