Rearranging Equations to Find x in Terms of a and b | Homework Help

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To rearrange the equation a(a+x)=b(b-x) for x in terms of a and b, first expand the equation to get a^2 + ax = b^2 - bx. Next, isolate the x terms by moving them to one side, resulting in ax + bx = b^2 - a^2. Factor out x to obtain x(a + b) = b^2 - a^2. Finally, solve for x, yielding x = (b^2 - a^2) / (a + b).
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Homework Statement



I need to rearrange the following equation to find x in terms of a and b, assuming that a≠-b

Homework Equations



a(a+x)=b(b-x)


The Attempt at a Solution



So far I have multiplied out the brackets and came up with

a^{}2+ax=b^{}2-bx

but I don't really know where to go from there to find x in terms of a and b.Would be really grateful if someone could help. Thanks.
 
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Try adding and subtracting terms from each side so you get each x term on the same side.
 
you can move things over from the left to the right or vice versa i.e.

a + b = d - e
can become by moving things around:

a + b - d = -e

notice how we change the sign when things move over!

for ur question u want to get the x's on one side as the next step so move things over to get the x's together and the a's and b's together.
 
So would this be right?:

ax-bx=a2-b2
x(a-b)=a2-b2
x=(a2-b2)/(a-b)
 
Note ax+bx=b2-a2 and you have to correct your steps
 
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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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