Having trouble with one of my math homework problems. I need to find the least common denominator (LCD) to solve the problem. I'm not sure how to figure this out one. Thanks in advance.
$$ \frac{3}{j^2+6j} + \frac{2j}{j+6} – \frac{2}{3j} .$$
[Math] Find the least common denominator of algebraic fractions
algebra-precalculus
Related Solutions
We want to simplify the following:
$$\frac{(x-1)(7x + 6)}{(x-1)(x+1)^2} - \frac {7}{(x+1)}$$
First, we cancel the common factor $(x-1)$ in the left-hand fraction:
$$\frac{(x-1)(7x + 6)}{(x-1)(x+1)^2} - \frac {7}{(x+1)} = \frac{(7x + 6)}{(x+1)^2} - \frac {7}{(x+1)}$$
Now we find a common denominator to add the fractions, and see that we want both denominators to be $(x+1)^2$. To accomplish this, we can multiply the numerator and the denominator of the second fraction by the factor of $(x+1)$:
$$\frac{(7x + 6)}{(x+1)^2} - \frac{7}{(x+1)} \cdot \frac{(x+1)}{(x+1)} = \frac{(7x + 6)}{(x+1)^2} - \frac {7(x+1)}{(x+1)^2}$$ $$ = \frac{(7x + 6 - 7(x+1))}{(x+1)^2} \;=\;\frac{ -1}{(x+1)^2} $$ $$\;=\;\frac{-1}{x^2 + 2x + 1}$$
We may suppose that Fiona, George and Henry each think of $$\frac{1}{n_1a^{a_1}b^{b_1}},\quad \frac{1}{n_2a^{a_2}b^{b_2}},\quad \frac{1}{n_3a^{a_3}b^{b_3}}$$ respectively where $n_1,n_2,n_3$ are constants.
Then, we have $$10ab^2=\text{lcm}(n_1,n_2)a^{\max(a_1,a_2)}b^{\max(b_1,b_2)}$$ $$20a^3b^2=\text{lcm}(n_2,n_3)a^{\max(a_2,a_3)}b^{\max(b_2,b_3)}$$ $$4a^3b=\text{lcm}(n_3,n_1)a^{\max(a_3,a_1)}b^{\max(b_3,b_1)}$$ i.e. $$\text{lcm}(n_1,n_2)=10,\qquad \text{lcm}(n_2,n_3)=20,\qquad \text{lcm}(n_3,n_1)=4$$ $$\max(a_1,a_2)=1,\qquad \max(a_2,a_3)=3,\qquad \max(a_3,a_1)=3$$ $$\max(b_1,b_2)=2,\qquad \max(b_2,b_3)=2,\qquad \max(b_3,b_1)=1$$
(a)
From $\max(b_3,b_1)=1$, we see that $b_3\le 1$ and $b_1\le 1$. From $\max(b_1,b_2)=2$, we get $b_2=2$.
So, George's fraction has the highest power of $b$, and the power is $b^2$.
(b)
From $\text{lcm}(n_3,n_1)=4$, we get $n_3\le 4$ and $n_1\le 4$. It follows that $n_3$ is not divisible by $5$. From $\text{lcm}(n_2,n_3)=20=2^2\cdot 5$, we see that $n_2$ has to be divisible by $5$, so we get $n_2\ge 5$.
So, George's fraction has the largest constant.
(c)
Using
$$\max(A,B,C)=\max(\max(A,B),\max(B,C))\tag1$$
$$\text{lcm}(n_1,n_2,n_3)=\text{lcm}(\text{lcm}(n_1,n_2),\text{lcm}(n_2,n_3))\tag2$$
(The proof is written at the end of this answer)
we get $$\begin{align}&\text{lcm}(n_1,n_2,n_3)a^{\max(a_1,a_2,a_3)}b^{\max(b_1,b_2,b_3)} \\\\&=\text{lcm}(\text{lcm}(n_1,n_2),\text{lcm}(n_2,n_3))a^{\max(\max(a_1,a_2),\max(a_2,a_3))}b^{\max(\max(a_1,a_2),\max(a_2,a_3))} \\\\&=\text{lcm}(10,20)a^{\max(1,3)}b^{\max(2,2)} \\\\&=20a^3b^2\end{align}$$
Proof for $(1)$ :
We may suppose that $A\le C$.
If $A\le C\le B$, then we have $\max(A,B,C)=B$ and $$\max(\max(A,B),\max(B,C))=\max(B,B)=B$$
If $A\le B\le C$, then we have $\max(A,B,C)=C$ and $$\max(\max(A,B),\max(B,C))=\max(B,C)=C$$
If $B\le A\le C$, then we have $\max(A,B,C)=C$ and $$\max(\max(A,B),\max(B,C))=\max(A,C)=C$$
Proof for $(2)$ :
One can consider the prime factorization of $n_1,n_2,n_3$, and use $(1)$ for the exponent of each prime factor.
Best Answer
Hint: First try finding the LCD of some integers. For example, evaluate $$\frac{3}{4} + \frac{7}{10} + \frac{13}{25}$$ and be very conscious of how you're doing it when you get a common denominator. Next try adding these: $$\frac{1}{x^2} + \frac{1}{x}$$ and $$\frac{1}{x+3} + \frac{1}{x^2-9}$$ It's the same thing each time, but you need to use a bit of algebra. Now try your homework problem. Good luck!