Inequalities – dual inequality with one variable – Exercise 5690

Exercise

Solve the inequality:

2(x+3)+x4x+6<x+72(x+3)+x\leq 4x+6<x+7

Final Answer


0x<130\leq x<\frac{1}{3}

Solution

2(x+3)+x4x+6<x+72(x+3)+x\leq 4x+6<x+7

Split the dual inequality into two inequalities:

2(x+3)+x4x+62(x+3)+x\leq 4x+6

4x+6<x+74x+6<x+7

Solve the first inequality:

2(x+3)+x4x+62(x+3)+x\leq 4x+6

2x+6+x4x+62x+6+x\leq 4x+6

2x+4x+x6+62x+4x+x\leq -6+6

7x07x\leq 0

x0x\leq 0

Solve the second inequality:

4x+6<x+74x+6<x+7

4xx<764x-x<7-6

3x<13x<1

x<13x<\frac{1}{3}

Finally, intersect both results:

x0x\leq 0

and

x<13x<\frac{1}{3}

Therefore, final answer is

0x<130\leq x<\frac{1}{3}

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