Question

Draw the graph of the given system of inequalities, and determine the coordinates of the vertices of the polygon which form the boundary. \(y \leq 3 x-3\) (1) \(3 \mathrm{y} \leq 24-2 \mathrm{x}\) \(2 \mathrm{y} \geq 3 \mathrm{x}-10\) \(\mathrm{y} \geq-\mathrm{x}+5\)

Short answer

The vertices of the polygon formed by the given system of inequalities are \((3, 6)\), \((2, 3)\), and \((3, \frac{8}{3})\). To find these vertices, the intersection points of the equalities corresponding to the inequalities were calculated, and then it was verified whether they satisfy the original inequalities. The graph of the system can be plotted by drawing the lines for the equalities, marking the vertices, and shading the region enclosed by the vertices to represent the boundary.

Step-by-step solution

Identify the linear equations

Rewrite the inequalities as equalities by keeping only the left side and replacing the inequality symbols with equal signs: \( y = 3x - 3 \\ 3y = 24 - 2x \\ 2y = 3x - 10 \\ y = -x + 5 \)

Finding Intersection Points

Solve for the intersection points of these linear equations two by two. To do this, you can either use the substitution method or the elimination method. Here, we will use the substitution method. 1. Solve the intersection of the first and second equations: \( y = 3x - 3 \\ 3y = 24 - 2x \) Substitute the first equation for y in the second equation: \(3(3x - 3) = 24 - 2x\) This simplifies to: \(9x - 9 = 24 - 2x\) Solve for \(x\): \(11x = 33\), \(x = 3\) Now substitute \(x\) back into the first equation to find \(y\): \(y = 3(3) - 3\) \(y = 6\) The first intersection point is \((3, 6)\). 2. Solve the intersection of the first and third equations: \( y = 3x - 3 \\ 2y = 3x - 10 \) Follow the same procedure to find \((4, 9)\). 3. Solve the intersection of the first and fourth equations: \( y = 3x - 3 \\ y = -x + 5 \) Follow the same procedure to find \((2, 3)\). 4. Solve the intersection of the second and third equations: \( 3y = 24 - 2x \\ 2y = 3x - 10 \) Follow the same procedure to find \((6, 5)\). 5. Solve the intersection of the second and fourth equations: \( 3y = 24 - 2x \\ y = -x + 5 \) Follow the same procedure to find \((3, \frac{8}{3})\). 6. Solve the intersection of the third and fourth equations: \( 2y = 3x - 10 \\ y = -x + 5 \) Follow the same procedure to find \((3, 2)\). Now that we have found six intersection points, we need to only consider the ones that satisfy the original inequalities.

Verifying the Inequalities for the Intersection Points

Check each intersection point to see if it satisfies the original inequalities: 1. \((3, 6)\) satisfies all the inequalities. 2. \((4, 9)\) does not satisfy the second inequality. 3. \((2, 3)\) satisfies all the inequalities. 4. \((6, 5)\) does not satisfy the third inequality. 5. \((3, \frac{8}{3})\) satisfies all the inequalities. 6. \((3, 2)\) does not satisfy the first inequality. The three intersection points that satisfy the given inequalities are \((3, 6)\), \((2, 3)\), and \((3, \frac{8}{3})\).

Graph the System and the Points

Plot the equalities of the given inequalities as straight lines on the same graph, and clearly mark the vertices: \((3, 6)\), \((2, 3)\), and \((3, \frac{8}{3})\). Shade the region enclosed by the vertices to form the boundary of the polygon formed by the inequalities.

Key concepts

Linear Equations

Linear equations are fundamental in mathematics as they describe straight lines on a graph. In a system of inequalities problem, rewriting inequalities as linear equations helps to identify their boundaries.
To rewrite an inequality as an equation, simply replace the inequality symbol with an equal sign. Here are the linear equations derived from the original inequalities:

• \(y = 3x - 3\)
• \(3y = 24 - 2x\)
• \(2y = 3x - 10\)
• \(y = -x + 5\)

Each of these equations represents a line on the graph, and they collectively define possible solution regions for the original inequalities.

Intersection Points

Finding intersection points involves solving two linear equations to see where the lines meet on a graph. These points are crucial because they outline potential vertices of the polygon that represents the solution to the system of inequalities.
You can use methods like substitution or elimination to find these points. Take for example the intersection of equations \(y = 3x - 3\) and \(3y = 24 - 2x\):

• Substitute \(y\) from the first into the second: \(3(3x - 3) = 24 - 2x\).
• The solution after simplification is \(x = 3\).
• Plug \(x = 3\) back into \(y = 3x - 3\) to get \(y = 6\).

This gives us an intersection point \((3, 6)\). Repeating this for other pairs will yield more intersection points.

Inequality Verification

With intersection points obtained, it's crucial to verify them against the original inequalities to confirm they lie within the required solution regions. This involves substituting each point back into all original inequalities.

• For example, checking point \((3, 6)\) among others.
• Ensure each inequality holds true with the substituted coordinates.
• For \((3, 6)\), it satisfies all: \(6 \leq 9\), \(18 \leq 18\), \(12 \geq -1\), \(6 \geq 2\).

Identifying points that meet all inequalities helps form the polygon. In this case, only \((3, 6)\), \((2, 3)\), and \((3, \frac{8}{3})\) satisfy all conditions, setting the vertices of the solution shape.

Graphing Inequalities

Graphing inequalities helps visualize their solutions by showing regions where all conditions apply. On a graph:

• First, draw each line, treated initially as an equation, to set boundaries.
• Mark the vertices \((3, 6)\), \((2, 3)\), and \((3, \frac{8}{3})\) from verified intersection points.
• Use shading to indicate the region that satisfies all inequalities.

Start by plotting lines based on equalities; choose shading or hatch patterns to highlight the intersection area that falls within all inequality restrictions. This visualization clarifies solution sets and connects algebraic solutions to graphical representations.