# How do you solve for y in #x=(y+2)/(5y-1)#?

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To solve for ( y ) in the equation ( x = \frac{y + 2}{5y - 1} ), first, cross multiply to get rid of the fraction, then isolate ( y ) on one side of the equation. The steps are as follows:

- Multiply both sides of the equation by ( 5y - 1 ): ( x(5y - 1) = y + 2 ).
- Expand the left side: ( 5xy - x = y + 2 ).
- Move all terms involving ( y ) to one side of the equation: ( 5xy - y = x + 2 ).
- Factor out ( y ) from the left side: ( y(5x - 1) = x + 2 ).
- Divide both sides by ( 5x - 1 ): ( y = \frac{x + 2}{5x - 1} ).

So, the solution for ( y ) in terms of ( x ) is ( y = \frac{x + 2}{5x - 1} ).

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To solve for y in the equation x = (y + 2) / (5y - 1), you can cross multiply and then isolate y. The steps are as follows:

- Multiply both sides of the equation by (5y - 1) to eliminate the denominator.
- Distribute x to both terms within the parentheses on the right side of the equation.
- Move all terms involving y to one side of the equation.
- Combine like terms.
- Solve for y by isolating it.

The solution steps are as follows:

- (x(5y - 1) = y + 2)
- (5xy - x = y + 2)
- (5xy - y = x + 2)
- (y(5x - 1) = x + 2)
- (y = \frac{x + 2}{5x - 1})

Therefore, the solution for y is (y = \frac{x + 2}{5x - 1}).

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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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