মূলদ রাশির যোগ ও বিয়োগ

$$\textbf{Adding and Subtracting Rational Expressions}$$

Let's do another example, this time with subtracting rational expressions.

Step One

First we convert the two rational expressions to equivalent ones with a common denominator. $$\begin{align} \dfrac{x+1}{3}-\dfrac{14x+5}{9x}&=\bigg(\dfrac{9x}{9x}\bigg) \cdot \dfrac{x+1}{3}-\dfrac{14x+5}{9x} \cdot \bigg(\dfrac{3}{3}\bigg)\\ \\ &=\dfrac{(9x)(x+1)}{3(9x)}-\dfrac{3(14x+5)}{3(9x)} \\ \end{align}$$

Step Two

Next, we combine these new rational expressions. $$\begin{align} \dfrac{(9x)(x+1)}{3(9x)}-\dfrac{3(14x+5)}{3(9x)}&=\dfrac{(9x)(x+1)- 3(14x+5)}{3(9x)}\\ \end{align}$$

Step Three

Lastly, we distribute and combine like terms until the expression is simplified. $$\begin{align} \dfrac{(9x)(x+1)- 3(14x+5)}{3(9x)}&=\dfrac{(9x^2+9x)- (42x+15)}{3(9x)}\\ \\ &=\dfrac{9x^2+9x -42x-15}{3(9x)}\\ \\ &=\dfrac{9x^2-33x-15}{3(9x)}\\ \\ &=\dfrac{3(3x^2-11x-5)}{3(9x)}\\ \\ &=\dfrac{\cancel{3}(3x^2-11x-5)}{\cancel{3}(9x)}\\ \\ &=\dfrac{3x^2-11x-5}{9x}\\ \end{align}$$

Notice that we could have recognized $9x$ as the Least Common Denominator of $3$ and $9x,$ which would have saved us some steps.

(Videos) Rational Expressions Arithmetic

by ThinkwellVids

In this video by ThinkwellVids, you learn how to add and subtract rational expressions.

$\large{\text{Summary}}$

Adding and subtracting rational expressions can be a pain. The reason is, you need to make sure they have common denominators before you can add or subtract.

Let's look at an example and see how we do this. $$\frac{1}{x+3}+\frac1x$$ We can't add these together yet because they have different denominators. To find a common denominator, you have to multiply both the top and bottom by the denominator of the other fraction.

So, since $\frac{1}{x+3}$ has a denominator of $x+3$, we need to multiply $\frac1x$ by $\frac{x+3}{x+3}$. $$\frac{1}{x+3}+\frac1x\cdot\color{#8925AE}{\frac{x+3}{x+3}}$$ Similarly, since $\frac1x$ has a denominator of $x$, we need to multiply $\frac{1}{x+3}$ by $\frac{x}x$. $$\color{#8925AE}{\frac xx}\cdot\frac{1}{x+3}+\frac1x\cdot\color{#8925AE}{\frac{x+3}{x+3}}$$

Simplifying each fraction, we get: $$\frac{x}{x(x+3)}+\frac{x+3}{x(x+3)}$$ Now, we have two fractions with the same denominator and we can add. Remember, we only add the numerators. $$=\frac{x+x+3}{x(x+3)}$$ Finally, we simplify as much as possible. $$\boxed{=\frac{2x+3}{x^2+3x}}$$

Interactive Example

In this video by ThinkwellVid, you can walk through an example.

Let's work through this problem together. $$\frac{1}{x+z}+\frac{1}{x-z}$$

They don't! One has addition and the other has subtraction. Thus, our next step is to multiply each fraction by something in order to make their denominators equal.

We want to multiply the top and bottom of this fraction by the denominator of the other fraction. $$\color{#8925AE}{\frac{x-z}{x-z}}\cdot\frac{1}{x+z}+\frac{1}{x-z}$$

We want to multiply the top and bottom of this fraction by the denominator of the other fraction. $$\color{#8925AE}{\frac{x-z}{x-z}}\cdot\frac{1}{x+z}+\frac{1}{x-z}\cdot\color{#8925AE}{\frac{x+z}{x+z}}$$

Multiplying each fraction, we get: $$\frac{x-z}{(x+z)(x-z)}+\frac{x+z}{(x+z)(x-z)}$$

Adding the numerators, we get: $$\frac{x-z+x+z}{(x+z)(x-z)}=\frac{2x}{(x+z)(x-z)}$$

The final step is to multiply the terms in the denominator. $$\boxed{=\frac{2x}{x^2-z^2}}$$

Three Rational Expressions

In this video by ThinkwellVid, the equation has three rational expressions. $$\frac{1}{t^2+t-12}-\frac{1}{t^2-7t+12}+\frac{1}{t^2-16}$$ Clearly none of these have the same denominator. You could multiply each fraction by the denominators of the other two but that would get ugly real fast. Instead, we can take a shortcut, and start by factoring each denominator. $$\frac{1}{(t+4)(t-3)}-\frac{1}{(t-3)(t-4)}+\frac{1}{(t+4)(t-4)}$$ We can now find the Least Common Multiple or LCM. Start by looking at each term in each denominator. Add it to the LCM if it isn't there before.

Starting with the first rational, there's the term $(t+4)$. We don't have that so we add it to our LCM. $$\text{LCM}=(t+4)$$ We also have the term $(t-3)$. This also isn't into the LCM so we place that in there too. $$\text{LCM}=(t+4)(t-3)$$

Moving to the second rational, there's the term $(t-3)$. We already have this in our LCM so we skip it. The next is $(t-4)$. We don't have this so we can place it in. $$\text{LCM}=(t+4)(t-3)(t-4)$$

The final rational has the terms $(t+4)$ and $(t-4)$. Both of these are in the LCM so we can skip them. Thus, our final LCM is: $$\boxed{\text{LCM}=(t+4)(t-3)(t-4)}$$ This is what we want each denominator to be. For example, the first fraction, $\frac{1}{(t+4)(t-3)}$, has $(t+4)$ and $(t-3)$ but needs a $(t-4)$ to equal the LCM. Thus, we multiply the top and bottom by $(t-4)$.

Doing this for all the fractions, we get: $$\color{#8925AE}{\frac{t-4}{t-4}}\cdot\frac{1}{(t+4)(t-3)}-\color{#8925AE}{\frac{t+4}{t+4}}\cdot\frac{1}{(t-3)(t-4)}+\color{#8925AE}{\frac{t-3}{t-3}}\cdot\frac{1}{(t+4)(t-4)}\\ =\frac{t-4}{(t+4)(t-3)(t-4)}-\frac{t+4}{(t+4)(t-3)(t-4)}+\frac{t-3}{(t+4)(t-3)(t-4)}$$ Since they all have the same denominator, we can finally combine them. Remember to only add and subtract the numerators. $$=\frac{t-4-(t+4)+t-3}{(t+4)(t-3)(t-4)}$$ Simplifying, we get: $$\begin{align} &=\frac{t-4-t-4+t-3}{(t+4)(t-3)(t-4)}\\ &\boxed{=\frac{t-11}{(t+4)(t-3)(t-4)}} \end{align}$$

Adding and Subtracting Rational Expressions with Uncommon Denominators

Now that we've covered the basics of simplifying rational expressions, we can move on. Let's learn how to add or subtract rational expressions with uncommon denominators.

We want to add $\frac76+\frac34$. Since they don't have common denominators we can't simply add them. Instead we multiply both fractions by the least common multiple of the denominators.

So in the problem above we have a denominator with $6$ and one with $4$. To figure out the common denominator we break the denominators down into their prime factorizations. $$6=2\cdot3\\4=2\cdot2$$

Now we ask, "Is there a factor missing from the first that the second has?" and "Is there a factor missing from the second that the first has?" Well the $4$ has two 2's and the $6$ only has one. So the $6$ is missing a factor of $2.$ The $4$ has zero factors of $3$ whereas the $6$ has one factor of $3.$ So the $4$ is missing a factor of $3.$

We can then multiply the numerator and denominator of each fraction by the missing factor. So, $$\begin{align} \dfrac76+\dfrac34&=\frac76\color{#8925AE}{\ \cdot\frac22} \ + \ \frac34\color{#8925AE}{\ \cdot \ \frac33}\\ &=\frac{14}{12}+\frac9{12}\\ &=\frac{23}{12} \end{align}$$

As you can see, when we multiply to get the least common denominator, it gives both fractions the same bottom number. This makes it easier to add or subtract them. Just like with regular fractions, we need the bottoms to match before we can combine the tops.

Let's look at an example with rational expressions: $\frac{3}{x^2+7x+10}+\frac{4}{x^2+5x+6}$. So the first step is to factor the polynomials in the denominator. $$\begin{align} &\dfrac{3}{x^2+7x+10}+\dfrac{4}{x^2+5x+6}\\ &=\frac{3}{(x+5)(x+2)}+\frac{4}{(x+2)(x+3)} \end{align}$$ So just like the problem above, we see that the first fraction is missing a factor of $x+3$. The second fraction is missing a factor of $x+5$. $$\begin{align} &\dfrac{3}{x^2+7x+10}+\dfrac{4}{x^2+5x+6}\\ &=\frac{3}{(x+5)(x+2)}+\frac{4}{(x+2)(x+3)}\\ &=\frac{3}{(x+5)(x+2)}\color{#8925AE}{\cdot\frac{x+3}{x+3}}\color{black}+\frac{4}{(x+2)(x+3)}\color{#8925AE}{\cdot\frac{x+5}{x+5}}\\ &=\frac{3(x+3)}{(x+5)(x+3)(x+2)}+\frac{4(x+5)}{(x+5)(x+3)(x+2)}\\ &=\frac{3x+9}{(x+5)(x+3)(x+2)}+\frac{4x+20}{(x+5)(x+3)(x+2)}\\ &=\frac{7x+29}{(x+5)(x+3)(x+2)} \end{align}$$ The result cannot be factored anymore so it is our final answer. Sometimes, after adding or subtracting, you can reduce it further.

Looking Further

Let's continue to build on this idea. In the next lessons, you will learn how to:

• Multiply rational expressions
• Divide rational expressions

(Video) Adding and Subtracting Rational Expressions

by yaymath

Here's a video by yaymath showing you how to add and subtract rational expressions (or functions).

$\large{\text{Summary}}$

To do this the denominators must be the same. In the video, he goes over examples showing you how to get like denominators and then how to add or subtract the terms.

(Videos) Adding and Subtracting Rational Expressions with Common Denominators

by mathman1024

Here's a video by mathman1024 showing how to add and subtract rational expressions with common denominators.

$\large{\text{Summary}}$

To add or subtract rational expressions you must have common denominators. $$\frac{a}c+\frac{b}c=\frac{a+b}c$$

He looks at the example $\frac{x}{x+3}+\frac{7}{x+3}$. $$\dfrac{x}{x+3}+\dfrac{7}{x+3}=\frac{x+7}{x+3}$$

Try these next ones on your own!

1. $\frac{x+5}{x-2}+\frac{3x+1}{x-2}$
2. $\frac{4}{x-7}-\frac{8}{x-7}$
3. $\frac{3x+1}{x^2-4}-\frac{7}{x^2-4}$

In this video, mathman1024 looks at some problems where you have to remove a negative factor from the denominator before you can add or subtract.

The first one he looks at is $\frac{5}{x-6}+\frac{7}{6-x}$. Since these two rational expressions have different denominators we can't just simply add them. But we can factor out a negative one from the denominator of one. $$\begin{align} &\dfrac{5}{x-6}+\dfrac{7}{6-x}\\ &=\dfrac{5}{x-6}+\dfrac{7}{\color{#8925AE}{-x+6}}\\ &=\dfrac{5}{x-6}+\dfrac{7}{\color{#8925AE}{-(x-6)}}\\ &=\dfrac{5}{x-6}\color{#8925AE}{-}\dfrac{7}{x-6}\\ &=\frac{-2}{x-6} \end{align}$$

Try this next one on your own!

1. $\frac{x}{x-5}+\frac{9}{5-x}$