So I have this equation:
$$\frac{2}{3}a^2-\frac{4}{9}a^2 = 8a$$
So this is a really easy problem, I could just multiply
$$\frac{2}{3}*\frac{3}{3} = \frac{6}{9}$$
Then subtract
$$\frac{6}{9}a^2 - \frac{4}{9}a^2 = \frac{2}{9}a^2=8a$$ $$36a=a^2$$ $$36=a$$
However, I want to solve the equation by getting rid of the fractions right at the beginning:
$$\frac{2}{3}a^2-\frac{4}{9}a^2 = 8a$$
So I thought it'd be much simpler if I could get rid of these fractions by multiplying everything by a single value. Therefore, I thought what value can I multiply 2 and 4 so it gives me a divisible value by 3 and 9? It took me some time but I came up with 9
$$\frac{2}{3}a^2-\frac{4}{9}a^2 = 8a$$ $$9(\frac{2}{3}a^2-\frac{4}{9}a^2) = 9(8a)$$ $$6a^2-9a^2 = 8a$$
My question is if there is an easier way to find this value that when multiplied it eliminates the fractions? It took a few valuable seconds to figure out it was 9, so I was wondering if this process has a name or any rules to find it quicker?
$\endgroup$ 24 Answers
$\begingroup$Method 1: $\frac ab x + \frac cd x = gy$
Multiple by both $b$ and $d$. Multiplying by $bd$ will always eleminate everything. so $bd\frac abx + bd \frac cd x = adx + bc x = bdgy$.
Ex: $\frac 23 a^2 - \frac 49 a^2 = 8a$
$3*9\frac 23 a^2 -3*9\frac 49 a^2 = 3*9*8a$
$18a^2 - 12a^2 = 216 a$
But notice that was over kill. Every term is a multiple of 3.
Method 2: Do it one term at a time:
$\frac ab x + \frac cd x = gy$
$ ax + \frac {bc}dx = by$
$ dax + bcx = bdy$.
Ex:
$\frac 23 a^2 - \frac 49 a^2 = 8a$
$3\frac 23 a^2 - 3\frac 49a^2 = 2a^2 - \frac 43 a^2 = 3*8a = 24a$
$3*2a^2 - 3\frac 43 a^2 = 6a^2 - 4a^2 = 3*24a = 72a$
Notice in this case you avoided the overkill by factoring out the excessive $3$ as it appeared. Had you started with $9$ it wouldn't have come up at all.
$\frac 23 a^2 - \frac 49 a^2 = 8a$
$9*\frac 23 a^2 - 9*\frac 49 a^2 = 9*8a$
$3*2a^2 - 4a^2 = 72a$.
Method 3: The conclusion we can reach. Multiply by the least common multiple.
$\frac ab x + \frac cd x = gy$
$\text{lcm}(bd)\frac ab x + \text{lcm}(bd) \frac cd = \text{lcm}(bd)gy$
$d'a x + b'c x = \text{lcm}(bd) gy$.
Example: $\frac 23 a^2 - \frac 49 a^2 = 8a$. $3=3$ and $9 = 3^2$ so $\text{lcm}(3,9) = 9$.
$9 \frac 23 a^2 + 9\frac 49 a^2 = 9*8a$
$6a^2 - 4a^2 = 9*8a$.
But notice: You not only should get rid of the denominaters you should also get rid of the common factor two.
Final $\frac ab x + \frac cd x = \frac gh y$. Multiply both sides by $\frac {\text{lcm}(b,d,h)}{\gcd(a,c,g)}$
So for $\frac 23 a^2 - \frac 49 a^2 = 8a$. $\text{lcm}(3,9) = 3$ and $\gcd(2,4,8) = 2$.
So multiply everything by $\frac 92$.
$\frac 92\frac 23 a^2 - \frac 92\frac 29 a^2 = \frac 928a$
$3a^2 - 2a^2 = 36a$.
$\endgroup$ $\begingroup$$$9\left(\frac{2}{3}a^2-\frac{4}{9}a^2\right)=8a\cdot9$$ it's $$6a^2-4a^2=72a$$ and it's not $6a^2-9a^2=8a$, which you wrote.
The continue is $$a^2-36a=0$$ or $$a(a-18)=0,$$ which gives $a=0$ or $a=18$.
$\endgroup$ 0 $\begingroup$Just take the least common multiple of all your denominators.
$\endgroup$ $\begingroup$You're simply finding a common denominator to work with. If you have an expression $\frac{n_1}{d_1}+\frac{n_2}{d_2}+...+\frac{n_k}{d_k}$, then you want the denominator to be $\text{lcm}(d_1,d_2,...,d_k)$.
This now gives for your expression
$$\frac{\frac{n_1}{d_1}\text{lcm}(d_1,d_2,...,d_k)+\frac{n_2}{d_2}\text{lcm}(d_1,d_2,...,d_k)+...+\frac{n_k}{d_k}\text{lcm}(d_1,d_2,...,d_k)}{\text{lcm}(d_1,d_2,...,d_k)}$$
The terms in the numerator are all integers which is what you're after.
In your case, the lcm is $9$, so you have
$$\frac{2}{3}a^2-\frac{4}{9}a^2 = 8a$$
$\implies$
$$\frac{1}{9}\left(9\times\frac{2}{3}a^2-9\times\frac{4}{9}a^2\right) = 8a$$
$\implies$
$$\left(6a^2-4a^2\right) = 9\times8a$$
(Also, as pointed out, you have a typo on the bottom line)
$\endgroup$
