How many ordered pairs of real numbers satisfy the following system of equations?
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x = 3 − 3 y from the first equation. We can substitute this into the second: ∣ ∣ 3 − 3 y ∣ − ∣ y ∣ ∣ = 1 ⟹ ∣ 3 − 3 y ∣ − ∣ y ∣ = ± 1 . Now we just need to find the number of unique solutions for y .
I decided to do some case testing, though there may be a faster method:
Case 1: ∣ 3 − 3 y ∣ − ∣ y ∣ = 1
y ≤ 0 ⟹ 3 − 3 y + y = 1 ⟹ y = 1 . Extraneous since 1 > 0 .
0 < y ≤ 1 ⟹ 3 − 3 y − y = 1 ⟹ y = 2 1 .
y > 1 ⟹ 3 y − 3 − y = 1 ⟹ y = 2 .
Case 2: ∣ 3 − 3 y ∣ − ∣ y ∣ = − 1
y ≤ 0 ⟹ 3 − 3 y + y = − 1 ⟹ y = 2 . Extraneous since 2 > 0 .
0 < y ≤ 1 ⟹ 3 − 3 y − y = − 1 ⟹ y = 1 .
y > 1 ⟹ 3 y − 3 − y = − 1 ⟹ y = 1 . Extraneous since 1 = 1 .
There are clearly 3 solutions to the equation.