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Question
which graph represents the solutions to this inequality? 6g + 7 ≤ 55 or -7g + 22 < -62
Step1: Solve \(6g + 7 \leq 55\)
Subtract 7 from both sides: \(6g \leq 55 - 7 = 48\)
Divide by 6: \(g \leq \frac{48}{6} = 8\)
So, \(g \leq 8\) (closed circle at 8, arrow left? Wait, no, wait the graph has 8 and 12. Wait, maybe I misread. Wait, let's solve the second inequality.
Step2: Solve \(-7g + 22 < -62\)
Subtract 22: \(-7g < -62 - 22 = -84\)
Divide by -7 (reverse inequality): \(g > \frac{-84}{-7} = 12\)
So, \(g > 12\) (open circle at 12? Wait no, wait the first inequality: \(6g +7 \leq 55\) gives \(g \leq 8\) (closed circle at 8, arrow towards lower numbers? But the graphs have 8 and 12. Wait, maybe the first graph: top graph has closed at 8 (arrow up) and open at 12 (arrow down)? Wait no, let's re-express.
Wait, the compound inequality is \(6g +7 \leq 55\) OR \(-7g +22 < -62\).
First inequality: \(6g \leq 48\) → \(g \leq 8\) (closed circle at 8, arrow in the direction of numbers less than 8? But the top graph has a closed circle at 8 (arrow up) and open at 12 (arrow down). Wait, maybe the second inequality: \(-7g < -84\) → \(g > 12\) (open circle at 12, arrow in direction of numbers greater than 12? But the top graph: closed at 8 (arrow up: meaning \(g \geq 8\)?) Wait, no, the red dot is closed, so inclusive. The open circle is exclusive.
Wait, let's re-express the solutions:
For \(6g +7 \leq 55\): \(g \leq 8\) (closed circle at 8, arrow to the left (smaller numbers)). But the top graph has closed at 8 (arrow up: maybe \(g \geq 8\)?) Wait, maybe I made a mistake. Let's recalculate \(6g +7 \leq 55\):
\(6g \leq 55 -7 = 48\) → \(g \leq 8\). So \(g\) is less than or equal to 8: closed circle at 8, arrow to the left (towards 0, etc.). But the top graph has closed at 8 (arrow up: maybe \(g \geq 8\)) and open at 12 (arrow down: \(g < 12\))? No, that doesn't match. Wait the second inequality: \(-7g +22 < -62\):
\(-7g < -84\) → divide by -7 (reverse inequality): \(g > 12\) (open circle at 12, arrow to the right (larger numbers)).
Now, the compound inequality is OR, so solutions are \(g \leq 8\) OR \(g > 12\).
Now, let's check the graphs:
Top graph: closed at 8 (so \(g \geq 8\)?) No, closed circle means inclusive. Wait, maybe the first graph: closed at 8 (arrow up: \(g \geq 8\)) and open at 12 (arrow down: \(g < 12\)) → that would be \(8 \leq g < 12\), which is not our solution.
Middle graph: open at 8 (so \(g > 8\)) and closed at 12 (arrow down: \(g \leq 12\)) → \(8 < g \leq 12\), not our solution.
Wait, maybe I messed up the direction. Wait, the third graph: open at 8 (arrow up: \(g > 8\)) and closed at 12 (arrow down: \(g \leq 12\))? No, the third graph has open at 8 (red circle with plus) and closed at 12 (red dot), arrow up from 8 and down from 12? Wait, no, the third graph's arrow is up from 8 and down from 12? Wait, the original problem's graphs:
Top graph: closed at 8 (red dot), arrow up; open at 12 (red circle with plus), arrow down.
Middle graph: open at 8 (red circle with plus), arrow up; closed at 12 (red dot), arrow down.
Third graph: open at 8 (red circle with plus), arrow up; closed at 12 (red dot), arrow down? Wait, no, the third graph's arrow is up from 8 and down from 12? Wait, maybe the correct graph is the top one? Wait no, let's re-express the solutions:
\(g \leq 8\) (closed at 8, arrow left) OR \(g > 12\) (open at 12, arrow right). But the graphs given have 8 and 12, with arrows between? Wait, maybe the first inequality is \(6g +7 \leq 55\) → \(g \leq 8\) (closed at 8, arrow up: maybe the number line is vertical? Wait, the graph is vertical? So 8 is above 12? Wait, the numbers 8 and 12: 8 is abo…
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The top graph (with closed circle at 8 and open circle at 12, arrows up from 8 and down from 12)