How do you solve the system #1/4x+y=3/4# and #3/2xy=5/2#?
These fractions are a pain, so let's eliminate them by multiplying through by 4 to get:
Now add 3 to both sides to get
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To solve the system ( \frac{1}{4}x + y = \frac{3}{4} ) and ( \frac{3}{2}x  y = \frac{5}{2} ):
 Solve one of the equations for one variable.
 Substitute the expression found in step 1 into the other equation.
 Solve for the remaining variable.
 Once you have one variable's value, substitute it back into one of the original equations to find the other variable's value.
Let's solve it:

From the first equation: ( \frac{1}{4}x + y = \frac{3}{4} ) ( y = \frac{3}{4}  \frac{1}{4}x )

Substitute ( y = \frac{3}{4}  \frac{1}{4}x ) into the second equation: ( \frac{3}{2}x  (\frac{3}{4}  \frac{1}{4}x) = \frac{5}{2} )

Solve for ( x ): ( \frac{3}{2}x  \frac{3}{4} + \frac{1}{4}x = \frac{5}{2} ) ( \frac{7}{4}x  \frac{3}{4} = \frac{5}{2} ) ( \frac{7}{4}x = \frac{5}{2} + \frac{3}{4} ) ( \frac{7}{4}x = \frac{10}{4} + \frac{3}{4} ) ( \frac{7}{4}x = \frac{7}{4} ) ( x = 1 )

Substitute ( x = 1 ) into the first equation to find ( y ): ( \frac{1}{4}(1) + y = \frac{3}{4} ) ( \frac{1}{4} + y = \frac{3}{4} ) ( y = \frac{3}{4} + \frac{1}{4} ) ( y = 1 )
Therefore, the solution to the system is ( x = 1 ) and ( y = 1 ).
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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