How do you solve for y in #Ax - By = C#?

Answer 1

#y=(-C+Ax)/B#

We begin with #Ax-By=C#. Our goal is to isolate #y#, and the step I would do is to subtract #Ax# on both sides. If we do that, we have #-By=C-Ax#. From here we need to undo the #-B*y#, so we should divide by #-B#. Now, I like to break this up into two steps just so I don't make any silly mistakes. So first I divide by #-1# on both sides, which gives us #By=(C-Ax)/-1# or #Bx=-C+Ax#. Now I just divide by #B# on both sides, which leaves us with #y=(-C+Ax)/B#.
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Answer 2

To solve for ( y ) in the equation ( Ax - By = C ), you would first isolate ( y ) by moving the term involving ( y ) to one side of the equation and then divide both sides by ( -B ). The steps are as follows:

  1. Subtract ( Ax ) from both sides to isolate the term involving ( y ): ( -By = C - Ax ).
  2. Divide both sides by ( -B ) to solve for ( y ): ( y = \frac{C - Ax}{-B} ).
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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