How do you solve for c in #a(c - b) = d#?

Answer 1

#c = d/a+b# unless #a=0#...

Case #bb(a = 0)#
If #a=0# then #a(c-b) = 0# regardless of the value of #c#.

Hence:

Case #bb(a != 0)#
If #a!=0# then we can divide both sides of the equation by #a# to get:
#c-b = d/a#
Then add #b# to both sides to find:
#c = d/a+b#
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Answer 2

To solve for ( c ) in the equation ( a(c - b) = d ), you first distribute the ( a ) across the parentheses:

( ac - ab = d )

Then, add ( ab ) to both sides to isolate the term involving ( c ):

( ac = d + ab )

Finally, divide both sides by ( a ) to solve for ( c ):

( c = \frac{d + ab}{a} )

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Answer 3

To solve for ( c ) in the equation ( a(c - b) = d ), you would first distribute ( a ) across the parentheses. Then, you would isolate ( c ) by dividing both sides of the equation by ( a ). The solution would be ( c = \frac{d}{a} + 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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