How do you solve the following system: #3x  2y = 9, 2x + 3y = 34#?
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To solve the system: (3x  2y = 9) (2x + 3y = 34)
You can use either the substitution method or the elimination method. Here, I'll use the elimination method:

Multiply the first equation by 2 and the second equation by 3 to make the coefficients of (x) in both equations the same: (6x  4y = 18) (6x + 9y = 102)

Now, subtract the first equation from the second equation: ((6x + 9y)  (6x  4y) = 102  18) (6x + 9y  6x + 4y = 84) (13y = 84)

Solve for (y): (y = \frac{84}{13})

Substitute the value of (y) into either of the original equations. Let's use the first equation: (3x  2\left(\frac{84}{13}\right) = 9)

Solve for (x): (3x  \frac{168}{13} = 9) (3x = 9 + \frac{168}{13}) (3x = \frac{117}{13}) (x = \frac{117}{39}) (x = 13)
So, the solution to the system is (x = 13) and (y = \frac{84}{13}).
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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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