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Which of the following is the correct graph of the compound inequality [tex]4p + 1 \ \textgreater \ -7[/tex] or [tex]6p + 3 \ \textless \ 33[/tex]?
Let's solve the given compound inequality step-by-step.
### Inequality 1: [tex]\( 4p + 1 > -7 \)[/tex]
1. Subtract 1 from both sides: [tex]\[
4p + 1 - 1 > -7 - 1
\][/tex] This simplifies to: [tex]\[
4p > -8
\][/tex]
2. Divide by 4: [tex]\[
\frac{4p}{4} > \frac{-8}{4}
\][/tex] This simplifies to: [tex]\[
p > -2
\][/tex]
So the solution for the first inequality is: [tex]\[ p > -2 \][/tex]
### Inequality 2: [tex]\( 6p + 3 < 33 \)[/tex]
1. Subtract 3 from both sides: [tex]\[
6p + 3 - 3 < 33 - 3
\][/tex] This simplifies to: [tex]\[
6p < 30
\][/tex]
2. Divide by 6: [tex]\[
\frac{6p}{6} < \frac{30}{6}
\][/tex] This simplifies to: [tex]\[
p < 5
\][/tex]
So the solution for the second inequality is: [tex]\[ p < 5 \][/tex]
### Combine the Solutions Using OR
We now need to find the union of the two solutions [tex]\( p > -2 \)[/tex] or [tex]\( p < 5 \)[/tex].
When combining these solutions with OR, we see that the resulting solution set covers all real numbers because every number [tex]\( p \)[/tex] satisfies at least one of the inequalities [tex]\( p > -2 \)[/tex] or [tex]\( p < 5 \)[/tex]. This compound inequality doesn't exclude any real number because the union of these two intervals [tex]\((-2, \infty) \cup (-\infty, 5)\)[/tex] actually covers the entire real line.
Thus, the correct graph of the compound inequality [tex]\( 4 p + 1 > -7 \)[/tex] or [tex]\( 6 p + 3 < 33 \)[/tex] would be the entire number line.
The graph would look like this: [tex]\[
\text{<----|====================================|---->}
\text{ -2 5 }
\][/tex]
Each point on the number line satisfies the compound inequality.
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