Question

student name date pd letters of correct graphs: letters of incorrect graphs and explanation of the error made: possible error(s) or look for(s): is the correct type of boundary line used (dashed or solid)? is the boundary line graphed correctly? is the correct side shaded? graph a: inequality: 2x - 3y ≥ -6 graph b: inequality: 2x + 4y > 8 graph c: inequality: 6x + 4y ≤ 24 graph d: inequality: x - 2y < -2

Explanation:

Step1: Analyze Graph A

The inequality is \(2x - 3y \geq - 6\). For inequalities with \(\geq\) or \(\leq\), the boundary line should be solid. But Graph A has a dashed line, so the boundary line type is incorrect. Also, let's rewrite the inequality in slope - intercept form (\(y=mx + b\)):
\(2x-3y\geq - 6\)
\(-3y\geq - 2x - 6\)
\(y\leq\frac{2}{3}x + 2\)
The slope of the line should be \(\frac{2}{3}\), but we need to check the graph. However, the first error is the boundary line type (dashed instead of solid).

Step2: Analyze Graph B

The inequality is \(2x + 4y>8\). Rewrite it as \(y>-\frac{1}{2}x + 2\). The boundary line should be dashed (since the inequality is \(>\)). Let's check the shading. Let's test the point \((0,0)\): \(2(0)+4(0)=0 ot>8\), so the side not containing \((0,0)\) should be shaded. But we need to check the graph. Wait, the inequality \(2x + 4y>8\) in slope - intercept form is \(y>-\frac{1}{2}x + 2\). Let's check the boundary line equation. If \(x = 0\), \(y = 2\); if \(y=0\), \(x = 4\). The graph of Graph B: let's see the line. The line in Graph B: when \(x = 0\), \(y = 2\); when \(y = 0\), \(x=4\)? Wait, no, the line in Graph B seems to have a different slope. Wait, the inequality is \(2x + 4y>8\), dividing by 2: \(x + 2y>4\), or \(y>-\frac{1}{2}x + 2\). The slope is \(-\frac{1}{2}\). The line in Graph B: let's check the slope. From the graph, the line goes from \((-3,4)\) to \((4,0)\), the slope is \(\frac{0 - 4}{4+3}=-\frac{4}{7} eq-\frac{1}{2}\). Also, the shading: the inequality is \(y>-\frac{1}{2}x + 2\), so above the line. But the graph is shaded below? Wait, no, the graph of Graph B is shaded above? Wait, the grid: let's re - express. Maybe the error in Graph B is the boundary line equation (incorrect slope) or shading. But let's go back to the original problem's correct graphs.

Step3: Analyze Graph C

The inequality is \(6x + 4y\leq24\). Rewrite as \(y\leq-\frac{3}{2}x + 6\). The boundary line should be solid (since \(\leq\)). Let's check the line: when \(x = 0\), \(y = 6\); when \(y = 0\), \(x = 4\). The graph of Graph C: when \(x = 0\), \(y = 6\); when \(y = 0\), \(x = 4\) (the line goes from \((0,6)\) to \((4,0)\)), slope is \(-\frac{6}{4}=-\frac{3}{2}\), correct. The boundary line is solid (correct for \(\leq\)), and the shading: let's test \((0,0)\): \(6(0)+4(0)=0\leq24\), so the side containing \((0,0)\) should be shaded. In Graph C, the shading is on the side of \((0,0)\) (since \((0,0)\) is in the shaded region), and the boundary line is solid. So Graph C is correct.

Step4: Analyze Graph D

The inequality is \(x-2y<-2\). Rewrite as \(-2y<-x - 2\), then \(y>\frac{1}{2}x + 1\) (remember to reverse the inequality when dividing by a negative number). The boundary line should be dashed (since the inequality is \(>\)). But Graph D has a solid line, so the boundary line type is incorrect. Also, the slope of the line: the inequality \(y>\frac{1}{2}x + 1\) has a slope of \(\frac{1}{2}\). Let's check the graph: the line in Graph D has a slope of \(\frac{1 - (-1)}{0+3}=\frac{2}{3} eq\frac{1}{2}\). So the boundary line is incorrect (solid instead of dashed, and incorrect slope).

Now, the correct graphs: Graph C is correct. Let's check Graph A again. The inequality is \(2x-3y\geq - 6\), or \(y\leq\frac{2}{3}x + 2\). The boundary line should be solid (since \(\geq\)), but Graph A has a dashed line. So Graph A is incorrect. Graph D: boundary line is solid (should be dashed) and incorrect slope. Graph B: incorrect boundary line equation (slope) and maybe shading. So the correct graphs are Graph C (and maybe another? Wait, the "Letters of Correct Graphs" has two blanks filled: B and C? Wait, no, the original paper has "Letters of Correct Graphs" with A and C? Wait, no, the user's paper: "Letters of Correct Graphs:" has A and C written? Wait, no, the image shows "Letters of Correct Graphs:" with two blanks filled, maybe A and C? Wait, no, let's re - check.

Wait, the problem is to find correct graphs. Let's re - evaluate each graph:

Graph A: Inequality \(2x-3y\geq - 6\)

Rewrite as \(y\leq\frac{2}{3}x + 2\). The boundary line should be solid (because of \(\geq\) in the original inequality, when we solve for \(y\), it's \(\leq\), so the boundary line for \(y\leq\) is solid). But Graph A has a dashed line. So Graph A is incorrect.

Graph B: Inequality \(2x + 4y>8\)

Rewrite as \(y>-\frac{1}{2}x + 2\). The boundary line should be dashed (because of \(>\)). Let's check the line: when \(x = 0\), \(y = 2\); when \(y = 0\), \(x = 4\) (since \(2x+4(0)=8\Rightarrow x = 4\); \(2(0)+4y=8\Rightarrow y = 2\)). The line in Graph B: let's check the points. The line in Graph B goes from \((-3,4)\) to \((4,0)\), which is not the line \(2x + 4y = 8\) (since when \(x=-3\), \(2(-3)+4y=8\Rightarrow - 6 + 4y=8\Rightarrow 4y = 14\Rightarrow y = 3.5 eq4\); when \(x = 4\), \(2(4)+4y=8\Rightarrow 8 + 4y=8\Rightarrow y = 0\), so the point \((4,0)\) is correct, but \((-3,4)\) is not. So the boundary line is graphed incorrectly. Also, the shading: for \(y>-\frac{1}{2}x + 2\), we test \((0,0)\): \(2(0)+4(0)=0 ot>8\), so we shade above the line. The graph of Graph B is shaded above? Wait, the grid: the graph of Graph B is shaded above the line? Maybe, but the boundary line is incorrect. So Graph B is incorrect.

Graph C: Inequality \(6x + 4y\leq24\)

Rewrite as \(y\leq-\frac{3}{2}x + 6\). The boundary line: when \(x = 0\), \(y = 6\); when \(y = 0\), \(x = 4\) (since \(6x+4(0)=24\Rightarrow x = 4\); \(6(0)+4y=24\Rightarrow y = 6\)). The line in Graph C goes from \((0,6)\) to \((4,0)\), which is correct. The boundary line is solid (correct for \(\leq\)). The shading: test \((0,0)\): \(6(0)+4(0)=0\leq24\), so we shade below the line (since \(y\leq-\frac{3}{2}x + 6\)). The graph of Graph C is shaded below the line, which is correct. So Graph C is correct.

Graph D: Inequality \(x - 2y<-2\)

Rewrite as \(y>\frac{1}{2}x + 1\) (divide both sides by \(-2\), reverse inequality). The boundary line should be dashed (because of \(>\)). But Graph D has a solid line. Also, rewrite the inequality as \(x-2y + 2<0\), or \(y=\frac{1}{2}x + 1\) is the boundary line. The slope of the line in Graph D: from the graph, the line goes from \((-3,-1)\) to \((3,2)\), slope is \(\frac{2+1}{3 + 3}=\frac{3}{6}=\frac{1}{2}\), which is correct. But the boundary line is solid (should be dashed), and the shading: for \(y>\frac{1}{2}x + 1\), we shade above the line. But the graph of Graph D is shaded below? Wait, no, the graph of Graph D: the line is solid, and the shading is on the side. Wait, the inequality is \(x-2y<-2\), let's test \((0,0)\): \(0-0=-2 ot<-2\), so we should shade the side not containing \((0,0)\). The line \(y=\frac{1}{2}x + 1\) passes through \((0,1)\) and \((2,2)\). The point \((0,0)\) is below the line, so we should shade above the line. But Graph D's line is solid (should be dashed) and maybe shading is incorrect.

So the correct graphs are Graph C (and maybe Graph B? Wait, no, Graph B's boundary line: \(2x + 4y>8\), the line \(2x + 4y = 8\) has \(x\) - intercept 4 and \(y\) - intercept 2. The line in Graph B: when \(x = 0\), \(y = 2\); when \(y = 0\), \(x = 4\)? Wait, the line in Graph B: from the graph, the line goes from \((-3,4)\) to \((4,0)\), which is not the line \(2x + 4y = 8\) (since at \(x=-3\), \(2(-3)+4y=8\Rightarrow y = 3.5\), not 4). So Graph B's boundary line is incorrect.

So the correct graphs are Graph C (and maybe Graph A? No, Graph A has a dashed line). Wait, the "Letters of Correct Graphs" in the image has A and C written? Maybe the student made a mistake, but according to our analysis, Graph C is correct.

Answer:

Correct Graphs: C (and maybe B? Wait, no, based on analysis, Graph C is correct. If we consider the "Letters of Correct Graphs" in the image has A and C, but our analysis shows Graph C is correct. So the answer is Graph C (and maybe another, but based on the problem, the correct graph is C (and possibly B, but more likely C)).