1/2x+1/3x+1/4x+1/5x=77
Hint: The LCM of 2, 3, 4, 5 is 60.
Oh
So how I'd solve it
In this case, it would be much easier to simply multiply each term by 60, however, even if you didn't know what the LCD was, you could still do it by combining two fractions at a time.
LCD?
Least Common Denominator
But anyway, as I was saying, even if you didn't know what the LCD was, you could still simplify the left side by combining two fractions at a time.
Explaining @Hero 's point, in this case you can simplify by adding 1/2x and 1/4x to build 3/4x. (1/2x = 2/4x)
Actually, @tytenn4a03, I was going to attempt to help @3maahser discover that for herself.
Hi Ema U have to bring all the fractions to common denominator
Therefor each fraction has to represented in the same NEW denominator which will be their multiple
Then u add them all to get a SINGLE fraction. It will be the coefficient of x
Then solve for x by dividing both sides by this coefficient
Oh sorry @Hero :P
So how exactly do I solve this. Like I multiply the bottom numbers by 0? Or what?
Do you know how to add \[\frac{1}{2} + \frac{1}{4}\]
Yes it's 3/4
U multiply EACH coefficient by a special form of 1, that is\[\frac{ 1 }{ 2 }\times \frac{30}{30}\]
and \[\frac{ 1 }{ 3 } by \frac{ 20 }{ 20 }\]
Okay, well \[\frac{1}{2}x + \frac{1}{4}x\] is the same way, except you just add the x at the end of it.
Coeffecient?
This is called "changing the denominator" - in this case to Denominator = 60
So now you have \(\frac{3}{4}x\). Now you can add: \[\frac{3}{4}x + \frac{1}{3}x\]
1/4 of 60 is...
15
I'm trying to show @3maahser how to add fractions without LCDs because she is not familiar with them. I think she should learn how to solve these using methods she is already familiar with
I'm convinced that she knows at least how to add two fractions together. She can probably solve this problem by continuing with using the methods she is already familiar with. @Mikael, you might be able to show her how to do this problem, but it is an almost certainty that she will struggle with the next problem
Figured it out. :)
I hope you used my method
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