How do you solve #17=5y-3#?

Answer 1

#y = 4#

#17 = 5y - 3#
To solve for the variable #y#, we have to make it by itself.
First, add #color(blue)3# to both sides of the equation: #17 quadcolor(blue)(+quad3) = 5y - 3 quadcolor(blue)(+quad3)#
#20 = 5y#
Now divide both sides by #color(blue)5#: #20/color(blue)5 = (5y)/color(blue)5#
#4 = y#
Therefore, #y = 4#.

I hope this is useful.

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Answer 2

To solve the equation 17 = 5y - 3, you need to isolate the variable y. First, add 3 to both sides of the equation to get 20 = 5y. Then, divide both sides by 5 to find y = 4. So, the solution to the equation is y = 4.

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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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