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# Algebra/Alg2 solving inequalities and absolute values

Question
2 |9t-3| + 4t ≤ 12

To solve 2|9t-3| + 4t ≤ 12, what matters is whether 9t-3 is positive or negative.

If 9t-3 is positive, the equation is 2(9t-3) + 4t ≤ 12.
If 9t-3 is negative, the equation is 2(-9t+3) + 4t ≤ 12.

For 9t-3 to be positive, solve 9t-3 >= 0.
That is the same as 9t >= 3, so we need t >= 3/9,  which is t >= 1/3.
In this case, our equation is 2(9t-3) + 4t ≤ 12.
This is the same as 18t - 6 + 4t ≤ 12.  Noting that 18t + 4t is 22t and adding 6 to both sides gives 22t <= 18, so t <= 18/22 => t <= 9/11.
This implies we need 1/3 <= t <= 9/11.

For 9t-3 being negative, this means 9t-3 < 0 => 9t < 3 => t < 1/3.
In this case, the equation to solve is  2(-9t+3) + 4t ≤ 12.
That multiplies into -18t + 6 + 4t ≤ 12.
Combining the t terms and subtracting 6 from both sides gives -14t ≤6.
We need to multiply both sides by -1/14, and since it is negative,
it changes the sign to a greater than or equal to.
That gives t >= 6/14 => t >= 3/7.
Since we need both t to be less than a third and greater than 3/7, this is not possible.

Algebra

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#### Scott A Wilson

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