How do you solve for x in #5ax-c=4c+ax #?

Answer 1

#x=(5c)/(4a#

#1#. Group all like terms together. Start by subtracting #ax# from both sides of the equation.
#5ax-c=4c+ax#
#5ax# #color(red)(-ax)-c=4c+ax# #color(red)(-ax)#
#4ax-c=4c#
#2#. Continue grouping all like terms together by adding #c# to both sides.
#4ax-c# #color(blue)(+c)=4c# #color(blue)(+c)#
#4ax=5c#
#3#. Isolate for #x# by dividing both sides by #4a#.
#(4ax)/color(darkorange)(4a)=(5c)/color(darkorange)(4a)#
#color(green)(|bar(ul(color(white)(a/a)x=(5c)/(4a)color(white)(a/a)|)))#
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Answer 2

To solve for ( x ) in the equation ( 5ax - c = 4c + ax ), first, gather all terms containing ( x ) on one side and constants on the other side. Then, combine like terms and isolate ( x ) by dividing both sides by the coefficient of ( x ). The solution is ( x = \frac{5c}{4a + 1} ).

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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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