How do you solve for x in #ax+bx-c=0#?

Answer 1
#x(a+b)-c=0# #x(a+b)=c# #x=(c)/(a+b)#
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Answer 2

To solve for ( x ) in the equation ( ax + bx - c = 0 ), you can use the following steps:

  1. Factor out the common factor ( x ) from the left side of the equation: ( x(a + b) - c = 0 ).
  2. Add ( c ) to both sides of the equation: ( x(a + b) = c ).
  3. Divide both sides by ( a + b ): ( x = \frac{c}{a + b} ).

So, the solution for ( x ) is ( x = \frac{c}{a + b} ).

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Answer 3

To solve for ( x ) in the equation ( ax + bx - c = 0 ), you can use the quadratic formula or factor out ( x ) and then solve for ( x ).

Using the quadratic formula:

[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} ]

Factoring out ( x ):

[ x(a + b) = c ]

[ x = \frac{c}{{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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