to view answer.A rational expression is the quotient of two polynomials (see simplifying rational expressions for more explanation and examples). It is common in mathematical situations that we will have to perform the operations of addition, subtraction, multiplication, and division on rational expressions. We use the same techniques for rational expressions as for numerical fractions.
Adding and subtracting rational expressions is like adding and subtracting fractions. In order to be able to add or subtract rational expressions, we need them to have common denominators.
Find the sum of the two rational expressions. Simplify your result as much as possible:
$${\Large\frac{{x + 3}}{{x - 2}} }+ {\Large\frac{{x - 8}}{{x + 6}}}$$
1. First, we need to have common denominators. To do this, we need to find the least common multiple of the polynomials in the denominators. In many cases with polynomials, the least common denominator is simply the product of the two polynomials. This just means that we need to multiply the left rational expression by $${\Large\frac{{x + 6}}{{x + 6}}}$$ (which is just the number 1 in a different form, so it does not change the value of the expression), and the right rational expression by $${\Large\frac{{x - 2}}{{x - 2}}}$$. This way, both denominators will be: $$\left( {x - 2} \right)\left( {x + 6} \right)$$.
2. Now let’s perform these multiplications on the two terms in the expression:
$${\Large\frac{{x + 6}}{{x + 6}}} \cdot{\Large \frac{{x + 3}}{{x - 2}}} + {\Large\frac{{x - 8}}{{x + 6}}} \cdot{\Large \frac{{x - 2}}{{x - 2}}}$$
which multiplies to be:
$$ = {\Large\frac{{\left( {x + 6} \right)\left( {x + 3} \right)}}{{\left( {x + 6} \right)\left( {x - 2} \right)}}} + {\Large\frac{{\left( {x - 8} \right)\left( {x - 2} \right)}}{{\left( {x + 6} \right)\left( {x - 2} \right)}}}$$
3. Now, the two additive terms have common denominators, so we can add across the top and keep this denominator (to see more explanation why, refer to adding and subtracting fractions):
$${\Large\frac{{\left( {x + 6} \right)\left( {x + 3} \right) + \left( {x - 8} \right)\left( {x - 2}\right)}}{{\left( {x + 6} \right)\left( {x - 2} \right)}}}$$
4. Finally, we would like to simplify this expression by multiplying out the numerator and combining like terms:
$${\Large\frac{{x^2 + 9x + 18 + x^2 - 10x + 16}}{{\left( {x + 6} \right)\left( {x - 2} \right)}}}$$
By combining like-terms, we get:
$${\Large\frac{{2x^2 - x + 34 \;}}{{\left( {x + 6} \right)\left( {x - 2} \right)}}}$$
5. At this point, we should try to simplify the expression. If we could factor the top into two binomials, one of which would be either $$
\left( {x + 6} \right)$$ or $$\left( {x - 2} \right)$$, we could write a simpler expression. However, there’s no way that the numerator can factor with either of those two factors. So, there is nothing else to factor here. For a complete answer, then, our sum of rational expressions is:
$${\Large\frac{{x + 3}}{{x - 2}} }+{\Large \frac{{x - 8}}{{x + 6}}} = {\Large\frac{{2x^2 - x + 34}}{{\left( {x + 6} \right)\left( {x - 2} \right)}}};\;\;x \ne - 6,\;\;x \ne 2$$
Find the difference of the two rational expressions. Simplify your result as much as possible:
$${\Large\frac{{2 - x}}{{x^2 + 2x - 3}}} - {\Large\frac{{4 - x}}{{x - 1}}}$$
1. Just as with addition, when we subtract fractions, we need the denominators to be the same. It will always work the way we did this above, by multiplying the first fraction by the denominator of the second fraction over itself (thus, by multiplying the first fraction by one). So, for example, in this problem, we could multiply the first rational expression by $${\Large\frac{{x - 1}}{{x - 1}}}$$ and multiply the second rational expression by $${\Large\frac{{x^2 + 2x - 3 \;}}{{x^2 + 2x - 3 \;}}}$$. However, we can make this problem a little less messy by first factoring the denominators.
2. When we factor the quadratic in the denominator of the first term, we get:
$$x^2 + 2x - 3 = \left( {x + 3} \right)\left( {x - 1} \right)$$
So, we can rewrite the problem as:
$${\Large\frac{{2 - x}}{{\left( {x + 3} \right)\left( {x - 1} \right)}}} - {\Large\frac{{4 - x}}{{x - 1}}}$$
3. Now, we notice that the denominator of the second term is a factor of the denominator of the first term. Thus, instead of multiplying each of the rational expressions, we only need to multiply the second one by $$\frac{{x + 3}}{{x + 3}}$$ and we will have common denominators:
$${\Large\frac{{2 - x}}{{\left( {x + 3} \right)\left( {x - 1} \right)}}} - {\Large\frac{{4 - x}}{{x - 1}}} ={\Large \frac{{2 - x}}{{\left( {x + 3} \right)\left( {x - 1} \right)}}} -{\Large \frac{{4 - x}}{{x - 1}}} \cdot {\Large\frac{{x + 3}}{{x + 3}}}$$
Now our problem is:
$${\Large\frac{{2 - x}}{{\left( {x + 3} \right)\left( {x - 1} \right)}}} - {\Large\frac{{\left( {4 - x} \right)\left( {x + 3}\right)}}{{\left( {x - 1} \right)\left( {x + 3} \right)}}}$$
4. Since we have common denominators, we can now subtract the terms on the top, to get:
$${\Large\frac{{\left( {2 - x} \right) - \left( {4 - x} \right)\left( {x + 3} \right)}}{{\left( {x + 3} \right)\left( {x - 1} \right)}}}$$
5. We can further simplify the top, by using the distributive property:
$${\Large\frac{{\left( {2 - x} \right) - \left( {4x + 12 - x^2 - 3x} \right)}}{{\left( {x + 3} \right)\left( {x - 1} \right)}}}$$
Then, distributing the negative:
$${\Large\frac{{\left( {2 - x} \right) - 4x - 12 + x^2 + 3x \;}}{{\left( {x + 3} \right)\left( {x - 1} \right)}}}$$
Finally, combining like terms on the top, we get:
$${\Large\frac{{2 - x - 4x - 12 + x^2 + 3x \;}}{{\left( {x + 3} \right)\left( {x - 1} \right)}} }={\Large \frac{{x^2 - 2x - 10\;}}{{\left( {x + 3} \right)\left( {x - 1} \right)}}}$$
There is no way that we are going to be able to reduce this further, since the numerator does not have either $$\left( {x + 3} \right)$$ or $$\left( {x - 1} \right)$$ as a factor. (To check this, try to find integers that add to –2 and multiply to be –10 – there are not any).
So, the final form of the answer is:
$${\Large\frac{{2 - x}}{{x^2 + 2x - 3}}} -{\Large \frac{{4 - x}}{{x - 1}}} = {\Large\frac{{x^2 - 2x - 10 \;}}
{{\left( {x + 3} \right)\left( {x - 1} \right)}}};\;\;x \ne - 3,\;\;x \ne 1$$
Multiply the following rational expressions. Simplify your result as much as possible:
$${\Large\frac{{x^2 + 2}}{{\left( {x - 1} \right)^2 }}} \cdot {\Large\frac{{x - 1}}{x}}$$
1. Just with multiplying any other fractions, to multiply rational expressions, we just multiply the numerators and multiply the denominators:
$${\Large\frac{{x^2 + 2}}{{\left( {x - 1} \right)^2 }}} \cdot {\Large\frac{{x - 1}}{x}} = {\Large\frac{{\left( {x^2 + 2} \right)\left( {x - 1} \right)}}{{x\left( {x - 1} \right)^2 }}}$$
2. At this point, do not use the distributive property to multiply out the top and bottom. It is better to leave them in factored form, to see if we can divide out any factors that appear in both the numerator and denominator. In this case, we see that the numerator and the denominator both have $$\left( {x - 1} \right)$$ multiplied in. So, we can divide one factor of this from both the denominator and the numerator:
$${\Large\frac{{\left( {x^2 + 2} \right)\left( {x - 1} \right)}}{{x\left( {x - 1} \right)^2 }} }={\Large \frac{{\left( {x^2 + 2} \right)}}{{x\left( {x - 1} \right)}}} \cdot {\Large\frac{{\left( {x - 1}\right)}}{{\left( {x - 1} \right)}}}$$
Because $${\Large\frac{{x - 1}}{{x - 1}}} = 1$$, we can write the fraction in simplified form.
Therefore, our simplified and final answer is:
$${\Large\frac{{x^2 + 2}}{{x\left( {x - 1} \right)}}};\;\;x \ne 0,\;\;x \ne 1$$
Divide the following rational expressions. Simplify your result as much as possible:
$${\Large\frac{{x - 2}}{{x + 3}}} \div {\Large\frac{{x - 2}}{5}}$$
1. Remember that, when dividing fractions, we multiply the reciprocal of the second fraction. So, this division problem is rewritten as a multiplication problem:
$${\Large\frac{{x - 2}}{{x + 3}}} \cdot {\Large\frac{5}{{x - 2}}}$$
2. Now, multiplying the numerators and denominators, we get:
$${\Large\frac{{5\left( {x - 2} \right)}}{{\left( {x + 3} \right)\left( {x - 2} \right)}}}$$
3. Now, before we divide out any factors that are in the numerator and denominator, note that this is our answer (it is just not simplified yet). Thus, whichever values of x cannot be entered into this expression should also not be entered into any simplified version. Thus, we note here that $$x \ne - 3,\;\;x \ne 2$$.
4. Since the factor $$\left( {x - 2} \right)$$ in both the numerator and denominator divides out to 1:
$${\Large\frac{{5\left( {x - 2} \right)}}{{\left( {x + 3} \right)\left( {x - 2} \right)}}} ={\Large\frac{5}{{x + 3}}} \cdot{\Large\frac{{x - 2}}{{x - 2}} }= {\Large\frac{5}{{x + 3}}} \cdot 1$$
And thus, our final answer is:
$${\Large\frac{5}{{x + 3}}};\;\;x \ne - 3,\;\;x \ne 2$$
Some practice problems to check your skills:
Find the sum, difference, product, or quotient of the following rational expressions. Simplify your result as much as possible:
1. $${\Large\frac{x}{{x - 2}}} +{\Large \frac{{5 - x}}{{x - 7}}}$$
Think about it for a moment and then access this link to view answer.
2. $${\Large\frac{{3x}}{{x - 1}} }+ {\Large\frac{1}{x}}$$
Think about it for a moment and then access this link to view answer.
3. $${\Large\frac{{2x^2 - 1}}{{x - 7}}} - {\Large\frac{{x + 7}}{3}}$$
Think about it for a moment and then access this link to view answer.
4. $${\Large\frac{{x^2 - 2}}{{2\left( {2x + 7} \right)^3 }}} \cdot{\Large \frac{{\left( {2x + 7} \right)^3 }}{{\left( {x - 2} \right)^2 }}}$$
Think about it for a moment and then access this link to view answer.
5. $${\Large\frac{{x^2 + 7x + 12}}{{x - 3}}} \div {\Large\frac{{x + 3}}{{x - 3}}}$$
Think about it for a moment and then access this link to view answer.