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Maths help - them tricky, TRICKY algebra questions. (not for the faint hearted)

Alright, here it is:

7/x+2 + 1/x-1 = 4

I got ONE. Photomath told me the answer was -1/2. I don't understand why. I don't understand how.

I used to be an algebra genius, but my powers appear to be failing me. If anyone could help, it would be greatly appreciated - thanks.

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Reply 1

_Xenon_

Original post by ZiggyStarDust_

I LOVE ALGEBRA. I will test it to see.

Reply 2

username2174765

ooooh that looks tricky ill have a go

Reply 3

the bear

did you mean

7/(x+2) + 1/(x-1) = 4

?

:holmes:

Reply 4

Math12345

\frac{7}{x+2}+\frac{1}{x-1}=4

\frac{7(x-1)+1(x+2)}{(x+2)(x-1)} = 4

7(x-1)+1(x+2) = 4(x+2)(x-1)

7x-7+x+2 = 4(x^2+2x-x-2)

8x-5 = 4x^2+4x-8

4x^2-4x-3=0

(2x-3)(2x+1)=0

x=-\frac{1}{2}, \frac{3}{2}

(edited 7 years ago)

Reply 5

there'snoneed

Yeah that's what I got.

Reply 6

username2281303

Original post by the bear

did you mean

7/(x+2) + 1/(x-1) = 4

?

:holmes:

yep

Reply 7

username2281303

Original post by Math12345

\frac{7}{x+2}+\frac{1}{x-1}=4

\frac{7(x-1)+1(x+2)}{(x+2)(x-1)} = 4

7(x-1)+1(x+2) = 4(x+2)(x-1)

7x-7+x+2 = 4(x^2+2x-x-2)

8x-5 = 4x^2+4x-8

4x^2-4x-3=0

(2x-3)(2x+1)=0

x=-1/2, 3/2

okay, thank you for this

this makes more sense!

Reply 8

Chlktrain135

Alright so this is my working:

-7/(x+2) + 1/(x-1)=4

-Cross multiply to get:
[7(x-1)+1(x+2)]/(x-1)(x+2)=4
-Expand brackets to get:
(7x-7+x+2)/(x^2+2x-x-2)=4
-Simplify further to get:
(8x-5)/(x^2+x-2)=4
-Bring denominator to the top to get:
(8x-5)=4(x^2+x-2)
-Expand brackets to form a quadratic equation:
4x^2-4x-3=0
-Factorise quadratic equation to get:
(2x+1)(2x-3)=0
At which point you should have two solutions, x=-1/2 and x=3/2

Hope this can help :smile:

I love algebra too!! (And any form of math, really).

Edit: oops, looks like I was a bit too late to the party :smile:

(edited 7 years ago)

Reply 9

_Xenon_

x = -1/2 and x = 3/2

You could factorise or use quadratic formula whichever you prefer later in the question. :wink:

Would you like detailed explanation ? :smile:

Reply 10

username2281303

Original post by Chlktrain135

I love algebra too!! (And any form of math, really).

Edit: oops, looks like I was a bit too late to the party :smile:

Haha, no that's okay

This is still a useful answer actually, because you explained the steps.

so thank you ! (:

Reply 11

ihatehannah

here's one nice algebra question http://prntscr.com/b6fl58

(edited 7 years ago)

Reply 12

username2281303

Original post by _Xenon_

x = -1/2 and x = 3/2

You could factorise or use quadratic formula whichever you prefer later in the question. :wink:

Would you like detailed explanation ? :smile:

the quadratic formula is my main bae, i'll definitely use that in the future

Nope, that's okay I think I'm on track for this question now, but thank you! :-)

Reply 13

username2174765

Original post by Math12345

\frac{7}{x+2}+\frac{1}{x-1}=4

\frac{7(x-1)+1(x+2)}{(x+2)(x-1)} = 4

7(x-1)+1(x+2) = 4(x+2)(x-1)

7x-7+x+2 = 4(x^2+2x-x-2)

8x-5 = 4x^2+4x-8

4x^2-4x-3=0

(2x-3)(2x+1)=0

x=-\frac{1}{2}, \frac{3}{2}

yeah i got this too

Reply 14

Math12345

Original post by ihatehannah

here's one nice algebra question http://prntscr.com/b6fl58

x^{-\frac{2}{3}}=\frac{64}{9}

x = (\frac{64}{9})^{-\frac{3}{2}}

x = (\frac{9}{64})^{\frac{3}{2}}

x = \frac{27}{512}

I'll explain the first to second step:

(x^{-\frac{2}{3}})^{-\frac{3}{2}} = x^{\frac{6}{6}} = x

. The rest are just indice rules

(edited 7 years ago)

Reply 15

_Xenon_

Original post by ZiggyStarDust_

the quadratic formula is my main bae, i'll definitely use that in the future

Nope, that's okay I think I'm on track for this question now, but thank you! :-)