How do you solve #3(1+x)=2[3(x+2)(x+1)]#?
Start by removing the brackets  in the term on the right, start inside the square brackets first.
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To solve the equation 3(1+x)=2[3(x+2)(x+1)], you first distribute the terms inside the brackets and then combine like terms on each side of the equation. Here are the steps:

Distribute 2 inside the brackets on the right side: 3(1+x) = 2[3(x+2)  (x+1)] becomes: 3(1+x) = 2[3x + 6  x  1]

Simplify inside the brackets on the right side: 3(1+x) = 2[2x + 5]

Distribute 3 and 2 inside the brackets: 3(1+x) = 4x + 10

Distribute 3 to 1 and x: 3 + 3x = 4x + 10

Subtract 3x from both sides: 3 = x + 10

Subtract 10 from both sides: 7 = x
So the solution to the equation is x = 7.
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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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