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# Equation with the variable in the denominator

Sal solves the equation 7 - 10/x = 2 + 15/x. Created by Sal Khan.

## Want to join the conversation?

• What if you have a fraction in the equation that's like : 7/x-9 and the other side of the equation has -2/x, how would you work that out?
• You will cross multiply and then solve. To cross multiply, multiply the denominator on the left by the numerator on the right. Then do the same for the denominator on the right and the numerator on the left. These new expressions will be equal to each other. For your example, it will look like:

(x-9)(-2)=(x)(7)
Distribute/multiply to get:
-2x + 18 = 7x
18 = 9x
Divide both sides by 9.
2 = x
• In the video, Sal minused the 2x, but couldn't you also minus the 7x to? I
• Yes, you could have subtracted 7x rather than the 2x. If you prefer to work with positive numbers, then moving the 2x keeps the coefficient of x a positive value.
• Why does multiplying x by 10/x get you 10? Is it because if u multiply you get 10x over x which simplifies into 10?
• Remember that "x" is really "x/1," just like 2 is 2/1.
Let's take a similar problem without variables: 2 &bull; 3/2.
When you multiply 2 (or 2/1) by 3/2, you multiply numerator by numerator, and denominator by denominator. You end up with 6/2. When you reduce (or simplify), you divide both the numerator and the denominator by their GCF (greatest common factor). 6/2 = 3, and 2/2 = 1.
So you're left with 3/1, or 3.

Now look back at your original problem, x &bull; 10/x. When you multiply (remember that x = x/1), you end up with 10x/x. Now we need to simplify. Obviously the only factor between the top and bottom is x, so we divide both the numerator and the denominator by x.
10x/x = 10, and x/x = 1,
so we're left with 10/1, or 10.
• Wait but when you multiply the problem through by x like Sal did at , this is the same thing as (x/1) which is not equivalent to 1--so Sal is changing the equation. How does this work?
• The properties of equality let us change equations into equivalent equations as long as we do the same operation to both side. Sal is multiplying both sides of the equation by "x". So, his version of the equation is equivalent to the original.
Hope this helps.
• Around , Sal says this simplifies to a linear equation. Doesn't a linear equation require that there be two variables?
• A linear equation can have 1 variable. It then creates a horizontal or vertical line.
For example, in the video, Sal gets "x=5". This creates a vertical line. It is interpreted as x=5 for all values of Y. So, the line contains points like: (5, 0); (5,-3); (5,11); etc.
Hope this helps.

``6/x + 3 = 12/2x + 1``

We can try to solve it as usual:

``Let's multiply both sides on x6 + 3x = 12/2 + xThen isolate x (subtract it from the both sides)6 + 2x = 12/2This is equals6 + 2x = 6Subtract 6 from the both sides2x = 0Divide by 2 both sidesx = 0``

But we can't substitute zero instead of `x` in the equation because it's impossible to have zero as denominator.

So should we care about exclusions for `x` when it resides in denominator?
• Yes, you should care about exclusions for X. While your work created x=0, it is not a valid solution because it doesn't make the equation be true (both sides equal). Division by 0 is undefined.
• What is 7Xsomething is7
• Do you mean 7 * x = 7, if so, then x = 1
• how do you solve 6=a/4+2
• subtract 2 from both sides, then multiply result by 4
• so all we really have to do is distribute the x and then go from there?
• Hi, was wondering, why can't we just subtract both fractions by 10/x or add 15/x. Is it an option? Thank you.
• That would have worked. Most math problems have multiple options like that

## Video transcript

So I have the equation 7 minus 10/x is equal to 2 plus 15/x. And so this isn't the type of equation that you might think that you're used to solving. But I'll give you a few moments to see if you can solve it on your own. Well, what we'll see is we can do a quick multiplication of both sides to actually simplify this to a form that we are more used to looking at. So what's probably bothering you, because it's bothering me, is these x's that we have in the denominators right over here. We're like, well, how do we deal with that? Well, whenever we see an x in the denominator, the temptation is to multiply it by x. But we can't just multiply one of the terms by x. We have to multiply the entire side by x. So we could multiply this entire side by x. But we can't just multiply the left-hand side by x. We'd also want to multiply the right-hand side by x. And so what will that give us? Well, we distribute the x. We get x times 7 is 7x. And then x times negative 10/x, well, that's just going to be negative 10. So you get negative 10 right over there. So the left-hand side simplifies to 7x minus 10. And then your right-hand side, once again, distribute the x. x times 2 is 2x. x times 15/x, well, x times something over x is just going to be the something. x times 15/x is just going to be 15-- plus 15. So now we've simplified this to a linear equation. We have the variable on both sides. So we just have to do some of the techniques that we already know. So the first thing that I like to do is maybe get all my x's on the left-hand side. So I want to get rid of this 2x right over here. So I subtract 2x from the right-hand side. Now, and I always remind you, I can't do that just to the right-hand side. If I did it just to the right-hand side, it wouldn't be an equality anymore. You have to do that to the left-hand side as well. And so we are left with-- let me get that pink color again. On the left-hand side, 7x, 7 of something minus 2 of something, well, you're going to have 5 of that something, minus 10. These two x's negate each other. And you're left with equals 15. Now we can get rid of this negative 10 by adding 10 to both sides. You know, I like that green color when I do stuff to both sides. So I can add 10 to both sides. And I'm left with 5x-- these negate each other-- is equal to 25. And this is the home stretch. You see where this is going. We can divide both sides by 5. And we are left with x is equal to 5. Now let's verify that this actually worked. So let's go back to the original equation. We have 7 minus 10/5. This needs to be equal to-- I'm just taking our 5 and substituting it back here. This needs to be equal to 2 plus 15/5. So this is 7 minus 10/5. This is just 2. It needs to be equal to 2 plus 15/5, which is just 3. So 2 plus 3, 7 minus 2 is 5, 2 plus 3 is 5, 5 is indeed equal to 5. And we are done.