Algebra

Here I have a question..

72
______ = 1

x^2 - 9 (...)

I know but I can't just give up on the question. I have to have a little bash at it

I think MrM is suggesting that you need to get the basics in place before you tackle more advanced questions that depend on basic skills.

But just for information, when you have a problem like this where several operations take place, you have to reverse them in the correct order.

Your original problem is basically 72/something = 1. So what would you do to find "something"?

1 multiplied by 72?

72 multiplied by 9? 648

Or am I looking at the wrong calculation?

Thanks for the explanation. I feel much better informed now.

I live very close to langley grammar school , slough grammar school (my school) , and other grammer schools , sorry o.e

I will explain you the steps to help you to solve this task. Pay attention!

$\displaystyle \dfrac{72}{x^2 - 9} = 1$

x² - 9 is the denominator of the term on the left side. That's why you have to multiply the equation by the denominator, so the denominator 'disappears' on the left side. You get:

$\displaystyle 72 = x^2 - 9$.

-9 is is the subtrahend on the right side of the equation. You have to add the equation by 9, if -9 should dissapear on the right side. You get:

$\displaystyle 81 = x^2$.

And now you take the root to get your result. It is x = 9.

Put x = 9 in your equation to check your result:

$\displaystyle \dfrac{72}{9^2 - 9} = \dfrac{72}{81 - 9} = \dfrac{72}{72} = 1$

As you can see, the result is right. Any questions?