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How do you combine \[\dfrac{{6b}}{{b - 4}} - \dfrac{1}{{b + 1}}\] ?

Answer
VerifiedVerified
530.4k+ views
Hint: To solve this question there is a simple way that is taking the LCM. Combining it means forming a single fraction instead of the two. So taking LCM will help in combining them. After finding the LCM we will try to simplify the equation till we can.

Complete step by step solution:
Given that,
 \[\dfrac{{6b}}{{b - 4}} - \dfrac{1}{{b + 1}}\]
These are two separate fractions. We will find their LCM first simply by cross multiplying,
 \[ = \dfrac{{6b\left( {b + 1} \right) - 1\left( {b - 4} \right)}}{{\left( {b - 4} \right)\left( {b + 1} \right)}}\]
Now multiply the terms outside the brackets with the terms inside. In denominator we split the terms as shown below.
 \[ = \dfrac{{6{b^2} + 6b - b + 4}}{{b\left( {b + 1} \right) - 4\left( {b + 1} \right)}}\]
Now repeat above step for denominator,
 \[ = \dfrac{{6{b^2} + 6b - b + 4}}{{{b^2} + b - 4b - 4}}\]
Now perform the operation on the terms with same coefficient, like here the terms with coefficient b
 \[ = \dfrac{{6{b^2} + 5b + 4}}{{{b^2} - 3b - 4}}\]
This is the simplified form.
Now if we can find the roots or factors, we say then it can be solved further. But since the roots so found are also not giving simplified output so we will consider this as the final answer.
So, the correct answer is “ \[ \dfrac{{6{b^2} + 5b + 4}}{{{b^2} - 3b - 4}}\] ”.

Note: Note that combining is not like directly adding or subtracting the terms here in this case because the denominators of the terms or fractions are different.
 \[\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{{a + c}}{{b + d}} \times \] this is not correct way. So we need to take the LCM.
Rather if the terms had same denominator we can proceed very simply like,
 \[\dfrac{a}{b} + \dfrac{c}{b} = \dfrac{{a + c}}{b}\]
We can further simplify them if there is any hope with the factors or roots.
Roots of numerator are: \[b = \dfrac{{ - 5 \pm \sqrt { - 71} }}{{12}}\]
Roots of denominator are: \[b = 4, - 1\]
So we cannot solve this further.
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