How many integer solutions are there to the equation x + y + z = 20 with x ≥ −2, y > 2, and z ≥ 0?
Added by Alyssa N.
Step 1
For x ≥ -2, we can set x = x' - 2, where x' ≥ 0. For y > 2, we can set y = y' + 3, where y' ≥ 0. For z ≥ 0, we can leave it as it is. Substituting these into the original equation, we get: (x' - 2) + (y' + 3) + z = 20 Simplifying, we get: Show more…
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