Welcome to the world of mathematics, where Big Ideas Math Chapter 6 Answer Key serves as your ultimate guide. This comprehensive resource is designed to empower you with the knowledge and skills necessary to conquer the challenges of Chapter 6.

Delve into the intricacies of solving equations and inequalities with variables on both sides, unravel the mysteries of substitution and elimination methods for systems of equations, and master the art of tackling systems of inequalities. With clear explanations, illustrative examples, and practice problems, this answer key transforms mathematical concepts into accessible and manageable building blocks.

Overview of Big Ideas Math Chapter 6

Chapter 6 of Big Ideas Math delves into the intriguing world of geometry, focusing on the fundamentals of shapes, their properties, and their applications in real-world scenarios.

The chapter comprehensively covers various geometric concepts, including:

Polygons

• Definition and classification of polygons based on the number of sides and angles.
• Properties of different types of polygons, such as triangles, quadrilaterals, and pentagons.

Triangles

• Classification of triangles based on side lengths (scalene, isosceles, equilateral) and angle measures (acute, obtuse, right).
• Properties of triangles, including angle sum property, exterior angle property, and triangle inequality theorem.

Quadrilaterals

• Classification of quadrilaterals based on side lengths and angle measures (parallelograms, rectangles, squares, trapezoids).
• Properties of different types of quadrilaterals, such as opposite sides being parallel, diagonals bisecting each other, and angle sum property.

Circles

• Definition and properties of circles, including radius, diameter, and circumference.
• Relationships between the radius, diameter, and circumference of a circle.

Area and Perimeter, Big ideas math chapter 6 answer key

• Formulas for calculating the area and perimeter of polygons, including triangles, quadrilaterals, and circles.
• Applications of area and perimeter in real-world scenarios, such as calculating the area of a garden or the perimeter of a fence.

Section 1: Solving Equations with Variables on Both Sides

Solving equations with variables on both sides involves isolating the variable on one side of the equation to determine its value. The steps to solve such equations include:

1. Simplify both sides of the equation by performing basic operations (adding, subtracting, multiplying, or dividing) to isolate the variable on one side.
2. Combine like terms on both sides to simplify the equation further.
3. Check the solution by substituting the obtained value back into the original equation to ensure it holds true.

Example 1:

Solve for x: 3x + 5 = 2x – 1

Solution:

1. Subtract 2x from both sides: 3x
   • 2x + 5 = 2x
   • 1
   • 2x
2. Simplify: x + 5 =
   

   1

3. Subtract 5 from both sides: x + 5
   • 5 =
   • 1
   • 5
4. Simplify: x =
   

   6

Checking:

Substituting x = -6 back into the original equation:

3(-6) + 5 = 2(-6) – 1

-18 + 5 = -12 – 1

-13 = -13

Since the equation holds true, the solution x = -6 is correct.

Section 2: Solving Inequalities with Variables on Both Sides

Solving inequalities with variables on both sides is similar to solving equations with variables on both sides. However, there are a few key differences to keep in mind.First, when solving inequalities, we are not looking for an exact solution. Instead, we are looking for a range of values that satisfy the inequality.Second,

when we multiply or divide both sides of an inequality by a negative number, we need to reverse the direction of the inequality.Finally, when we graph an inequality, we use a shaded region to represent the range of values that satisfy the inequality.

Solving Inequalities with Variables on Both Sides

To solve an inequality with variables on both sides, we can use the same steps that we use to solve an equation with variables on both sides.

• Simplify both sides of the inequality.
• Isolate the variable term on one side of the inequality.
• Solve for the variable.
• Check your solution by plugging it back into the original inequality.

Example

Solve the inequality:“`

• x
• 5 > x + 3

“`

1. Simplify both sides of the inequality

“`

• x
• 5 > x + 3
• x
• x > 3 + 5

x > 8“`

2. Check your solution by plugging it back into the original inequality

“`

• (8)
• 5 > 8 + 3
• 16
• 5 > 8 + 3
• > 11

“`The solution checks, so x > 8 is the solution to the inequality.

Checking the Solution to an Inequality

To check the solution to an inequality, we can plug it back into the original inequality. If the inequality is true, then the solution is correct.If the inequality is not true, then the solution is incorrect.In the example above, we plugged x = 8 back into the original inequality and found that the inequality was true.

Therefore, x = 8 is the correct solution to the inequality.

Section 3: Solving Systems of Equations by Substitution

Solving systems of equations by substitution involves replacing one variable in one equation with an equivalent expression from another equation. This allows us to solve for one variable and then substitute that value back into the other equation to find the value of the remaining variable.

To solve a system of equations by substitution, follow these steps:

1. Solve one of the equations for one variable.
2. Substitute the expression from step 1 into the other equation.
3. Solve the resulting equation for the remaining variable.
4. Substitute the value from step 3 back into one of the original equations to find the value of the first variable.
5. Check your solution by substituting both values back into both original equations.

Example

Solve the system of equations:

• x + y = 5
• x – y = 1

Step 1: Solve the first equation for x.

x = 5

y

Step 2: Substitute the expression for x into the second equation.

(5

• y)
• y = 1

Step 3: Solve the resulting equation for y.

y = 2

Step 4: Substitute the value of y back into the first equation to find x.

x + 2 = 5

x = 3

Step 5: Check the solution by substituting both values back into both original equations.

+ 2 = 5 (True)

• 3
• 2 = 1 (True)

Therefore, the solution to the system of equations is x = 3 and y = 2.

Section 4: Solving Systems of Equations by Elimination

Solving systems of equations by elimination involves adding or subtracting equations to eliminate one of the variables. This process is used to find the values of the variables that satisfy both equations simultaneously.

Steps Involved in Solving Systems of Equations by Elimination

1. Write the system of equations in standard form (ax + by = c).
2. If the coefficients of one of the variables are opposites, add the equations together to eliminate that variable.
3. If the coefficients of one of the variables are not opposites, multiply one or both equations by a constant to make the coefficients opposites.
4. Add or subtract the equations to eliminate one of the variables.
5. Solve the resulting equation for the remaining variable.
6. Substitute the value of the variable found in step 5 back into one of the original equations to solve for the other variable.

Example

Solve the following system of equations by elimination:

“`x + y = 5

• x
• y = 1

“`

Solution:

Multiply the first equation by 2 to make the coefficients of y opposites:

“`

• (x + y) = 2(5)
• x + 2y = 10

“`

Add the two equations together:

“`(2x + 2y) + (2x

• y) = 10 + 1
• x + y = 11

“`

Solve the resulting equation for x:

“`

• x = 11
• y

x = (11

y)/4

“`

Substitute the value of x back into the first equation to solve for y:

“`(11

y)/4 + y = 5

(11

• y + 4y)/4 = 5
• y = 19

y = 19/3“`

Therefore, the solution to the system of equations is (x, y) = ((11 – 19/3)/4, 19/3).

Checking the Solution

To check the solution, substitute the values of x and y back into both original equations:

“`x + y = 5((11

• 19/3)/4) + 19/3 = 5
• /4 + 19/3 = 5
• = 5 (True)
• x
• y = 1
• ((11
• 19/3)/4)
• 19/3 = 1
• /2
• 19/3 = 1
• = 1 (True)

“`

Since both equations are true, the solution is correct.

Section 5: Solving Systems of Inequalities

Solving systems of inequalities involves finding the values of variables that satisfy all the inequalities in the system. The solution region is the set of all points that satisfy the inequalities.

To solve a system of inequalities, we can follow these steps:

• Graph each inequality on the same coordinate plane.
• Identify the region that satisfies all the inequalities.
• Check the solution by substituting a point from the solution region into each inequality.

Checking the Solution

To check the solution to a system of inequalities, we can substitute a point from the solution region into each inequality. If the point satisfies all the inequalities, then the solution is correct.

For example, consider the system of inequalities:

“`y > x + 1y < 2x - 1 ```

The solution region is the shaded region in the graph below:

[Image of a graph with two lines and a shaded region]

To check the solution, we can substitute the point (2, 3) from the solution region into each inequality:

“`

> 2 + 1 (true)

3 < 2(2) - 1 (true) ```

Since the point (2, 3) satisfies both inequalities, the solution is correct.

Chapter Review: Big Ideas Math Chapter 6 Answer Key

In Chapter 6, we delved into various techniques for solving equations and inequalities with variables on both sides, as well as systems of equations and inequalities.

Key Concepts

*

-*Solving Equations with Variables on Both Sides

We learned methods such as isolating the variable, combining like terms, and using inverse operations to find the solution.

• -*Solving Inequalities with Variables on Both Sides

  We explored the concepts of isolating the variable and using the properties of inequalities to find the solution set.

-*Solving Systems of Equations by Substitution

This method involves solving one equation for a variable and substituting its expression into the other equation to solve for the remaining variable.

-*Solving Systems of Equations by Elimination

This method involves adding or subtracting the equations to eliminate one variable and solve for the other.

-*Solving Systems of Inequalities

We studied the concepts of graphing inequalities and finding the region that satisfies both inequalities simultaneously.

Practice Problems

1. Solve for x

3×5 = 2x + 1

2. Solve the inequality

2(x- 3) < 5x + 1 3. Solve the system of equations: x + y = 5 x - y = 1 4. Solve the system of inequalities: y > x

2

y < 2x + 1

Important Vocabulary

*

-*Equation

A mathematical statement that two expressions are equal.

• -*Inequality

  A mathematical statement that two expressions are not equal.

-*Solution

A value that makes an equation or inequality true.

-*System of Equations

A set of two or more equations that are solved simultaneously.

-*System of Inequalities

A set of two or more inequalities that are solved simultaneously.

FAQ Section

What is the Big Ideas Math Chapter 6 Answer Key?

It is a comprehensive resource that provides step-by-step solutions to all exercises and problems in Big Ideas Math Chapter 6.

How can I use the Big Ideas Math Chapter 6 Answer Key?

Use it as a study guide to check your answers, reinforce your understanding, and identify areas where you need additional practice.

Is the Big Ideas Math Chapter 6 Answer Key available online?

Yes, it is often available on the publisher’s website or other educational platforms.