Answer: Check out the diagram below
[tex]1.5 \le x \le 5.5[/tex]
The lengths can be between 4 and 12 units
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Explanation:
Let's find an expression that represents the area.
The length is 2x+1 and the width is 3. Multiply those two and we get
3(2x+1) = 6x+3
The area is 6x+3 square units.
In other words, A = 6x+3 represents the area.
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We're then told that [tex]12 \le A \le 36[/tex] saying that the area is between 12 and 36 square units (including both endpoints).
We'll replace the A with 6x+3 and solve for x like so:
[tex]12 \le A \le 36\\\\12 \le 6x+3 \le 36\\\\12-3 \le 6x+3-3 \le 36-3 \ \text{ .... see note1}\\\\9 \le 6x \le 33\\\\\frac{9}{6} \le \frac{6x}{6} \le \frac{33}{6} \ \text{ .... see note2}\\\\\frac{3}{2} \le x \le \frac{11}{2}\\\\1.5 \le x \le 5.5\\\\[/tex]
note1: I subtracted 3 from all sides to undo the "plus 3"
note2: I divided all sides by 6 to undo the multiplication
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We just found that [tex]1.5 \le x \le 5.5\\\\[/tex] so it says that x is between 1.5 and 5.5 including both endpoints.
Plug in the smallest possible x value to find the smallest possible length
length = 2x+1
L = 2(1.5)+1
L = 3+1
L = 4
The smallest possible length is 4 units. Multiply this with the height 3 and we get L*W = 4*3 = 12 which is the smallest possible area.
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Now use x = 5.5 to find the largest value of L possible
L = 2x+1
L = 2(5.5)+1
L = 11+1
L = 12
Then note how L*W = 12*3 = 36 is the max area possible, which matches with that your teacher wrote. So this confirms we have the correct value.
If you're not sure which values go where, check out the diagram below.
