MHB Solving quadratics with fractions

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The discussion focuses on solving quadratic equations involving fractions, specifically the equation 2/(3x + 1) + 3/(1 - x) = 1/2. Participants emphasize the importance of multiplying through by the product of the denominators to eliminate fractions and simplify the equation. One user demonstrates the correct approach by rewriting the fractions with a common denominator and cross-multiplying to derive a solvable quadratic equation. The conversation highlights common pitfalls, such as sign errors and miscalculations, while also expressing appreciation for the collaborative problem-solving process. Overall, the thread serves as a resource for those struggling with similar quadratic equations involving fractions.
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Sorry if this has been covered elsewhere but I'm having problems with equations such as this which are quadratics with fractions:

2 over 3x + 1 PLUS 3 over 1-x = 1/2.

I know you are supposed to multiply each term by the product of the denominators but i keep getting weird results -so if someone could show me their reliable method, I'd be really grateful.

Thanks a lot!
 
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Simonio said:
Sorry if this has been covered elsewhere but I'm having problems with equations such as this which are quadratics with fractions:

2 over 3x + 1 PLUS 3 over 1-x = 1/2.

I know you are supposed to multiply each term by the product of the denominators but i keep getting weird results -so if someone could show me their reliable method, I'd be really grateful.

Thanks a lot!

do you mean

$\frac{2}{3x+1} + \frac{3}{1-x} = \frac{1}{2}$
 
Yes!
 
then you multiply by (3x+1)(1-x) on both sides what do you get ?
 
kaliprasad said:
then you multiply by (3x+1)(1-x) on both sides what do you get ?
Well, I get 7x=3 (on the left hand side) = 12x squared-8x + 4

So: 12x squared - 8x +4 = 7x + 3

So: 12x squared -15x -1 = 0 which I then can't seem to factorize(Angry) I've obviously done something fundamentally wrong---help!:confused:
 
Simonio said:
Well, I get 7x=3 (on the left hand side) = 12x squared-8x + 4

So: 12x squared - 8x +4 = 7x + 3

So: 12x squared -15x -1 = 0 which I then can't seem to factorize(Angry) I've obviously done something fundamentally wrong---help!:confused:
both are wrong. Could you show all the steps
 
Last edited:
kaliprasad said:
both are wrong. Could you show all the steps

I've tried it again:

(3x + 1)(1-x)2/3x+1 = 2 - 2x
(3x+1)(1-x)3/1-x = 9x+3
(3x+1)(1-x)1/2 = (6x=2)(2-2x) = -12x squared + 8x + 4

So: -12x squared +8x +4 = 7x = 5

This doesn't look right! Where am I going wrong?
 
Simonio said:
I've tried it again:

(3x + 1)(1-x)2/3x+1 = 2 - 2x
(3x+1)(1-x)3/1-x = 9x+3
(3x+1)(1-x)1/2 = (6x=2)(2-2x) = -12x squared + 8x + 4

So: -12x squared +8x +4 = 7x = 5

This doesn't look right! Where am I going wrong?

I am unable to follow your steps

$\frac{2}{3x+1} + \frac{3}{1-x} = \frac{1}{2}$

shall become

$2(1-x) + 3(3x+1) =\frac{(3x+1)(1-x)}{2}$

can you proceed from here
 
kaliprasad said:
I am unable to follow your steps

$\frac{2}{3x+1} + \frac{3}{1-x} = \frac{1}{2}$

shall become

$2(1-x) + 3(3x+1) =\frac{(3x+1)(1-x)}{2}$

can you proceed from here

So: 7x+5 = (Multiply by 2) (6x squared + 2)(1-x) = 6x squared + 4x + 2

So: 6x squared +4x + 2= 7x + 5

How does this look now?? Thanks
 
  • #10
Simonio said:
So: 7x+5 = (Multiply by 2) (6x squared + 2)(1-x) = 6x squared + 4x + 2

So: 6x squared +4x + 2= 7x + 5

How does this look now?? Thanks

you should divide RHS by 2 but you multiplied
 
  • #11
kaliprasad said:
you should divide RHS by 2 but you multiplied

I multiplied by 2 to cancel out the fraction.
 
  • #12
When I do these types of problems, I have to remember to be very careful, as it is easy to make a mistake.

I treat "rational functions" (fractions where we have variables in the numerator and denominator) just like "ordinary fractions".

As you may recall, if we want to add 2 fractions, we need to get a common denominator, in order to add the numerators.

So, we turn:

$\dfrac{2}{3x + 1}$ into:

$\dfrac{2(1 - x)}{(3x + 1)(1 - x)}$ and:

$\dfrac{3}{1-x}$ into:

$\dfrac{3(3x + 1)}{(3x + 1)(1 - x)}$.

Now we can "add the tops" while "keeping the bottoms", that is:

$\dfrac{2}{3x + 1} + \dfrac{3}{1-x} = \dfrac{2(1 - x)}{(3x + 1)(1 - x)} + \dfrac{3(3x + 1)}{(3x + 1)(1 - x)}$

$ = \dfrac{2(1 - x) + 3(3x + 1)}{(3x + 1)(1 - x)}$.

This is a bit messy, so let's clean it up a little bit:

The bottom is:

$(3x + 1)(1 - x) = (3x + 1)(1) - (3x + 1)(x) = 3x + 1 - 3x^2 - x = -3x^2 + 2x + 1$

The top is:

$2(1 - x) + 3(3x + 1) = 2 - 2x + 9x + 3 = 7x + 5$

So our original equation:

$\dfrac{2}{3x + 1} + \dfrac{3}{1 - x} = \dfrac{1}{2}$

is equivalent to:

$\dfrac{7x + 5}{-3x^2 + 2x + 1} = \dfrac{1}{2}$

To "un-fraction-ize", we "cross-multiply" so we get:

$(2)(7x + 5) = -3x^2 + 2x + 1$, which we can then simplify to:

$3x^2 + 12x + 9 = 0$

You should be able to factor this fairly easily.

Do you understand the steps I have taken?
 
  • #13
Thanks a lot, that's great. You are quite right to say that it is easy to make a mistake! One can easy make a discrepancy with the signs and sometimes very basic addition and subtraction! I'm a 54 year-old dad who is re-learning some of this stuff to help his son, so some of this maybe to do with my age -but I really appreciate the help, this is a really great site.
 
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