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# Equation with two rational expressions (old example 2)

Sal solves the equation 5/(2x)-4/(3x)=7/18 by first finding the LCM (least common multiple) of 2x and 3x. Created by Sal Khan and Monterey Institute for Technology and Education.

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• How can I find out what x cannot be? •  It's a mathematical rule that you can't have a 0 in the denominator because any number divided by 0 is undefined.
In general that's the logic you have to follow, " is there any number in this equation that would make it undefined (and thus impossible to solve) ?
"if yes, what would x had to be to get that number?"
And that is your "X that can't be"

Example given in the previous video:
5/(x-1)= 3/(x+3)

You know that you can't have 0 in the denominator.
In the first fraction you have the denominator x-1. What would x had to be for the denominator be 0?

it had to be 1; 1-1=0. So X can't be 1.

By the same logic:

x+3
what number would make x+3=0.
-3 so -3+3=0

You can also rule out -3.

For this example you already know that X cannot be 1 or -3.
• At the example is dividing x by x and 18 by 2.
I thought we were supposed to multiply the 18x by the denominator as we do in the
"Solving Rational Equations 1" example (can be seen at onwards)? • He's basically doing just that. When you are multiplying 18x by 5 over 2x, the x can be "cancelled out", leaving a 2, and that 2 can also be cancelled out as well. Basically, cancelling out is just division, so he is using those interchangeably.

In other words, what you are supposed to do is multiply every term in both sides of the equation by the LCD (lowest common denominator) of the fractions, and that will always cancel out the denominators. (When you set it up as a fraction multiplication, you might notice that you'll be able to cancel factors out diagonally to simplify. That's where the cancelling out happens. )

It isn't specifically multiplying by the denominators, but it's actually dividing into the denominators, putting a one in the denominator afterwards, and taking the result and multiplying that by the numerator (You can think of it that way too). This always works because you are taking the LCD of all of the fractions in the rational equation, and the whole point in using the LCD is that it has factors that will divide into the factors of each of the denominators evenly.

Just don't forget to multiply every term in the equation, on both sides, by this LCD. That includes non-fractions on either side of the equation. That's why it's most practical to do this when your equation is mostly fractions, although you can do whatever you'd like to solve your equations.
• what happens if there is a variable in the numerator?
ex. x-3/(3x+2)=1/5 • At , 18 is divided by the denominator (2) and then the quotient (9) is multiplied by the numerator.
I'm pretty fuzzy on the logic behind this. If anyone can explain why this works I think I'd retain this a lot better than I am now. • The key for this problem (and many problems involving fractions) is to ELIMINATE the fractions:) If you can find a number into which all the denominators are divisible, all you need to do is multiply that number by each term, and then when you simply the fractions the denominators will cancel out.
The specific issue you seem to have is the order in which Sal is simplifying. It would be just as correct to multiply the 18x by 5 (the numerator) to get 90x, and then divide that by 2x (the denominator) to get 45. See? Same result. Sal did it in smaller steps to avoid having to multiply 18x by 5, which some people can't do in their heads. Remember the 18x is really 18x/1 so you are just multiplying fractions; Sal chose to do some cancellation/simplification before he multiplied. Did this make sense or did I make it worse?
• I want to know the answer and know how to check it. Here's my problem: 3/3x+4=2/5x-6 • Since you are adding fractions, the first thing you should do is convert everything to a common denominator. You didn't use parentheses so I can't tell if you mean
'3/(3x+4) = 2/(5x-6)' or '3/(3x) + 4 = 2/(5x) - 6'.
My guess is you meant '3/(3x+4) = 2/(5x-6)'. So multiple the left side by (5x-6)/(5x-6), and the right side by (3x+4)/(3x+4). You're really just multiplying both sides by 1, so this is perfectly valid. That gives you a numerator of 3*(5x-6) on the left side and 2*(3x+4) on the right side. The denominator on both sides is (3x+4)(5x-6). Now you only have to compare the numerators to solve for x.
3*(5x-6) = 2*(3x+4) ... 15x - 18 = 6x + 8 ... 9x = 26 ... x = 26/9. To check if this is correct, substitute the solution back into the original expression and verify that both sides are equal. Doing that I get 9/38 = 9/38.
• How do you solve a quadratic equation with multiple different denominators? • I was lost after the first 1 minute and 12 seconds. I'm in college remedial math, and I still don't understand equations. • Its like saying 4=4. We know that is true... but there is a bunch of ways ways we can write that same thing....
ex. 2x2=4, 2+2=4,6-2=4 etc..... correct?
now, if I say, some number 'x' + 2 gives me 3, then we know that 'x' must be 1..... correct?
So writing that in a bit more mathematical fashion we can say
x + 2 = 3.
now if you go back to the 4=4 equation, if i subtract 1 from the right side, we would get 4=3... that's not true. but if i subtracted 1 from both sides, i get 3=3 which is true. so in general the equality is maintained as long as we do the same operation on both sides of the equation. now back to x + 2 = 3... subtract both sides by 2 you get x = 1..... its the same with division and multiplication.... if you have doubts mail me or add me as your coach... my email id is informjaka@gmail.com
• how do I solve the rational equation 3x=12+15/x   