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Bunuel
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Bunuel
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MBAToronto2024
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MBAToronto2024
KarishmaB
Isn’t it mistake here?
If y>0, we’re getting y = -12. Which is not valid solution.
­
Where is it stated that y is positive? Why should it be positive? If you substitute y = −12 and y = −16, you'll see that both values satisfy 22 − |y + 14| = 20, so there is no mistake in the solution at all.
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MBAToronto2024
KarishmaB
Isn’t it mistake here?
If y>0, we’re getting y = -12. Which is not valid solution.
­
The point of transition here is y = -14, not y = 0. We get y = -12 when y >= -14
Look at what Bunuel has done above.

|y+14| = 2

The two cases will be either y + 14 >= 0 or y + 14 < 0

Alternatively, y will take 2 values which are a distance of 2 away from -14. So y will be -16 and -12.
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I like the solution - it’s helpful.
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