How do you solve #2x + 3y = 12# and #x + 4y = 11#?
Combining like terms, we get
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To solve the system of equations (2x + 3y = 12) and (x + 4y = 11), you can use either the substitution method or the elimination method. Let's use the elimination method:

Multiply the second equation by 2 to make the coefficients of (x) in both equations equal: (2(x + 4y) = 2(11) \Rightarrow 2x + 8y = 22).

Now, subtract the first equation from the modified second equation: ((2x + 8y)  (2x + 3y) = 22  12).

Simplify the equation: (2x + 8y  2x  3y = 22  12 \Rightarrow 5y = 10).

Solve for (y): (y = \frac{10}{5} \Rightarrow y = 2).

Substitute (y = 2) into one of the original equations. Let's use the second equation: (x + 4(2) = 11 \Rightarrow x + 8 = 11).

Solve for (x): (x = 11  8 \Rightarrow x = 3).
So, the solution to the system of equations is (x = 3) and (y = 2).
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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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