Solving quadratics with fractions

[Simonio]

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!

 

[kaliprasad]

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

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

 

[Simonio]

Yes!

 

[kaliprasad]

then you multiply by (3x+1)(1-x) on both sides what do you get ?

 

[Simonio]

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!

 

[kaliprasad]

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!

both are wrong. Could you show all the steps

 

[Simonio]

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?

 

[kaliprasad]

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

 

[Simonio]

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

 

[kaliprasad]

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

 

[Simonio]

you should divide RHS by 2 but you multiplied

I multiplied by 2 to cancel out the fraction.

 

[Deveno]

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?

 

[Simonio]

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.