Question

all possible solutions.

1. \\(\

$$\begin{cases} y \\leq -\\dfrac{2}{3}x - 3 \\\\ y \\geq 3x + 8 \\end{cases}$$

\\)

1. \\(\

$$\begin{cases} 7x + 3y \\leq -24 \\\\ x + 3y \\leq -6 \\end{cases}$$

\\)

1. \\(\

$$\begin{cases} y < 2x + 5 \\\\ 2x - y \\leq 3 \\end{cases}$$

\\)

Explanation:

Problem 2:

Step 1: Analyze \( y \leq -\frac{2}{3}x - 3 \)

The inequality is in slope - intercept form \( y=mx + b \), where the slope \( m =-\frac{2}{3}\) and the y - intercept \( b=-3 \). Since the inequality is \( y\leq\), we draw a solid line (because the inequality includes equality) and shade the region below the line.

Step 2: Analyze \( y\geq3x + 8 \)

This is also in slope - intercept form with slope \( m = 3\) and y - intercept \( b = 8\). Since the inequality is \( y\geq\), we draw a solid line and shade the region above the line.

Step 3: Find the intersection

To find if there is a solution, we can try to find the intersection of the two lines \( y=-\frac{2}{3}x - 3\) and \( y = 3x+8\). Set \( -\frac{2}{3}x-3=3x + 8\). Multiply through by 3 to get \( - 2x-9 = 9x+24\). Then, \( -2x-9x=24 + 9\), \( - 11x=33\), so \( x=-3\). Substitute \( x = - 3\) into \( y=3x + 8\), we get \( y=3\times(-3)+8=-9 + 8=-1\). Now, check the inequalities at the point of intersection. For \( y\leq-\frac{2}{3}x - 3\), when \( x=-3,y=-1\), \( -\frac{2}{3}\times(-3)-3=2 - 3=-1\), so \( y=-1\leq - 1\) (true). For \( y\geq3x + 8\), when \( x=-3,y=-1\), \( 3\times(-3)+8=-1\), so \( y=-1\geq - 1\) (true). But we need to check the regions. The region below \( y =-\frac{2}{3}x-3\) and above \( y = 3x + 8\). Let's take a test point, say \( x = 0\). For \( y\leq-\frac{2}{3}(0)-3=-3\) and \( y\geq3(0)+8 = 8\). A number can't be both less than or equal to - 3 and greater than or equal to 8. So the solution set is empty? Wait, no, the lines intersect at \( (-3,-1)\), but the regions defined by the inequalities: the region below \( y=-\frac{2}{3}x - 3\) and above \( y = 3x+8\). Let's check the slope: the first line has a negative slope, the second has a positive slope. The region below \( y =-\frac{2}{3}x-3\) is going down from left to right, and the region above \( y = 3x + 8\) is going up from left to right. The only point that is on both lines is \( (-3,-1)\), but does this point satisfy both inequalities? \( y=-1\leq-\frac{2}{3}\times(-3)-3=2 - 3=-1\) (yes) and \( y=-1\geq3\times(-3)+8=-9 + 8=-1\) (yes). But are there any other points? Let's take a point to the left of \( x=-3\), say \( x=-4\). For \( y\leq-\frac{2}{3}\times(-4)-3=\frac{8}{3}-3=-\frac{1}{3}\) and \( y\geq3\times(-4)+8=-12 + 8=-4\). A point like \( (-4,-2)\): \( -2\leq-\frac{1}{3}\) (true) and \( -2\geq - 4\) (true). Wait, I made a mistake earlier. Let's re - evaluate. The line \( y = 3x+8\) at \( x=-4\) is \( y=-12 + 8=-4\), and the line \( y=-\frac{2}{3}x-3\) at \( x = - 4\) is \( y=\frac{8}{3}-3=-\frac{1}{3}\). So the region between the two lines? Wait, no: \( y\leq-\frac{2}{3}x - 3\) (below the first line) and \( y\geq3x + 8\) (above the second line). The first line has a slope of \( -\frac{2}{3}\), the second has a slope of 3. The two lines intersect at \( (-3,-1)\). For \( x<-3\), the line \( y = 3x + 8\) is below the line \( y=-\frac{2}{3}x-3\)? Let's take \( x=-5\). \( y=3\times(-5)+8=-7\), \( y=-\frac{2}{3}\times(-5)-3=\frac{10}{3}-3=\frac{1}{3}\). So \( -7\leq\frac{1}{3}\) (true) and \( -7\geq - 7\) (if we use \( x=-5\) in \( y = 3x + 8\), \( y=-7\)). Wait, actually, to graph:

• Graph \( y =-\frac{2}{3}x-3\) (solid line), shade below.
• Graph \( y = 3x + 8\) (solid line), shade above.

The intersection of the two shaded regions is the set of points that are below \( y=-\frac{2}{3}x-3\) and above \( y = 3x + 8\). The lines intersect at \( (-3,-1)\), and for \( x<-3\), the region above \( y = 3x + 8\) and below \( y=-\frac{2}{3}x-3\) exists. Wait, when \( x=-3\), both lines meet at \( (-3,-…

Step 1: Rewrite the inequalities in slope - intercept form

For \( 7x+3y\leq - 24\), solve for \( y\): \( 3y\leq - 7x-24\), so \( y\leq-\frac{7}{3}x-8\). The slope \( m=-\frac{7}{3}\), y - intercept \( b=-8\). Since the inequality is \( y\leq\), we draw a solid line and shade below.
For \( x + 3y\leq - 6\), solve for \( y\): \( 3y\leq - x-6\), so \( y\leq-\frac{1}{3}x - 2\). The slope \( m=-\frac{1}{3}\), y - intercept \( b=-2\). Since the inequality is \( y\leq\), we draw a solid line and shade below.

Step 2: Find the intersection of the two lines

Set \( -\frac{7}{3}x-8=-\frac{1}{3}x-2\). Multiply through by 3: \( - 7x-24=-x - 6\). Then, \( -7x+x=-6 + 24\), \( - 6x=18\), so \( x=-3\). Substitute \( x = - 3\) into \( y=-\frac{1}{3}x-2\), we get \( y=-\frac{1}{3}\times(-3)-2=1 - 2=-1\).

Step 3: Determine the shaded region

We need to find the region that is below both \( y =-\frac{7}{3}x-8\) and \( y=-\frac{1}{3}x-2\). Take a test point, say \( (0,0)\). For \( y\leq-\frac{7}{3}(0)-8=-8\), \( 0\leq - 8\) (false). For \( y\leq-\frac{1}{3}(0)-2=-2\), \( 0\leq - 2\) (false). Let's take a point in the third quadrant, say \( x=-6\). For \( y\leq-\frac{7}{3}\times(-6)-8 = 14 - 8 = 6\) and \( y\leq-\frac{1}{3}\times(-6)-2=2 - 2 = 0\). A point like \( (-6,-5)\): \( -5\leq6\) (true for \( 7x + 3y\leq - 24\): \( 7\times(-6)+3\times(-5)=-42-15=-57\leq - 24\) (true)) and \( -5\leq0\) (true for \( x + 3y\leq - 6\): \( -6+3\times(-5)=-6 - 15=-21\leq - 6\) (true)). The two lines intersect at \( (-3,-1)\). The region of solution is the region that is below both lines, which is bounded by the two lines and the intersection point. To graph:

• Draw \( y=-\frac{7}{3}x-8\) (solid, slope - 7/3, y - intercept - 8) and shade below.
• Draw \( y=-\frac{1}{3}x-2\) (solid, slope - 1/3, y - intercept - 2) and shade below.

The overlapping region (the solution set) is the region that is below both lines, with the boundary being the two lines and their intersection at \( (-3,-1)\).

Problem 6:

Step 1: Rewrite the inequalities

The first inequality is \( y<2x + 5\). It is in slope - intercept form with slope \( m = 2\) and y - intercept \( b = 5\). Since the inequality is \( y<\), we draw a dashed line (because the inequality does not include equality) and shade the region below the line.
The second inequality is \( 2x-y\leq3\). Solve for \( y\): \( -y\leq - 2x + 3\), multiply both sides by - 1 (and reverse the inequality sign) to get \( y\geq2x-3\). It is in slope - intercept form with slope \( m = 2\) and y - intercept \( b=-3\). Since the inequality is \( y\geq\), we draw a solid line and shade the region above the line.

Step 2: Analyze the slopes

Both lines \( y = 2x+5\) (dashed) and \( y=2x - 3\) (solid) have the same slope (\( m = 2\)), so they are parallel.

Step 3: Determine the solution region

We need to find the region that is below \( y = 2x+5\) (dashed line, shade below) and above \( y=2x - 3\) (solid line, shade above). Let's take a test point, say \( (0,0)\). For \( y<2x + 5\), \( 0<0 + 5\) (true). For \( y\geq2x-3\), \( 0\geq0 - 3\) (true). So the region between the two parallel lines (below \( y = 2x+5\) and above \( y=2x - 3\)) is the solution set. The distance between the two lines can be calculated, but for graphing:

• Draw \( y = 2x+5\) as a dashed line (slope 2, y - intercept 5) and shade below.
• Draw \( y=2x - 3\) as a solid line (slope 2, y - intercept - 3) and shade above.

The overlapping region (the solution set) is the region between these two parallel lines.

(Note: Since the problem says "graph all possible solutions", the above steps describe how to graph each system of inequalities. If you need to present the final answer as the description of the graph, for problem 2: The solution is the region bounded by the lines \( y =-\frac{2}{3}x-3\) (solid, shade below) and \( y = 3x + 8\) (solid, shade above) with intersection at \((-3,-1)\). For problem 4: The solution is the region bounded by the lines \( y=-\frac{7}{3}x-8\) (solid, shade below) and \( y=-\frac{1}{3}x-2\) (solid, shade below) with intersection at \((-3,-1)\). For problem 6: The solution is the region between the parallel lines \( y = 2x+5\) (dashed, shade below) and \( y=2x - 3\) (solid, shade above).)

Answer:

Step 1: Rewrite the inequalities

The first inequality is \( y<2x + 5\). It is in slope - intercept form with slope \( m = 2\) and y - intercept \( b = 5\). Since the inequality is \( y<\), we draw a dashed line (because the inequality does not include equality) and shade the region below the line.
The second inequality is \( 2x-y\leq3\). Solve for \( y\): \( -y\leq - 2x + 3\), multiply both sides by - 1 (and reverse the inequality sign) to get \( y\geq2x-3\). It is in slope - intercept form with slope \( m = 2\) and y - intercept \( b=-3\). Since the inequality is \( y\geq\), we draw a solid line and shade the region above the line.

Step 2: Analyze the slopes

Both lines \( y = 2x+5\) (dashed) and \( y=2x - 3\) (solid) have the same slope (\( m = 2\)), so they are parallel.

Step 3: Determine the solution region

We need to find the region that is below \( y = 2x+5\) (dashed line, shade below) and above \( y=2x - 3\) (solid line, shade above). Let's take a test point, say \( (0,0)\). For \( y<2x + 5\), \( 0<0 + 5\) (true). For \( y\geq2x-3\), \( 0\geq0 - 3\) (true). So the region between the two parallel lines (below \( y = 2x+5\) and above \( y=2x - 3\)) is the solution set. The distance between the two lines can be calculated, but for graphing:

• Draw \( y = 2x+5\) as a dashed line (slope 2, y - intercept 5) and shade below.
• Draw \( y=2x - 3\) as a solid line (slope 2, y - intercept - 3) and shade above.

The overlapping region (the solution set) is the region between these two parallel lines.

(Note: Since the problem says "graph all possible solutions", the above steps describe how to graph each system of inequalities. If you need to present the final answer as the description of the graph, for problem 2: The solution is the region bounded by the lines \( y =-\frac{2}{3}x-3\) (solid, shade below) and \( y = 3x + 8\) (solid, shade above) with intersection at \((-3,-1)\). For problem 4: The solution is the region bounded by the lines \( y=-\frac{7}{3}x-8\) (solid, shade below) and \( y=-\frac{1}{3}x-2\) (solid, shade below) with intersection at \((-3,-1)\). For problem 6: The solution is the region between the parallel lines \( y = 2x+5\) (dashed, shade below) and \( y=2x - 3\) (solid, shade above).)