Exercise
Solve the inequality:
2 ( x + 3 ) + x ≤ 4 x + 6 < x + 7 2(x+3)+x\leq 4x+6<x+7 2 ( x + 3 ) + x ≤ 4 x + 6 < x + 7
Final Answer
Show final answer
0 ≤ x < 1 3 0\leq x<\frac{1}{3} 0 ≤ x < 3 1
Solution
2 ( x + 3 ) + x ≤ 4 x + 6 < x + 7 2(x+3)+x\leq 4x+6<x+7 2 ( x + 3 ) + x ≤ 4 x + 6 < x + 7
Split the dual inequality into two inequalities:
2 ( x + 3 ) + x ≤ 4 x + 6 2(x+3)+x\leq 4x+6 2 ( x + 3 ) + x ≤ 4 x + 6
4 x + 6 < x + 7 4x+6<x+7 4 x + 6 < x + 7
Solve the first inequality:
2 ( x + 3 ) + x ≤ 4 x + 6 2(x+3)+x\leq 4x+6 2 ( x + 3 ) + x ≤ 4 x + 6
2 x + 6 + x ≤ 4 x + 6 2x+6+x\leq 4x+6 2 x + 6 + x ≤ 4 x + 6
2 x + 4 x + x ≤ − 6 + 6 2x+4x+x\leq -6+6 2 x + 4 x + x ≤ − 6 + 6
7 x ≤ 0 7x\leq 0 7 x ≤ 0
x ≤ 0 x\leq 0 x ≤ 0
Solve the second inequality:
4 x + 6 < x + 7 4x+6<x+7 4 x + 6 < x + 7
4 x − x < 7 − 6 4x-x<7-6 4 x − x < 7 − 6
3 x < 1 3x<1 3 x < 1
x < 1 3 x<\frac{1}{3} x < 3 1
Finally, intersect both results:
x ≤ 0 x\leq 0 x ≤ 0
and
x < 1 3 x<\frac{1}{3} x < 3 1
Therefore, final answer is
0 ≤ x < 1 3 0\leq x<\frac{1}{3} 0 ≤ x < 3 1
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