How would one go about solving this problem?

dripstxks

New member
Joined
Jan 20, 2019
Messages
1
2d^2 -d-10=d^2 -4d+3
d^2 +7d+10=d^2 +2-15

Find d
The line in the middle represents division im not exactly sure how to use this website just yet. And the answer is 4 but im not sure how they got the answer except for plugging in values till the right one popped up
 
2d^2 -d-10=d^2 -4d+3
d^2 +7d+10=d^2 +2-15

Find d
The line in the middle represents division im not exactly sure how to use this website just yet. And the answer is 4 but im not sure how they got the answer except for plugging in values till the right one popped up

It appears to me that substantial portions of the problem statement are missing.

For most such problems, one would normally collect all to one side and solve the resulting quadratic equation. (Guessing)
 
Hm... you mention something about "the line in the middle," but no such line appears in your picture. Am I correct in thinking that the problem should really be interpreted as such:

\(\displaystyle \dfrac{2d^2 - d - 10}{d^2 - 4d + 3} = \dfrac{d^2 +7d+10}{d^2 + 2d - 15}\)

In this case, d = 4 is an answer, but it's not the only answer. Or is it instead perhaps meant to be:

\(\displaystyle \dfrac{2d^2 - d - 10}{d^2 + 7d + 10} = \dfrac{d^2 -4d+3}{d^2 + 2d - 15}\)

In this case, d = 4 is the only answer, so I'm guessing you really meant this one. But I suppose it's still possible there's yet another interpretation out there... you'll have to tell us.
 
2d^2 -d-10=d^2 -4d+3
d^2 +7d+10=d^2 +2-15

Find d
The line in the middle represents division im not exactly sure how to use this website just yet. And the answer is 4 but im not sure how they got the answer except for plugging in values till the right one popped up

I'll assume that the interpretation given below is correct. You don't need to learn how to type nice equations (called LaTeX formatting); you can type the equation like this:

(2d^2 -d-10)/(d^2 +7d+10) = (d^2 -4d+3)/(d^2 +2d-15)

... Or is it instead perhaps meant to be:

\(\displaystyle \dfrac{2d^2 - d - 10}{d^2 + 7d + 10} = \dfrac{d^2 -4d+3}{d^2 + 2d - 15}\)

In this case, d = 4 is the only answer, so I'm guessing you really meant this one. But I suppose it's still possible there's yet another interpretation out there... you'll have to tell us.

Now we need to know what you have learned about the topic. Have you been taught to find the LCD and multiply both sides by that? Or perhaps to cross-multiply (which is often not the best way, but can be)? What have you tried, and what happened when you did?

It is not just a matter of trial and error!
 
Hm... you mention something about "the line in the middle," but no such line appears in your picture. Am I correct in thinking that the problem should really be interpreted as such:
\(\displaystyle \dfrac{2d^2 - d - 10}{d^2 - 4d + 3} = \dfrac{d^2 +7d+10}{d^2 + 2d - 15}\)
I'll assume that the interpretation given below is correct. You don't need to learn how to type nice equations (called LaTeX formatting); you can type the equation like this: (2d^2 -d-10)/(d^2 +7d+10) = (d^2 -4d+3)/(d^2 +2d-15)
Speaking of LaTeX formattingLaTeX it ain't hard to learn. Once learned it is done,
\(\displaystyle \dfrac{2d^2 -d-10}{d^2 +7d+10}=\dfrac{d^2 -4d+3}{d^2 +2d-15}\)
May I suggest factoring:
\(\displaystyle \begin{align*}2d^2 -d-10&=(2d-5)(d+2) \\d^2 +7d+10&=(d+2)(d+5)\\\\d^2 -4d+3&=(d-1)(d-3)\\d^2 +2d-15&=(d+5)(d-3) \end{align*}\)

Using the cancellation can you simplify the question?
 
And the winner of the most psychic helper of the day goes to tkhunny.

Good catch.

-Dan
 
Using PKA's suggestion means no quadratic needs to be solved:
leads to (2d - 5)/(d + 5) = (d - 1)/(d + 5).
 
Top