How do you solve for f in #2/3f + 5/12g=1- fg#?

Answer 1

See a solution process below:

First, subtract #color(red)(5/12g)# and add #color(blue)(fg)# to both sides of the equation to isolate the #f# terms on the left side of the equation:
#2/3f + color(blue)(fg) + 5/12g - color(red)(5/12g) = 1 - color(red)(5/12g) - fg + color(blue)(fg)#
#2/3f + fg + 0 = 1 - 5/12g - 0#
#2/3f + fg = 1 - 5/12g#
Next, factor an #f# out of each term on the left side of the equation:
#f(2/3 + g) = 1 - 5/12g#
Now, divide each side of the equation by #color(red)(2/3 + g)# to solve for #f# while keeping the equation balanced:
#(f(2/3 + g))/color(red)(2/3 + g) = (1 - 5/12g)/color(red)(2/3 + g)#
#(fcolor(red)(cancel(color(black)((2/3 + g)))))/cancel(color(red)(2/3 + g)) = (1 - 5/12g)/(2/3 + g)#
#f = (1 - 5/12g)/(2/3 + g)#
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Answer 2

To solve for ( f ), first, isolate the term involving ( f ) on one side of the equation and move all other terms to the opposite side. Then, simplify and solve for ( f ) by dividing both sides by the coefficient of ( f ).

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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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