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Solve the inequality.

[tex]\[
\left|\frac{1}{3} x + 3\right| \ \textless \ 15
\][/tex]

[tex]\[
-18 \ \textless \ x \ \textless \ 36
\][/tex]

Sagot :

GinnyAnsweridn
To solve the inequality

[tex]\[ \left| \frac{1}{3} x + 3 \right| < 15, \][/tex]

we need to consider the definition of absolute value:

[tex]\[ \left| A \right| < B \][/tex]

implies

[tex]\[ -B < A < B. \][/tex]

For our inequality, [tex]\( A = \frac{1}{3} x + 3 \)[/tex] and [tex]\( B = 15 \)[/tex]. Therefore, we have:

[tex]\[ -15 < \frac{1}{3} x + 3 < 15. \][/tex]

To remove the absolute value, we break this down into two separate inequalities:

1. [tex]\[ -15 < \frac{1}{3} x + 3 \][/tex]
2. [tex]\[ \frac{1}{3} x + 3 < 15 \][/tex]

Let's solve each part separately:

Solving the first inequality:

[tex]\[ -15 < \frac{1}{3} x + 3 \][/tex]

Subtract 3 from both sides to isolate the term with [tex]\( x \)[/tex]:

[tex]\[ -15 - 3 < \frac{1}{3} x \][/tex]
[tex]\[ -18 < \frac{1}{3} x \][/tex]

Multiply both sides by 3 to solve for [tex]\( x \)[/tex]:

[tex]\[ -18 \times 3 < x \][/tex]
[tex]\[ -54 < x \][/tex]

Solving the second inequality:

[tex]\[ \frac{1}{3} x + 3 < 15 \][/tex]

Subtract 3 from both sides:

[tex]\[ \frac{1}{3} x < 15 - 3 \][/tex]
[tex]\[ \frac{1}{3} x < 12 \][/tex]

Multiply both sides by 3 to solve for [tex]\( x \)[/tex]:

[tex]\[ x < 12 \times 3 \][/tex]
[tex]\[ x < 36 \][/tex]

Combining these results, we get:

[tex]\[ -54 < x < 36 \][/tex]

Thus, the solution to the inequality [tex]\( \left| \frac{1}{3} x + 3 \right| < 15 \)[/tex] is:

[tex]\[ -54 < x < 36. \][/tex]
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