$a^{ 2 }x+(a-1)=(a+1)x$; for $x$
I have been able to manipulate and solve for the indicated variable in these type of equations pretty easily until I came across this one. I know that I have to isolate $x$ on one side but I am having trouble with it. Would dividing both sides by $(a+1) $ be the proper first step?
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$\begingroup$Isolate all the terms that contain $x$ to one side to get $$ a^2 x - (a+1) x = 1 - a. $$Now, factor out the $x$ to get a linear equation: $$ x \cdot \left( a^2 - (a + 1) \right) = 1 - a. $$Now, divide out to isolate $x$: $$ x = \frac {1 - a}{a^2 - (a+1)} = \frac {1-a}{a^2-a-1}, $$ and you are done!
$\endgroup$ 2 $\begingroup$We are given $a^2 x +(a-1)= (a+1)$ and we must solve for $x$. The best first step is to subtract $(a+1)x$ from both sides, as well as $a-1$ from both sides, giving
$$ a^2 x -(a+1)x = -(a-1) $$
or, even better
$$a^2x-ax-x=1-a$$
Factoring the left side as $x(a^2-a-1)$ and dividing gives
$$ x = \frac{1-a}{a^2-a-1} $$
and we are done.
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