Inequalities

Compound Inequalities

Solve a compound inequality of the form a < bx + c ≤ d by applying the same operation to all three parts simultaneously. Because the coefficient is positive, no direction flip is needed.

Example

Apply the same operation to all three parts of a compound inequality.

highlighted = computed this step

Step 1 — Set up

Set up the expression.

−1<2x+1≤7-1 < 2x + 1 \le 7
Start with lower bound, middle expression, and upper bound.lower-1<middle2x + 1≤upper7subtractsubtract 1 from lower, middle, and upper--divide all parts by positive 2---13number lineshade betweenStart: -1 < 2x + 1 ≤ 7

Step 2 — Subtract all parts

Subtract 1 from all three parts.

−1−1<2x≤7−1-1 - \hl{1} < 2x \le 7 - \hl{1}
Subtract 1 from all three parts; both signs stay the same.lower-1<middle2x + 1≤upper7subtract-1 - 1 < 2x ≤ 7 - 1signs stay--divide all parts by positive 2---13number lineshade betweenStart: -1 < 2x + 1 ≤ 7

Step 3 — Simplify

Simplify the bounds to -2 and 6.

-2<2x≤6\hl{-2} < 2x \le \hl{6}
Simplify to -2 < 2x ≤ 6.lower-1<middle2x + 1≤upper7subtract-1 - 1 < 2x ≤ 7 - 1signs stay-2 < 2x ≤ 6divide all parts by positive 2---13number lineshade betweenStart: -1 < 2x + 1 ≤ 7

Step 4 — Divide all parts

Divide all three parts by positive 2.

−2/2<x≤6/2-2 / \hl{2} < x \le 6 / \hl{2}
Divide all three parts by positive 2.lower-1<middle2x + 1≤upper7subtract-1 - 1 < 2x ≤ 7 - 1signs stay-2 < 2x ≤ 6-2 / 2 < x ≤ 6 / 2---13number lineshade betweenStart: -1 < 2x + 1 ≤ 7

Step 5 — Solution

The solution bounds are -1 and 3.

-1<x≤3\hl{-1} < x \le \hl{3}
Solution: -1 < x ≤ 3.lower-1<middle2x + 1≤upper7subtract-1 - 1 < 2x ≤ 7 - 1signs stay-2 < 2x ≤ 6-2 / 2 < x ≤ 6 / 2-1 < x ≤ 3-13number lineshade betweenStart: -1 < 2x + 1 ≤ 7

Step 6 — Interval

Write the compound interval.

x∈(−1,3]x\in( -1 , 3 ]
Graph the interval (-1, 3].lower-1<middle2x + 1≤upper7subtract-1 - 1 < 2x ≤ 7 - 1signs stay-2 < 2x ≤ 6-2 / 2 < x ≤ 6 / 2-1 < x ≤ 3-13open at -1; closed at 3(-1, 3]Start: -1 < 2x + 1 ≤ 7
compound-inequality Apply each operation to ALL THREE parts of the compound inequality at once. The two relation symbols are preserved throughout because the divisor is positive.