You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
equation help
- Thread starter Cloverly
- Start date
HallsofIvy
Elite Member
- Joined
- Jan 27, 2012
- Messages
- 7,763
So \(\displaystyle \frac{x}{x+ 10}= \frac{8}{2x- 4}\)?
(For future reference, if you are using standard ASCII characters, write it using parentheses: x/(x+ 10)= 8/(2x- 4). Web readers do not keep spaces so such attempts to "draw" fractions just don't work.)
The first thing you should do is multiply both sides by (x+ 10)(2x- 4) to get rid of the denominators:
\(\displaystyle (x+10)(2x- 4)\frac{x}{x+ 10}= (x+ 10)(2x- 4)\frac{8}{2x- 4}\)
\(\displaystyle (2x- 4)(x)= (x+ 10)(8)\)
\(\displaystyle 2x^2- 4x= 8x+ 80\)
\(\displaystyle 2x^2- 12x- 80= 0\)
\(\displaystyle x^2- 6x- 40= 0\).
Can you solve that quadratic equation? What ever solutions you get, be sure to check them back into the original equation. Multiplying both sides of an equation by functions of the unknown (here x+ 10 and 2x- 4) can introduce "extraneous solutions" that satisfy the new equation but not the original.)
(For future reference, if you are using standard ASCII characters, write it using parentheses: x/(x+ 10)= 8/(2x- 4). Web readers do not keep spaces so such attempts to "draw" fractions just don't work.)
The first thing you should do is multiply both sides by (x+ 10)(2x- 4) to get rid of the denominators:
\(\displaystyle (x+10)(2x- 4)\frac{x}{x+ 10}= (x+ 10)(2x- 4)\frac{8}{2x- 4}\)
\(\displaystyle (2x- 4)(x)= (x+ 10)(8)\)
\(\displaystyle 2x^2- 4x= 8x+ 80\)
\(\displaystyle 2x^2- 12x- 80= 0\)
\(\displaystyle x^2- 6x- 40= 0\).
Can you solve that quadratic equation? What ever solutions you get, be sure to check them back into the original equation. Multiplying both sides of an equation by functions of the unknown (here x+ 10 and 2x- 4) can introduce "extraneous solutions" that satisfy the new equation but not the original.)
Thank you Very Much
I will try to solve from here.....not sure, but will try. Thanks again for your help and advice.
I will try to solve from here.....not sure, but will try. Thanks again for your help and advice.
So \(\displaystyle \frac{x}{x+ 10}= \frac{8}{2x- 4}\)?
(For future reference, if you are using standard ASCII characters, write it using parentheses: x/(x+ 10)= 8/(2x- 4). Web readers do not keep spaces so such attempts to "draw" fractions just don't work.)
The first thing you should do is multiply both sides by (x+ 10)(2x- 4) to get rid of the denominators:
\(\displaystyle (x+10)(2x- 4)\frac{x}{x+ 10}= (x+ 10)(2x- 4)\frac{8}{2x- 4}\)
\(\displaystyle (2x- 4)(x)= (x+ 10)(8)\)
\(\displaystyle 2x^2- 4x= 8x+ 80\)
\(\displaystyle 2x^2- 12x- 80= 0\)
\(\displaystyle x^2- 6x- 40= 0\).
Can you solve that quadratic equation? What ever solutions you get, be sure to check them back into the original equation. Multiplying both sides of an equation by functions of the unknown (here x+ 10 and 2x- 4) can introduce "extraneous solutions" that satisfy the new equation but not the original.)
Cloverly,
if you decide to factor the left-hand side and solve \(\displaystyle \ x^2 - 6x - 40 = 0, \)
you will have opposite signs in the factors, and the negative middle term
will have the form looking like this:
\(\displaystyle (x \ - \ \ larger \ factor \ of \ 40)(x \ + \ \ smaller \ factor \ of \ 40) \ = \ 0\)
Thus, you will choose one of the correct steps below to continue solving:
(x - 40)(x + 1) = 0
(x - 20)(x + 2) = 0
(x - 10)(x + 4) = 0
(x - 8)(x + 5) = 0
Again, check your solutions against the restrictions of the original equation.
(They are candidates at this point.)
if you decide to factor the left-hand side and solve \(\displaystyle \ x^2 - 6x - 40 = 0, \)
you will have opposite signs in the factors, and the negative middle term
will have the form looking like this:
\(\displaystyle (x \ - \ \ larger \ factor \ of \ 40)(x \ + \ \ smaller \ factor \ of \ 40) \ = \ 0\)
Thus, you will choose one of the correct steps below to continue solving:
(x - 40)(x + 1) = 0
(x - 20)(x + 2) = 0
(x - 10)(x + 4) = 0
(x - 8)(x + 5) = 0
Again, check your solutions against the restrictions of the original equation.
(They are candidates at this point.)