Answer :
Sure, let's solve the given problem step-by-step.
### 1. Form a System of Inequalities
Given the conditions in the problem:
1. The sum of two numbers (let's denote them as [tex]\( x \)[/tex] and [tex]\( y \)[/tex]) must be 24 or greater:
[tex]\[ x + y \geq 24 \][/tex]
2. The product of the two numbers must be less than 60:
[tex]\[ xy < 60 \][/tex]
So, our system of inequalities is:
[tex]\[ \begin{cases} x + y \geq 24 \\ xy < 60 \end{cases} \][/tex]
### 2. Graph the System of Inequalities
To graph these inequalities, we'll follow these steps:
#### Step 1: Graph the Inequality [tex]\( x + y \geq 24 \)[/tex]
This is a linear inequality. First, graph the boundary line [tex]\( x + y = 24 \)[/tex].
- Find the x-intercept: let [tex]\( y = 0 \)[/tex]
[tex]\[ x + 0 = 24 \Rightarrow x = 24 \][/tex]
- Find the y-intercept: let [tex]\( x = 0 \)[/tex]
[tex]\[ 0 + y = 24 \Rightarrow y = 24 \][/tex]
Now, plot the points (24, 0) and (0, 24) on the graph. Draw a straight line passing through these points.
- Since the inequality is [tex]\( x + y \geq 24 \)[/tex], shade the region above the line, including the line itself as it is 'greater than or equal to'.
#### Step 2: Graph the Inequality [tex]\( xy < 60 \)[/tex]
This is a non-linear inequality. First, graph the boundary line [tex]\( xy = 60 \)[/tex].
To get a sense of the curve:
- When [tex]\( x = 1 \)[/tex], [tex]\( y = 60 \)[/tex]
- When [tex]\( x = 2 \)[/tex], [tex]\( y = 30 \)[/tex]
- When [tex]\( x = 3 \)[/tex], [tex]\( y \approx 20 \)[/tex]
- Continue plotting points to sketch the curve of the hyperbola.
Now, draw the curve [tex]\( xy = 60 \)[/tex].
- Since the inequality is [tex]\( xy < 60 \)[/tex], shade the region below the curve.
#### Step 3: Intersection of the Two Inequalities
To find the solution to the system of inequalities, we identify the region that satisfies both inequalities simultaneously:
1. The region above and including the line [tex]\( x + y = 24 \)[/tex].
2. The region below the hyperbola [tex]\( xy = 60 \)[/tex].
### Graphical Representation
On a Cartesian plane:
- Line [tex]\( x + y = 24 \)[/tex]: Draw a straight line through points (24, 0) and (0, 24). Shade the half-plane above this line.
- Curve [tex]\( xy = 60 \)[/tex]: Draw a hyperbolic curve passing through points like (1, 60), (2, 30), etc. Shade the region under this curve.
The feasible solution area is where these two shaded regions overlap.
### Summary
By shading the regions mentioned above and looking at their intersection, you can find all the pairs of numbers (x, y) that satisfy both conditions:
[tex]\[ \begin{cases} x + y \geq 24 \\ xy < 60 \end{cases} \][/tex]
You should see that this region is where the sum of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] is at least 24 and their product is less than 60. The intersection will form a feasible region meeting both inequalities.
### 1. Form a System of Inequalities
Given the conditions in the problem:
1. The sum of two numbers (let's denote them as [tex]\( x \)[/tex] and [tex]\( y \)[/tex]) must be 24 or greater:
[tex]\[ x + y \geq 24 \][/tex]
2. The product of the two numbers must be less than 60:
[tex]\[ xy < 60 \][/tex]
So, our system of inequalities is:
[tex]\[ \begin{cases} x + y \geq 24 \\ xy < 60 \end{cases} \][/tex]
### 2. Graph the System of Inequalities
To graph these inequalities, we'll follow these steps:
#### Step 1: Graph the Inequality [tex]\( x + y \geq 24 \)[/tex]
This is a linear inequality. First, graph the boundary line [tex]\( x + y = 24 \)[/tex].
- Find the x-intercept: let [tex]\( y = 0 \)[/tex]
[tex]\[ x + 0 = 24 \Rightarrow x = 24 \][/tex]
- Find the y-intercept: let [tex]\( x = 0 \)[/tex]
[tex]\[ 0 + y = 24 \Rightarrow y = 24 \][/tex]
Now, plot the points (24, 0) and (0, 24) on the graph. Draw a straight line passing through these points.
- Since the inequality is [tex]\( x + y \geq 24 \)[/tex], shade the region above the line, including the line itself as it is 'greater than or equal to'.
#### Step 2: Graph the Inequality [tex]\( xy < 60 \)[/tex]
This is a non-linear inequality. First, graph the boundary line [tex]\( xy = 60 \)[/tex].
To get a sense of the curve:
- When [tex]\( x = 1 \)[/tex], [tex]\( y = 60 \)[/tex]
- When [tex]\( x = 2 \)[/tex], [tex]\( y = 30 \)[/tex]
- When [tex]\( x = 3 \)[/tex], [tex]\( y \approx 20 \)[/tex]
- Continue plotting points to sketch the curve of the hyperbola.
Now, draw the curve [tex]\( xy = 60 \)[/tex].
- Since the inequality is [tex]\( xy < 60 \)[/tex], shade the region below the curve.
#### Step 3: Intersection of the Two Inequalities
To find the solution to the system of inequalities, we identify the region that satisfies both inequalities simultaneously:
1. The region above and including the line [tex]\( x + y = 24 \)[/tex].
2. The region below the hyperbola [tex]\( xy = 60 \)[/tex].
### Graphical Representation
On a Cartesian plane:
- Line [tex]\( x + y = 24 \)[/tex]: Draw a straight line through points (24, 0) and (0, 24). Shade the half-plane above this line.
- Curve [tex]\( xy = 60 \)[/tex]: Draw a hyperbolic curve passing through points like (1, 60), (2, 30), etc. Shade the region under this curve.
The feasible solution area is where these two shaded regions overlap.
### Summary
By shading the regions mentioned above and looking at their intersection, you can find all the pairs of numbers (x, y) that satisfy both conditions:
[tex]\[ \begin{cases} x + y \geq 24 \\ xy < 60 \end{cases} \][/tex]
You should see that this region is where the sum of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] is at least 24 and their product is less than 60. The intersection will form a feasible region meeting both inequalities.
