Find the system of linear inequalities that corresponds to the system shown.


-12*x*

 + 

12*y*

  ---Select--- > = ≥ ≤ <  

0

-9*x*

 + 

14*y*

  ---Select--- = ≤ > ≥ <  

-20

3*x*

 + 

2*y*

  ---Select--- < ≥ ≤ > =  

40


Find all the corner points of the feasible region. (Order your answers from smallest to largest *x*, then from smallest to largest *y*.)
(*x*, *y*) = (  ,  ) (smallest *x*-value)
(*x*, *y*) = (  ,  )
(*x*, *y*) = (  ,  ) (largest *x*-value)

### Step 1: Write the system of inequalities

Given the equations with inequalities to be selected:

1. \(-12x + 12y \quad \text{select} \quad 0\)
2. \(-9x + 14y \quad \text{select} \quad -20\)
3. \(3x + 2y \quad \text{select} \quad 40\)

Let's assume plausible inequalities based on how feasible regions usually work (from least to greatest):

1. \(-12x + 12y \geq 0\)
2. \(-9x + 14y \leq -20\)
3. \(3x + 2y \leq 40\)

### Step 2: Simplify inequalities

1. \(-12x + 12y \geq 0 \quad \Rightarrow \quad -x + y \geq 0 \quad \Rightarrow \quad y \geq x\)
2. \(-9x + 14y \leq -20\)
3. \(3x + 2y \leq 40\)

So, the system of inequalities is:

\[
\begin{aligned}
    y &\geq x \\
    -9x + 14y &\leq -20 \\
    3x + 2y &\leq 40
\end{aligned}
\]

### Step 3: Find the corner points of the feasible region

We will find the intersection points of the lines by solving pairs of equations as equalities:

1. **Intersection of** \(y = x\) **and** \(-9x + 14y = -20\)

Substitute \(y = x\) into the second equation:

\[
-9x + 14x = -20 \quad \Rightarrow \quad 5x = -20 \quad \Rightarrow \quad x = -4
\]

Substitute \(x = -4\) into \(y = x\):

\[
y = -4
\]

Thus, one corner point is \((-4, -4)\).

2. **Intersection of** \(y = x\) **and** \(3x + 2y = 40\)

Substitute \(y = x\) into the third equation:

\[
3x + 2x = 40 \quad \Rightarrow \quad 5x = 40 \quad \Rightarrow \quad x = 8
\]

Substitute \(x = 8\) into \(y = x\):

\[
y = 8
\]

Thus, another corner point is \((8, 8)\).

3. **Intersection of** \(-9x + 14y = -20\) **and** \(3x + 2y = 40\)

Solve this system of two linear equations. First, multiply the second equation by 7 to eliminate \(y\):

\[
7(3x + 2y) = 7(40) \quad \Rightarrow \quad 21x + 14y = 280
\]

Now subtract the first equation from this:

\[
(21x + 14y) - (-9x + 14y) = 280 - (-20)
\]

\[
30x = 300 \quad \Rightarrow \quad x = 10
\]

Substitute \(x = 10\) into \(3x + 2y = 40\):

\[
3(10) + 2y = 40 \quad \Rightarrow \quad 30 + 2y = 40 \quad \Rightarrow \quad 2y = 10 \quad \Rightarrow \quad y = 5
\]

Thus, the third corner point is \((10, 5)\).

### Step 4: List the corner points in order

Order by smallest \(x\)-value, then smallest \(y\)-value:

- \((-4, -4)\) (smallest \(x\)-value)
- \((8, 8)\)
- \((10, 5)\) (largest \(x\)-value)

### Final Answer:
- \((x, y) = (-4, -4)\) (smallest \(x\)-value)
- \((x, y) = (8, 8)\)
- \((x, y) = (10, 5)\) (largest \(x\)-value)

Would you like more details on any step, or do you have questions?

---

**Related Questions:**
1. How do you graph systems of linear inequalities?
2. What is the role of feasible regions in linear programming?
3. How do you solve a system of equations algebraically?
4. What are common methods to find intersection points of two lines?
5. What are the applications of corner points in optimization problems?

**Tip:** Always graph the system of inequalities to visually confirm the feasible region and corner points.

Find the system of linear inequalities that corresponds to the system shown.

-12*x* + 12*y* ≥ 0
-9*x* + 14*y* ≤ -20
3*x* + 2*y* ≤ 40

Find all the corner points of the feasible region. (Order your answers from smallest to largest *x*, then from smallest to largest *y*.)
(*x*, *y*) = ( , ) (smallest *x*-value)
(*x*, *y*) = ( , )
(*x*, *y*) = ( , ) (largest *x*-value)

Learn how to find the system of linear inequalities corresponding to given equations and identify all the corner points of the feasible region. This problem covers key concepts such as linear inequalities, systems of equations, and solving for intersections. Suitable for students in Grades 9-12.

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