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Find all real numbers [tex]\(x\)[/tex] such that [tex]\(4x + 2 \leq 14\)[/tex] and [tex]\(-21x + 1 \ \textless \ 22\)[/tex].

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Sagot :

GinnyAnswer
Sure, let's solve the inequalities step-by-step.

First, we need to solve the inequality [tex]\(4x + 2 \leq 14\)[/tex].

1. Start with the inequality:

[tex]\[4x + 2 \leq 14\][/tex]

2. Subtract 2 from both sides of the inequality:

[tex]\[4x \leq 12\][/tex]

3. Divide both sides by 4:

[tex]\[x \leq 3\][/tex]

Next, we solve the inequality [tex]\(-21x + 1 < 22\)[/tex].

1. Start with the inequality:

[tex]\[-21x + 1 < 22\][/tex]

2. Subtract 1 from both sides of the inequality:

[tex]\[-21x < 21\][/tex]

3. Divide both sides by [tex]\(-21\)[/tex] (remember, when dividing by a negative number, the inequality sign reverses):

[tex]\[x > -1\][/tex]

Now, we combine the results of both inequalities.

From the first inequality, we have:

[tex]\[x \leq 3\][/tex]

From the second inequality, we have:

[tex]\[x > -1\][/tex]

Combining these results, we get:

[tex]\[-1 < x \leq 3\][/tex]

Hence, the solution to the inequality is:

[tex]\[-1 < x \leq 3\][/tex]
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