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Can someone help me solve this?

mellow2006

So I got a maths question that is:

Solve this equation
x ^-2/3 = 3x^-1

But I'm a bit stuck so can anyone help me answer this?

Reply 1

sillysillyman

Original post by mellow2006

So I got a maths question that is:

Solve this equation
x ^-2/3 = 3x^-1

But I'm a bit stuck so can anyone help me answer this?

you could cube both sides, leaving x^-2 = 27x^-3 . Take the reciprocals and then factorise. hope this helps

(edited 2 years ago)

Reply 2

mellow2006

Original post by sillysillyman

you could cube both sides, leaving x^-2 = 3x^-3 . Take the reciprocals and then factorise. hope this helps

Ok that helps :smile:

but how would I factorise the reciprocals?

Reply 3

mqb2766

Original post by mellow2006

Ok that helps :smile:

but how would I factorise the reciprocals?

If youre cubing both sides, you need to cube the whole of the right hand side.
Factorizing reciprocals is the same as normal, pull out the common term. You could write them as fractions to make it clearer.

Reply 4

sillysillyman

Original post by mellow2006

Ok that helps :smile:

but how would I factorise the reciprocals?

sorry, I wasn't clear at all. raise both sides by ^-3 (similar to cubing and then ^-1(taking reciprocal)). this would leave you with
(x^-2/3)^-3=(3x^-1)^-3
x^2=(3^-3)x^3 (we multiply the powers)
x^2=(x^3)/27 then take the terms to one side and equal to zero, then factorise to get the solutions.