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Can someone please tell how do you solve this i don't even know how to even start. I would really appreciate it.
if the image doesn't work then here is the link:
untitled_7.jpg - ImageHost.org
Can someone please tell how do you solve this i don't even know how to even start. I would really appreciate it.
" i don't even know how to even start"
Ah, my favorite words. If this is true, you should not be in this class and you should not have been given this assignment.
The idea is to use the same techniques you used to solve the 3000 other problems you have been given before you got to this point. Get x alone - simple as that.
Y = (b+d)/(1-x)
1) Notice that x = 1 would be bad. Why?
2) x in the denominator is no good. How do you get it out? Multiply by 1-x
Y(1-x) = (b+d)
3) Pesky Y is hanging around, attached by multiplication. Divide by it.
(1-x) = (b+d)/Y
You should notice that Y = 0 would be bad, as well. Why?
4) Simplfy
1-x = (b+d)/Y
5) Pesky 1 is hanging around, attached by addition. Subtract it.
-x = [(b+d)/Y]-1
5) Pesky "-" is hanging around, attached by multiplication. Divide by it.
x = -{[(b+d)/Y]-1}
6) Simplify
x = -{[(b+d)/Y]-1} = -[(b+d)/Y]+1 = 1-[(b+d)/Y]
That's about it. You must see that nothing went on here that you have not previously seen. Things are not where we want them, so get them there using appropriate arithmetic.
You do the next one and let's see what you get.
First get rid of the denominator terms >CAREFULLY<
You are missing the additive terms -d and c .
Let me do the first step (You can do it many different ways though)
$\displaystyle \frac{x+3}{a} - d = \frac{x-2}{b} + c$
Hammer this equation by multiplying a throughout, followed by b.
$\displaystyle bx + 3b - abd = ax - 2a + abc$
I thought you would say that.
What you did is not wrong, but not very useful!
Multiplying above and below as well generally does not help when there is equal to sign. It is more fruitful to multiply on both sides of equal to.
$\displaystyle \frac{x}{y} = \frac{a}{b} $
Mostly when people say multiply throughout by 'b', they mean
$\displaystyle \frac{bx}{y} = \frac{ba}{b} = a$