Question

svlc algebra 1a - standard (15260)
solving systems of linear inequalities
identifying the graph of a system of two - variable
linear inequalities
which is the graph of the system (x + 3y > - 3) and (y < \frac{1}{2}x + 1)?

Explanation:

Step1: Analyze \( x + 3y > -3 \)

Rewrite it in slope - intercept form (\(y=mx + b\)):
Subtract \(x\) from both sides: \(3y>-x - 3\)
Divide by 3: \(y>-\frac{1}{3}x - 1\)
The boundary line \(y =-\frac{1}{3}x - 1\) is dashed (because the inequality is \(>\), not \(\geq\)). To find the direction of the shading, we can test a point, say \((0,0)\). Plugging into \(x + 3y>-3\): \(0+3(0)>-3\) which is \(0 > - 3\), true. So we shade above the line \(y=-\frac{1}{3}x - 1\).

Step2: Analyze \(y<\frac{1}{2}x + 1\)

The boundary line \(y=\frac{1}{2}x + 1\) is dashed (because the inequality is \(<\), not \(\leq\)). Test the point \((0,0)\): \(0<\frac{1}{2}(0)+1\) which is \(0 < 1\), true. So we shade below the line \(y = \frac{1}{2}x+1\).

Step3: Match with the graphs

• The line \(y=-\frac{1}{3}x - 1\) has a slope of \(-\frac{1}{3}\) and y - intercept of \(- 1\). The line \(y=\frac{1}{2}x + 1\) has a slope of \(\frac{1}{2}\) and y - intercept of \(1\).
• The correct graph should have a dashed line for \(y>-\frac{1}{3}x - 1\) (shaded above) and a dashed line for \(y<\frac{1}{2}x + 1\) (shaded below). By looking at the slopes and intercepts and the shading regions, we can identify the correct graph.

Looking at the four graphs, the first graph (top - left) has the correct dashed lines and shading: the line with slope \(-\frac{1}{3}\) (dashed, shaded above) and the line with slope \(\frac{1}{2}\) (dashed, shaded below) intersecting in a way that the overlapping region is correctly shaded.

Answer:

The top - left graph (the first one among the four) is the graph of the system \(x + 3y>-3\) and \(y<\frac{1}{2}x + 1\).