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How do you solve for y #5x+6y-12=0#?

Answer 1

Avery Alexander

#y = -5/6 x + 2#

We begin with:

#5x+6y-12 = 0#

To isolate #y#, we begin by transferring terms that do not contain #y# to the opposite side of the equation.

First, let's subtract #5x# from both sides:

#5x color{orange}{-5x}+6y-12=0color{orange}{-5x}#

#=> 6y - 12 = -5x#

Now let's add #12# to both sides:

#6y - 12 color{orange}{+12}= -5x color{orange}{+12}#

#=> 6y = -5x+12#

Now that all the #y# terms and non #y# terms are on opposite sides, we can solve for #y#.

We just need to divide by #6# on both sides:

#6ycolor{orange}{-:6}= -5xcolor{orange}{-:6}+12color{orange}{-:6}#

#=> color{blue}{y = -5/6 x + 2}#

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Answer 2

Ethan Adkins

To solve for (y) in the equation (5x + 6y - 12 = 0), you need to isolate (y) on one side of the equation. First, move the term (5x) to the other side by subtracting (5x) from both sides of the equation. This leaves you with (6y - 12 = -5x). Then, to isolate (y), add 12 to both sides of the equation to get (6y = -5x + 12). Finally, divide both sides of the equation by 6 to solve for (y). So, (y = \frac{-5x + 12}{6}).

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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.