Finding LCD for Binomials (3x+4) and (2x-1)

[markl77]

I was given the binomials
(3x+4) and (2x-1), they are both denominators and I need to make them equal to eachother.
I've gone through many numbers but I still can't find how to make these two equal... Is there a simpler way to do it?

 

[ksdhart2]

I was given the binomials
(3x+4) and (2x-1), they are both denominators and I need to make them equal to eachother.
I've gone through many numbers but I still can't find how to make these two equal... Is there a simpler way to do it?

Erm... sorry, but I don't understand what you're talking about. Please post the full and exact text of the problem. Based on the vague description you've given and the title, my best guess is that you have something like this:

\(\displaystyle \dfrac{\text{something}}{3x+4} + \dfrac{\text{something else}}{2x-1}\)

and you need to find what that equals. If that's the case, one tactic might be to try a few test cases to get a feel for what's going on. Let's say x = 1. What would the two denominators be? What would their LCD be? How might you express that in terms of the original x? Now let's say x = 2. Same questions. What about for x = 3? Are you seeing a pattern? How does that help you solve the problem?

 

[markl77]

Erm... sorry, but I don't understand what you're talking about. Please post the full and exact text of the problem. Based on the vague description you've given and the title, my best guess is that you have something like this:

\(\displaystyle \dfrac{\text{something}}{3x+4} + \dfrac{\text{something else}}{2x-1}\)

and you need to find what that equals. If that's the case, one tactic might be to try a few test cases to get a feel for what's going on. Let's say x = 1. What would the two denominators be? What would their LCD be? How might you express that in terms of the original x? Now let's say x = 2. Same questions. What about for x = 3? Are you seeing a pattern? How does that help you solve the problem?

Sorry! I don't know how you do that kind of text. I just figured out that it was a rational expression ( I was thinking that the = was a -.)
The equation was:
3/(x+1)=5/(3x-1)
So I assume that if I multiply the left fraction by (3x-1) and the right fraction by (x+1) they will both be equal, correct?

 

[ksdhart2]

Sorry! I don't know how you do that kind of text. I just figured out that it was a rational expression ( I was thinking that the = was a -.)
The equation was:
3/(x+1)=5/(3x-1)
So I assume that if I multiply the left fraction by (3x-1) and the right fraction by (x+1) they will both be equal, correct?

The forum has a built in LaTeX parser which makes it easy to display math like that. For instance, if you type:

[tex]\dfrac{x}{5}[/tex]

it will render as \(\displaystyle \dfrac{x}{5}\). If you right click on any LaTeX display here on the forum and select "Show Math As -> TeX Commands," you'll get a pop-up window showing you the code you'd need to type between "tex" brackets.

As for the problem, the strategy you've outlined will not produce the correct answer. After doing the multiplications you suggest, we'd be left with:

\(\displaystyle \dfrac{3(3x-1)}{x+1}=\dfrac{5(x+1)}{3x-1}\)

This doesn't really get us any closer to solving the problem, but even if it did, it doesn't matter because it's an invalid step anyway. The solution to the above equation are:

\(\displaystyle \dfrac{7}{11} \pm \dfrac{2\sqrt{15}}{11}\)

Plugging in the first of these two values back into the given equation leaves us with:

\(\displaystyle \dfrac{3}{\dfrac{18}{11} + \dfrac{2\sqrt{15}}{11}}=\dfrac{5}{3 \left( \dfrac{7}{11} \pm \dfrac{2\sqrt{15}}{11} \right) -1}\)

In decimal form, this says that:

\(\displaystyle 3.21825... = -4.15474...\)

And we know that's not true. Therefore, the multiplication you suggested is invalid, because it changes the nature of the equation so that it has different solutions. In general, when you have two expressions that are equal, you can perform any operation you like to one side, as long as you do the SAME operation to the other side. In my third grade class, we learned this as "whatever you do to grandma, you have to do to grandpa."

In this specific case, you can multiply both sides of the equation by (3x - 1) or both sides by (x + 1). Either one will lead to the solution. If you multiply both by (3x - 1), you'll get:

\(\displaystyle \dfrac{3(3x-1)}{x + 1} = \dfrac{5(3x-1)}{3x-1}\)

Where does that lead you? What do you think you should do next?

 

[markl77]

The forum has a built in LaTeX parser which makes it easy to display math like that. For instance, if you type:

[tex]\dfrac{x}{5}[/tex]

it will render as \(\displaystyle \dfrac{x}{5}\). If you right click on any LaTeX display here on the forum and select "Show Math As -> TeX Commands," you'll get a pop-up window showing you the code you'd need to type between "tex" brackets.

As for the problem, the strategy you've outlined will not produce the correct answer. After doing the multiplications you suggest, we'd be left with:

\(\displaystyle \dfrac{3(3x-1)}{x+1}=\dfrac{5(x+1)}{3x-1}\)

This doesn't really get us any closer to solving the problem, but even if it did, it doesn't matter because it's an invalid step anyway. The solution to the above equation are:

\(\displaystyle \dfrac{7}{11} \pm \dfrac{2\sqrt{15}}{11}\)

Plugging in the first of these two values back into the given equation leaves us with:

\(\displaystyle \dfrac{3}{\dfrac{18}{11} + \dfrac{2\sqrt{15}}{11}}=\dfrac{5}{3 \left( \dfrac{7}{11} \pm \dfrac{2\sqrt{15}}{11} \right) -1}\)

In decimal form, this says that:

\(\displaystyle 3.21825... = -4.15474...\)

And we know that's not true. Therefore, the multiplication you suggested is invalid, because it changes the nature of the equation so that it has different solutions. In general, when you have two expressions that are equal, you can perform any operation you like to one side, as long as you do the SAME operation to the other side. In my third grade class, we learned this as "whatever you do to grandma, you have to do to grandpa."

In this specific case, you can multiply both sides of the equation by (3x - 1) or both sides by (x + 1). Either one will lead to the solution. If you multiply both by (3x - 1), you'll get:

\(\displaystyle \dfrac{3(3x-1)}{x + 1} = \dfrac{5(3x-1)}{3x-1}\)

Where does that lead you? What do you think you should do next?

thanks!
I realized that it was different with equations and you are supposed to get rid of the denominators.
I multiplied 3 to (3x-1) which was equal to 5(x+1)

3(3x-1)=5(x+1)
9x-3=5x+5
x=2