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solve the following inequality algebraically.\\(4|x + 1| + 6 < 54\\)

Question

solve the following inequality algebraically.\\(4|x + 1| + 6 < 54\\)

Explanation:

Step1: Isolate the absolute value term

Subtract 6 from both sides of the inequality:
$4|x + 1|+6 - 6<54 - 6$
$4|x + 1|<48$

Step2: Divide by the coefficient of the absolute value

Divide both sides by 4:
$\frac{4|x + 1|}{4}<\frac{48}{4}$
$|x + 1|<12$

Step3: Solve the compound inequality

The absolute value inequality $|a|0$) is equivalent to $-b < a < b$. So we have:
$-12

Step4: Isolate x

Subtract 1 from all parts of the compound inequality:
$-12-1$-13

Answer:

The solution to the inequality is $-13 < x < 11$