How do you solve #3x+3y=15# and #2x+3y=5#?
The solution for the system of equations is
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To solve the system of equations (3x + 3y = 15) and (2x + 3y = 5), you can use the method of substitution or elimination. Let's use the elimination method:

Multiply the second equation by 3 to make the coefficients of (y) the same in both equations: [6x + 9y = 15]

Subtract the second equation from the first equation: [3x  6x + 3y  9y = 15  (15)] [ 3x = 30]

Divide both sides by 3: [x = 10]

Substitute (x = 10) into either of the original equations. Let's use the first equation: [3(10) + 3y = 15]

Solve for (y): [30 + 3y = 15] [3y = 45] [y = 15]
So the solution to the system of equations is (x = 10) and (y = 15).
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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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