Solve the equation. 5 x − 2 3 x = 4 + 3 x \frac{5}{x}-\frac{2}{3x}=4+\frac{3}{x} x 5 − 3 x 2 = 4 + x 3

Hey, everyone. Let's take a look at this practice problem. So here I have 5 over x minus 2 over 3 x is equal to 4 +3x, And I wanna solve this rational equation. So let's go ahead and get started with step 1, which is to determine our restriction. Now my 3 denominators I have are x, 3x, and x. So by simply looking at these, I can see that if I were to plug in x equals 0 for any of them, that would make my denominator 0. So I know that x cannot be 0, and that is my restriction. So step 1 is done. I can go ahead and move on to step 2, which is to multiply by my least common denominator, my LCD, to eliminate fractions. Now, again, my 3 denominators are x, 3x, and x, and they all have the common factor of x. And then I have that extra 3 there, so my least common denominator is going to be 3x. Now remember when we multiply by our least common denominator, we want to make sure to distribute to every single term in our equation. So let's go ahead and expand this out. So I have 3x times 5 over x minus 3x times 2 over 3x. And that's equal to 3x times 4. And then finally, my last term, my least common denominator, 3x times 3 over x. Now it looks a little crazy right now. Let's go ahead and simplify a little bit. So in my very first term, this x is going to cancel, leaving me with 3 times 5, which is just 15. And then I have this 3x canceling here, leaving me with simply minus 2. And that's equal to 3x times 4, which just gives me positive 12 x. And then lastly, this x and x are going to cancel. I'm left with 3 times 3, which is simply a positive 9. So step 2 is done. Let's move on to step 3, solving the linear equation that we're left with. So I don't have anything that needs to get distributed, but I do have a like term right here, 15 and negative 2. So this will simplify to 13. I don't have any other like terms I need to get combined, so I just have 12 x plus 9. Now we wanna move all our x terms to one side, all of our constants to the other. Since I only have one x term and it's already on that right side, I'm going to keep it there and move my 9 over by subtracting it. So it will, of course, cancel there. And I'm left with 13 minuteus 9, which will give me a positive 4, and that's equal to 12 x. My last step is, of course, to isolate my x, which I can do by dividing both sides by 12. Now I can actually simplify that a little bit further, simplify that fraction 4 over 12. This simplifies to 1 third, and that's equal to x. So this is my solution, x is equal to 1 third. We can, of course, write it the other way if that makes you happy. So I have x equals 1 3rd. I've completed step 3. Now step 4 is, of course, to check my solution with my restriction. Now my restriction was that x could not be equal to 0, and x is not equal to 0. It's equal to 1 third, and that is my solution. That's all for this one. I'll see you in the next one.

1. Equations & Inequalities

Rational Equations

College Algebra

1. Equations & Inequalities

Rational Equations

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Multiple Choice

Solve the equation. $\frac{5}{x}-\frac{2}{3x}=4+\frac{3}{x}$

$x=0$

$x=1$

$x=\frac13$

No solution

Verified step by step guidance

Start by simplifying the given equation: $ \frac{5}{x} - \frac{2}{3x} = 4 + \frac{3}{x} $.

To eliminate the fractions, find a common denominator for the terms on the left side. The common denominator for $ x $ and $ 3x $ is $ 3x $.

Rewrite each term with the common denominator: $ \frac{15}{3x} - \frac{2}{3x} = 4 + \frac{3}{x} $.

Combine the fractions on the left side: $ \frac{15 - 2}{3x} = 4 + \frac{3}{x} $, which simplifies to $ \frac{13}{3x} = 4 + \frac{3}{x} $.

To solve for $ x $, multiply every term by $ 3x $ to clear the fractions: $ 13 = 12x + 9 $. Then, solve the resulting linear equation for $ x $.

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