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I Need Help On Algebraic Fractions!

Winch2012

Heya, im doing a past paper for the exam on Monday, it's a non calculator and I am stuck on a similar sort of question.

\frac{3}{x-2}+\frac{8}{x-3}=2

What I did was:

\frac{3(x-3)+8(x-2)}{(x-2)(x-3)}=2

I multiplied out the numerator and then cross multiplied the denominator onto the other side of the equals sign:

11x-25=2\times(x-2)(x-3)

11x-25=2(x^2-5x+6)

11x-25=2x^2-10x+12

2x^2-21x+37=0

Here is where i got stuck..
I had to options, either follow what Mathswatch had taught me and rearrange the whole equation to make 0 then factorise or use the quadratic formula..
Since this is a non calculator i attempted the rearranging and factorising:

2x^2-21x+37=0

I tried the ABC method ( a=2 .. b= -21 .. c= 37)
The rule is

A\times C

which would in this case

= 74

The factors of 74 are : 1 and 74, 2 and 37 and neither of these add or subtract to make B which is -21..

So I'm either lost, over complicating things or completely wrong!
Please, can someone help me?

Reply 1

Doglamlico

Original post by Winch2012

Heya, im doing a past paper for the exam on Monday, it's a non calculator and I am stuck on a similar sort of question.

\frac{3}{x-2}+\frac{8}{x-3}=2

What I did was:

\frac{3(x-3)+8(x-2)}{(x-2)(x-3)}=2

I multiplied out the numerator and then cross multiplied the denominator onto the other side of the equals sign:

11x-25=2\times(x-2)(x-3)

11x-25=2(x^2-5x+6)

11x-25=2x^2-10x+12

2x^2-21x+37=0

2x^2-21x+37=0

I tried the ABC method ( a=2 .. b= -21 .. c= 37)
The rule is

A\times C

which would in this case

= 74

I believe you're right - there's no simple solution, and you'd just have to give it in terms of the quadratic formula. http://www.wolframalpha.com/input/?i=3%28x-2%29%5E-1+%2B+8%28x-3%29%5E-1+%3D+2
My calculator came out with a solution of 8.26.... so yeah, there's no simple solution :smile:

Edit: Wolfram Alpha - the most amazing maths tool on the internet :biggrin:

it's soo good

Reply 2

ThatPerson

Original post by Junaid96

Edit: Wolfram Alpha - the most amazing maths tool on the internet :biggrin:

it's soo good

The question was non-calculator.

Reply 3

Winch2012

Yeah however, the mark scheme says that X = 5 or X = -0.5

The brief explanation is:
M1 for common denominator on LHS or clearing fractions
M1 for multiplying out brackets
A1 for

2^2 – 9x + 5 = 0

M1 for (2x ± 1)(x ± 5) or substitution into quadratic formula
A1 for 5 and - 0.5

(edited 11 years ago)

Reply 4

Icedstoat

Original post by ThatPerson

The question was non-calculator.

True, but that doesn't stop you using the formula:
(21+Sqrt137)/4 or (21-Sqrt137)/4 being the solutions as I get them

Reply 5

Doglamlico

Original post by ThatPerson

The question was non-calculator.

Yes, I know. I just used a calculator to confirm my suspicions.

Original post by Icedstoat

True, but that doesn't stop you using the formula:
(21+Sqrt137)/4 or (21-Sqrt137)/4 being the solutions as I get them

Yep, and as I said in my OP, "you'd just have to give it in terms of the quadratic formula"

Original post by Winch2012

Yeah however, the mark scheme says that X = 5 or X = -0.5

The brief explanation is:
M1 for common denominator on LHS or clearing fractions
M1 for multiplying out brackets
A1 for

2^2 – 9x + 5 = 0

M1 for (2x ± 1)(x ± 5) or substitution into quadratic formula
A1 for 5 and - 0.5

\frac{3}{x-2}+\frac{8}{x-3}=2

but.. 5 isn't a solution?

(edited 11 years ago)

Reply 6

raheem94

Original post by Junaid96

...

Original post by Icedstoat

...

Original post by Winch2012

Yeah however, the mark scheme says that X = 5 or X = -0.5

The brief explanation is:
M1 for common denominator on LHS or clearing fractions
M1 for multiplying out brackets
A1 for