QUESTION IMAGE
Question
solve the inequality.
$3x > -30$
\\(\circ\\) a \\(x < -10\\)
\\(\circ\\) b \\(x > -90\\)
\\(\circ\\) c \\(x > -10\\)
question 11 (1 point)
solve the inequality.
$3x - 5 > 10$
\\(\circ\\) a \\(x < 5\\)
\\(\circ\\) b \\(x > 5\\)
\\(\circ\\) c \\(x > \frac{5}{3}\\)
\\(\circ\\) d \\(x < \frac{5}{3}\\)
First Inequality ($3x > -30$)
Step1: Divide both sides by 3
To solve for \( x \), we divide both sides of the inequality \( 3x > -30 \) by 3. Since 3 is a positive number, the direction of the inequality sign remains the same.
\[
\frac{3x}{3} > \frac{-30}{3}
\]
Step2: Simplify both sides
Simplifying the left side gives \( x \), and simplifying the right side gives \( -10 \). So we have:
\[
x > -10
\]
Step1: Add 5 to both sides
To isolate the term with \( x \), we add 5 to both sides of the inequality \( 3x - 5 > 10 \).
\[
3x - 5 + 5 > 10 + 5
\]
Step2: Simplify both sides
Simplifying the left side gives \( 3x \), and simplifying the right side gives \( 15 \). So we have:
\[
3x > 15
\]
Step3: Divide both sides by 3
Now we divide both sides by 3 (a positive number, so the inequality sign remains the same) to solve for \( x \).
\[
\frac{3x}{3} > \frac{15}{3}
\]
Step4: Simplify both sides
Simplifying the left side gives \( x \), and simplifying the right side gives \( 5 \). So we have:
\[
x > 5
\]
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c. \( x > -10 \)