Depth · one concept from first principles · step 2 of 4
Match interval endpoints, splitting on the sign
The values of that satisfy the inequality are exactly . Find the integers and . (Assume .)
Depth · one concept from first principles · step 2 of 4
The values of that satisfy the inequality are exactly . Find the integers and . (Assume .)
|ax+1| ≤ b means ax+1 lies between -b and b. Subtract 1 throughout.
An absolute value at most b means the distance from 0 is at most b.
6.NS.C.7Convert To Algebrasays exactly the same thing as .
is how far sits from on the number line, so it is at most exactly when lies between and , one stretch on each side of .
Taking away from all three parts slides them the same distance along the line, so none of them changes its place in the order.
Dividing by a positive a keeps the signs. Matching endpoints and adding the two equations gives a=-1, which contradicts a > 0.
The only candidate breaks its own assumption, so this case is eliminated.
8.EE.C.8Eliminate PossibilitiesDividing by a negative a flips both signs. Matching endpoints and adding the equations gives a=-1, and then b=1-2a=3.
Here a=-1 < 0 and b=3 ≥ 0, so the case is consistent.
8.EE.C.8Identify SubproblemsSubstitute a=-1, b=3 back into the original inequality.
The solution set is exactly the given interval.
7.EE.B.4Convert To AlgebraWhen you divide by a letter, ask first whether it is positive or negative.
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