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Maths help-equations

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Original post by Andy98
correct.


Thank you so much for your help!
Original post by BlueBlueBells
Almost - remember, you have to work out:
x-6=0 and x+2=0


Thank you for your help! I really appreciate it!
Original post by isabel12144
Thank you for your help! I really appreciate it!


Glad to help :tongue:
Original post by SeanFM
Yep, so it almost looks like a quadratic that you can solve. What's next?


Thank you sooo much for your help! I really appreciate it!!!!
Original post by zooloowarrier
for (x-6)(x+2)=0
either x-6=0 or x+2=0
if you rearrange these you should get the correct values for x


Thank you so much for your help! I really appreciate it!!!
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Andy98
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Original post by isabel12144
Thank you so much for your help!


Not problem :hat2:
Sorry to ask again but I just want to check i'm doing it right.

1/x-5 +6/x =2
x/x(x-5) + 6(x-5)/x(x-5)=2
7x-30/x(x-5)=2
7x-30/x^2-5x=2
7x-30=2(x^2-5x)
7x-30=2x^2-10x
-30=2x^2-13x
2x^2-13x+30=0
Then you can't factorise so what do you do?
Original post by isabel12144
Sorry to ask again but I just want to check i'm doing it right.

1/x-5 +6/x =2
x/x(x-5) + 6(x-5)/x(x-5)=2
7x-30/x(x-5)=2
7x-30/x^2-5x=2
7x-30=2(x^2-5x)
7x-30=2x^2-10x
-30=2x^2-13x
2x^2-13x+30=0
Then you can't factorise so what do you do?


There's an error here - "-30=2x^2-13x"
it should be -30=2x^2-17x
2x^2-17x+30=0 is easily solvable, it might also be worth looking up the 'quadratic formula' for quadratics that aren't so straight forward to factor.

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