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Help with maths gcse question

Can somebody please tell me how to answer this?:

Solve:
2/(x+1) + x/(2x+3) = 1
Give solutions as surds


Thank you :smile:
Original post by u12davisd
Can somebody please tell me how to answer this?:

Solve:
2/(x+1) + x/(2x+3) = 1
Give solutions as surds


Thank you :smile:


Are these fractions?
Original post by u12davisd
Can somebody please tell me how to answer this?:

Solve:
2/(x+1) + x/(2x+3) = 1

Multiply through by (x+1)(2x+3) and solve the resulting quadratic.
Reply 3
Original post by Anonymous1502
Are these fractions?


Yepp sorry shouldve specified
Reply 4
Original post by RogerOxon
Multiply through by (x+1)(2x+3) and solve the resulting quadratic.


Wait so would it look like x+2/(x+1)(x+1)(2x+3) ? or multiply the numerator?
Solve:
2/(x+1) + x/(2x+3) = 1
Give solutions as surds

Ok.
So......

2(2x+3)+x(x+1)/(x+1)(2x+3)=1
4x+6+x^2+x/2x^2+3x+2x+3=1
x^2+5x+6/2x^2+5x+3=1
x^2+5x+6=2x^2+5x+3
0=x^2-3
x=root positive and negative 3

Hope it helped!:smile:
(edited 6 years ago)
2x+1+x2x+3=1\frac{2}{x+1} + \frac{x}{2x+3} = 1

Multiply both sides by (x+1)(2x+3)(x+1)(2x+3)

2(2x+3)+x(x+1)=(2x+3)(x+1)2(2x+3) + x(x+1) = (2x+3)(x+1)

Now expand it all out and solve.
So basically you'd get a common denominator:
2 + x/(x+1)(2x+3) = 1
So 2 + x = (x+1)(2x+3)
RS:
(x+1)(2x+3) = 2x^2 + 5x + 3
So 2x^2 + 4x + 1 = 0
2(x+1)^2 = 2x^2 + 4x + 2 (the plus 2 is one more than what we want)

So 2(x+1)^2 - 1 = 0
2(x+1)^2 = 1
(x+1)^2 = 1/2
x + 1 = ±√(1/2)
x = -1±√(1/2)

I rushed this a little bit so it's probably wrong lol sorry for wasting your time if it is wrong
Reply 8
Original post by BossThundero
Solve:
2/(x+1) + x/(2x+3) = 1
Give solutions as surds

Ok.
So......

2(2x+3)+x(x+1)/(x+1)(2x+3)=1
4x+6+x^2+x/2x^2+3x+2x+3=1
x^2+4x+6/2x^2+5x+3=1
x^2+4x+6=2x^2+5x+3
0=x^2+x-3

Put the equation x^2+x-3 into the quadratic equation in your calculator and you get the answer in surd form of -1plus or minus root 13 all over 2.
Hope it helped!:smile:


Thank you so much this was very helpful!! :biggrin:
Reply 9
Original post by RogerOxon
2x+1+x2x+3=1\frac{2}{x+1} + \frac{x}{2x+3} = 1

Multiply both sides by (x+1)(2x+3)(x+1)(2x+3)

2(2x+3)+x(x+1)=(2x+3)(x+1)2(2x+3) + x(x+1) = (2x+3)(x+1)

Now expand it all out and solve.



Ah okay got it now, I didn't get that you had to cross-multiply but thank you
Ah looking at the others and I'm probably wrong sorry aha
Original post by BossThundero
Put the equation x^2+x-3 into the quadratic equation in your calculator and you get the answer in surd form of -1plus or minus root 13 all over 2.
Hope it helped!:smile:


Original post by zainrazaa
x = -1±√(1/2)

I rushed this a little bit so it's probably wrong lol sorry for wasting your time if it is wrong


Sorry - both are wrong.
Original post by u12davisd
Ah okay got it now, I didn't get that you had to cross-multiply but thank you

So that you can check your answer, it is x=±3x=\pm \sqrt{3}
Original post by RogerOxon
Sorry - both are wrong.

Yes I understand that he said surd so there should be no fraction but I think he/she realised.:s-smilie::wink:
Original post by RogerOxon
So that you can check your answer, it is x=±3x=\pm \sqrt{3}


aha yeah I rushed it way too much. I know what I did wrong
I always make silly mistakes :facepalm:
Original post by BossThundero
Yes I understand that he said surd so there should be no fraction but I think he/she realised.:s-smilie::wink:

Yes. You lost an x in your working (4x+x=5x).
2(2x+3)+x(x+1)/(x+1)(2x+3)=1
4x+6+x^2+x/2x^2+3x+2x+3=1
x^2+5x+6/2x^2+5x+3=1
x^2+5x+6=2x^2+5x+3
0=x^2-3
x^2=3
x=±√3

x = √3
x = -√3
Original post by zainrazaa
aha yeah I rushed it way too much. I know what I did wrong
I always make silly mistakes :facepalm:

Everyone does, for quite a while. Maths takes a lot of practice to minimise silly mistakes.
pls use latex

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