Part A: Solve The Inequality And Graph The Solution On The Number Line: $ -6x + 1 \ \textgreater \ -4x + 7 $ Part B: Solve The System Of Inequalities And Identify Three Points In The Solution Set. $ \begin{align*} -4x + 6y & \

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Introduction


In this section, we will focus on solving a linear inequality and graphing the solution on the number line. The given inequality is βˆ’6x+1>βˆ’4x+7-6x + 1 > -4x + 7. We will use algebraic methods to solve the inequality and then represent the solution set on a number line.

Step 1: Subtract -4x from both sides of the inequality


To begin solving the inequality, we need to isolate the variable xx on one side of the inequality. We can do this by subtracting βˆ’4x-4x from both sides of the inequality.

βˆ’6x+1>βˆ’4x+7-6x + 1 > -4x + 7

Subtracting βˆ’4x-4x from both sides gives us:

βˆ’6x+4x+1>βˆ’4x+4x+7-6x + 4x + 1 > -4x + 4x + 7

Simplifying the left-hand side, we get:

βˆ’2x+1>7-2x + 1 > 7

Step 2: Subtract 1 from both sides of the inequality


Next, we need to isolate the term with the variable xx by subtracting 1 from both sides of the inequality.

βˆ’2x+1>7-2x + 1 > 7

Subtracting 1 from both sides gives us:

βˆ’2x>7βˆ’1-2x > 7 - 1

Simplifying the right-hand side, we get:

βˆ’2x>6-2x > 6

Step 3: Divide both sides of the inequality by -2


To solve for xx, we need to divide both sides of the inequality by βˆ’2-2. However, when we divide or multiply an inequality by a negative number, we need to reverse the direction of the inequality sign.

βˆ’2x>6-2x > 6

Dividing both sides by βˆ’2-2 gives us:

x<βˆ’3x < -3

Step 4: Graph the solution on the number line


The solution to the inequality x<βˆ’3x < -3 can be represented on a number line. We draw a closed circle at βˆ’3-3 and shade the region to the left of βˆ’3-3 to indicate that xx is less than βˆ’3-3.

Conclusion


In this section, we solved the linear inequality βˆ’6x+1>βˆ’4x+7-6x + 1 > -4x + 7 and graphed the solution on the number line. The solution set is x<βˆ’3x < -3, which can be represented on a number line by drawing a closed circle at βˆ’3-3 and shading the region to the left of βˆ’3-3.

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Introduction


In this section, we will focus on solving a system of linear inequalities and identifying three points in the solution set. The given system of inequalities is:

βˆ’4x+6y>12-4x + 6y > 12

2xβˆ’3y<βˆ’62x - 3y < -6

We will use algebraic methods to solve the system of inequalities and then identify three points in the solution set.

Step 1: Solve the first inequality for y


To begin solving the system of inequalities, we can solve the first inequality for yy.

βˆ’4x+6y>12-4x + 6y > 12

Dividing both sides by 6 gives us:

y>12+4x6y > \frac{12 + 4x}{6}

Simplifying the right-hand side, we get:

y>2+x3y > \frac{2 + x}{3}

Step 2: Solve the second inequality for y


Next, we can solve the second inequality for yy.

2xβˆ’3y<βˆ’62x - 3y < -6

Dividing both sides by βˆ’3-3 gives us:

y>2x+63y > \frac{2x + 6}{3}

Step 3: Find the intersection of the two solution sets


To find the solution set of the system of inequalities, we need to find the intersection of the two solution sets. We can do this by setting the two expressions for yy equal to each other and solving for xx.

2+x3=2x+63\frac{2 + x}{3} = \frac{2x + 6}{3}

Multiplying both sides by 3 gives us:

2+x=2x+62 + x = 2x + 6

Subtracting 2x2x from both sides gives us:

βˆ’x=4-x = 4

Dividing both sides by βˆ’1-1 gives us:

x=βˆ’4x = -4

Step 4: Find the corresponding value of y


Now that we have found the value of xx, we can find the corresponding value of yy by substituting x=βˆ’4x = -4 into one of the original inequalities.

y>2+x3y > \frac{2 + x}{3}

Substituting x=βˆ’4x = -4 gives us:

y>2βˆ’43y > \frac{2 - 4}{3}

Simplifying the right-hand side, we get:

y>βˆ’23y > -\frac{2}{3}

Step 5: Identify three points in the solution set


The solution set of the system of inequalities is the region in the xyxy-plane that satisfies both inequalities. We can identify three points in the solution set by choosing values of xx and yy that satisfy both inequalities.

One point in the solution set is (x,y)=(βˆ’4,βˆ’1)(x, y) = (-4, -1), which satisfies both inequalities.

Another point in the solution set is (x,y)=(βˆ’2,1)(x, y) = (-2, 1), which satisfies both inequalities.

A third point in the solution set is (x,y)=(0,2)(x, y) = (0, 2), which satisfies both inequalities.

Conclusion


In this section, we solved the system of linear inequalities βˆ’4x+6y>12-4x + 6y > 12 and 2xβˆ’3y<βˆ’62x - 3y < -6 and identified three points in the solution set. The solution set is the region in the xyxy-plane that satisfies both inequalities, and we can identify three points in the solution set by choosing values of xx and yy that satisfy both inequalities.

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Introduction


In this section, we will address some common questions and concerns related to solving inequalities and systems of inequalities. We will provide detailed explanations and examples to help clarify any confusion.

Q: What is the difference between a linear inequality and a system of linear inequalities?


A: A linear inequality is an inequality that can be written in the form ax+by>cax + by > c, where aa, bb, and cc are constants. A system of linear inequalities is a set of two or more linear inequalities that must be satisfied simultaneously.

Q: How do I solve a linear inequality?


A: To solve a linear inequality, you can use the following steps:

  1. Add or subtract the same value to both sides of the inequality to isolate the variable.
  2. Multiply or divide both sides of the inequality by a positive value to eliminate the coefficient of the variable.
  3. Reverse the direction of the inequality if you multiply or divide both sides by a negative value.

Q: How do I solve a system of linear inequalities?


A: To solve a system of linear inequalities, you can use the following steps:

  1. Solve each inequality separately using the steps outlined above.
  2. Find the intersection of the solution sets of the two inequalities.
  3. Identify three points in the solution set by choosing values of xx and yy that satisfy both inequalities.

Q: What is the difference between a solution set and a graph?


A: A solution set is the set of all possible values of xx and yy that satisfy a system of inequalities. A graph is a visual representation of the solution set, typically using a number line or a coordinate plane.

Q: How do I graph a solution set?


A: To graph a solution set, you can use the following steps:

  1. Draw a number line or a coordinate plane.
  2. Identify the boundary lines of the solution set.
  3. Shade the region that satisfies the inequality.
  4. Identify three points in the solution set by choosing values of xx and yy that satisfy the inequality.

Q: What are some common mistakes to avoid when solving inequalities and systems of inequalities?


A: Some common mistakes to avoid when solving inequalities and systems of inequalities include:

  • Not following the correct order of operations when solving inequalities.
  • Not reversing the direction of the inequality when multiplying or dividing by a negative value.
  • Not identifying the boundary lines of the solution set.
  • Not shading the correct region of the solution set.

Q: How can I practice solving inequalities and systems of inequalities?


A: You can practice solving inequalities and systems of inequalities by working through example problems and exercises. You can also use online resources and tools, such as graphing calculators and inequality solvers, to help you practice and improve your skills.

Conclusion


In this section, we addressed some common questions and concerns related to solving inequalities and systems of inequalities. We provided detailed explanations and examples to help clarify any confusion. By following the steps outlined above and practicing regularly, you can become proficient in solving inequalities and systems of inequalities.