In Unit 4 Lesson 7 of our math course, students were introduced to a set of practice problems aimed at reinforcing their understanding of the concepts covered in the preceding lessons. These practice problems covered a range of topics, including algebraic equations, graphing, and finding solutions. In this article, we will provide the answer key for these practice problems to help students check their work and assess their understanding of the material.
Starting with algebraic equations, students were tasked with solving equations with one variable. These problems required students to isolate the variable, perform the necessary operations, and arrive at the correct solution. By practicing these types of problems, students develop their algebraic problem-solving skills and enhance their ability to manipulate equations.
The practice problems also involved graphing linear equations on a coordinate plane. Students needed to determine the slope-intercept form of the equation, identify the slope and y-intercept, and plot corresponding points to create a line. Graphing equations helps students visualize the relationship between variables and strengthens their understanding of linear functions.
Lastly, students were presented with problems requiring them to find solutions to equations or word problems by applying the skills and concepts they have learned throughout the unit. These problems challenged students to think critically, analyze information, and determine the most appropriate solution strategy. By practicing with these problems, students gain confidence in their ability to solve real-world math problems.
Overall, the Unit 4 Lesson 7 practice problems provide students with an opportunity to reinforce and apply the skills and concepts covered in the unit. By reviewing the provided answer key, students can evaluate their progress and identify areas for improvement. Regular practice and reinforcement of mathematical concepts are crucial for students to build a strong foundation in math and succeed in more advanced topics.
Unit 4 Lesson 7 Practice Problems Answer Key
Here is the answer key for the practice problems in Unit 4 Lesson 7.
Problem 1:
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Answer: The equation 5x + 3 = 18 can be solved by subtracting 3 from both sides, giving us 5x = 15. Then, we divide both sides by 5, resulting in x = 3. Therefore, the solution is x = 3.
Problem 2:
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Answer: To solve the equation 2(x + 4) = 18, we first distribute the 2 to both terms inside the parentheses, giving us 2x + 8 = 18. Next, we subtract 8 from both sides, which gives us 2x = 10. Lastly, we divide both sides by 2, resulting in x = 5. Therefore, the solution is x = 5.
Problem 3:
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Answer: The equation 3(2x – 1) = 9 can be solved by first distributing the 3 to both terms inside the parentheses, giving us 6x – 3 = 9. Then, we add 3 to both sides, yielding 6x = 12. Finally, we divide both sides by 6, resulting in x = 2. Therefore, the solution is x = 2.
These are the answers to the practice problems in Unit 4 Lesson 7. Make sure to check your work and compare your answers to the answer key to ensure accuracy.
Problem 1: Solving Linear Equations
In algebra, linear equations are mathematical equations in which the highest power of the variable is 1. Solving linear equations is an essential skill in algebra as it allows us to find the value(s) of the variable that make the equation true.
The process of solving a linear equation involves isolating the variable on one side of the equation. This can be done by using inverse operations, which are operations that undo or reverse the effect of another operation. For example, if the equation contains addition, we can use subtraction to isolate the variable. Similarly, if the equation contains multiplication, we can use division.
Let’s consider an example to illustrate the process of solving linear equations. Suppose we have the equation 2x + 5 = 11. To solve for x, we need to isolate it on one side of the equation. We can start by subtracting 5 from both sides of the equation to eliminate the constant term. This gives us 2x = 6. Next, we divide both sides of the equation by 2 to isolate x. The solution is x = 3.
It is important to note that linear equations may have one solution, no solutions, or infinitely many solutions. If the equation simplifies to a false statement like 0 = 5, the equation has no solutions. If the equation simplifies to a true statement like 3 = 3, the equation has infinitely many solutions. In most cases, linear equations have exactly one solution.
Overall, solving linear equations requires using inverse operations to isolate the variable. By understanding the process and practicing various examples, one can become proficient in solving linear equations and apply this skill to solve real-life problems.
Graphing linear equations
Problem 2 in the Unit 4 lesson 7 practice problems involves graphing linear equations. Graphing linear equations is an important skill in algebra as it allows us to visually represent the relationship between two variables. The graph of a linear equation is a straight line, and its equation can be expressed in the form y = mx + b, where m is the slope of the line and b is the y-intercept.
In problem 2, students are given a linear equation and are asked to graph it on a coordinate plane. To graph a linear equation, we need to identify the slope and the y-intercept. The slope tells us how the line is slanting, whether it is going up or down. The y-intercept represents where the line crosses the y-axis.
Once we have identified the slope and y-intercept, we can plot the y-intercept on the y-axis and use the slope to find additional points on the line. For example, if the slope is 2/3, we can see that for every 3 units moved horizontally, the line goes up by 2 units vertically. This helps us to determine additional points on the line, which we can then connect to create the graph.
By graphing linear equations, we can easily see the relationship between the variables and make predictions about their values. It also helps us to solve problems that involve finding the solution to the equation. Overall, graphing linear equations is a valuable tool in understanding and analyzing algebraic relationships.
Problem 3: Finding the slope and y-intercept
In this problem, we are given a linear equation in the form y = mx + b, where m represents the slope and b represents the y-intercept.
To find the slope, we need to compare the change in y-coordinates to the change in x-coordinates between two points on the line. We can choose any two points, calculate the difference in their y-coordinates and x-coordinates, and divide the difference in y-coordinates by the difference in x-coordinates. The resulting quotient will be the value of the slope.
The y-intercept is the value of y when x is equal to 0. To find this, we substitute x = 0 into the equation and solve for y. The resulting value of y will be the y-coordinate of the point where the line intersects the y-axis.
To summarize, to find the slope and y-intercept of a linear equation, we compare the change in y-coordinates to the change in x-coordinates between two points on the line to find the slope, and substitute x = 0 into the equation to find the y-intercept. These values provide important information about the line and can be used to graph the equation or make predictions about the relationship between x and y.
Problem 4: Identifying parallel and perpendicular lines
In problem 4, we are tasked with identifying parallel and perpendicular lines. This requires an understanding of the properties and relationships between lines.
To determine if two lines are parallel, we need to compare their slopes. If the slopes are equal, then the lines are parallel. If the slopes are not equal, then the lines are not parallel.
On the other hand, to determine if two lines are perpendicular, we need to find the negative reciprocal of one line’s slope and compare it to the slope of the other line. If the slopes are negative reciprocals, then the lines are perpendicular. If the slopes are not negative reciprocals, then the lines are not perpendicular.
To solve problem 4, we need to examine the given lines and calculate their slopes. Once we have the slopes, we can compare them to determine if the lines are parallel or perpendicular. Remember to use the formulas for calculating slopes and negative reciprocals to accurately identify the relationships between the lines.
Problem 5: Solving systems of equations
Solving systems of equations is a fundamental concept in algebra. It involves finding the values of variables that satisfy two or more equations. In this problem, we will explore different methods for solving systems of equations.
One method for solving systems of equations is substitution. This involves solving one equation for one variable and substituting that expression into the other equation. By doing this, we can solve for the remaining variable. This method is useful when one equation can be easily solved for a variable.
Another method for solving systems of equations is elimination. This involves adding or subtracting the equations in such a way that one variable is eliminated. By doing this, we can solve for the remaining variable. This method is useful when the coefficients of one variable in both equations are the same.
It is also possible to solve systems of equations graphically. By plotting the equations on a coordinate plane, we can find the intersection point(s) of the lines. This intersection point represents the values of the variables that satisfy both equations.
Overall, solving systems of equations requires understanding various methods and choosing the most appropriate method for each problem. By practicing these methods, we can develop a strong foundation in algebra and problem-solving skills.
Problem 6: Word problems involving linear equations
In this section, we explored word problems that can be solved using linear equations. These problems involve finding unknown quantities by setting up and solving equations.
Throughout the practice problems, we encountered various types of word problems. We learned how to translate the given information into mathematical equations and solve for the unknown variables. Some of the types of word problems we tackled include:
- Finding the sum or difference of two numbers.
- Calculating the perimeter or area of geometric shapes.
- Determining the cost or profit in business scenarios.
- Solving age-related problems.
To solve these problems, we first identified the unknown variable and assigned a variable to represent it. Then, we used the given information to set up an equation using the known quantities. Finally, we solved the equation to find the value of the unknown variable.
This section provided us with useful strategies and techniques for approaching word problems involving linear equations. By practicing these problems, we have improved our problem-solving skills and gained a deeper understanding of how linear equations can be applied in real-life situations.
Overall, mastering word problems involving linear equations is essential for developing mathematical reasoning and problem-solving abilities. These skills will not only benefit us in the classroom but also in our everyday lives as we encounter various situations that require mathematical analysis and problem-solving.
Q&A:
Question 1: A clothing store sells t-shirts for $15 each and jeans for $40 each. Last month, the store made a total of $775 from selling t-shirts and jeans. If the store sold a total of 25 items, how many t-shirts and how many jeans did they sell?
Answer 1: Let’s assume that the store sold x t-shirts and y jeans. From the given information, we can set up two equations: 15x + 40y = 775 (equation 1) and x + y = 25 (equation 2). We can solve this system of linear equations to find the values of x and y, which represent the number of t-shirts and jeans sold, respectively.
Question 2: Julie has $500 in her savings account. She plans to save $50 each month. How many months will it take for her savings account to reach $1000?
Answer 2: Let’s assume that it takes x months for Julie’s savings account to reach $1000. From the given information, we can set up an equation: 500 + 50x = 1000. We can solve this linear equation to find the value of x, which represents the number of months it will take for Julie’s savings account to reach $1000.
Question 3: A rectangular garden is twice as long as it is wide. The perimeter of the garden is 54 feet. What are the dimensions of the garden?
Answer 3: Let’s assume that the width of the garden is x feet. From the given information, we can set up two equations: 2x + 2(2x) = 54 (equation 1) and 2x + 4x = 54 (equation 2). We can solve this system of linear equations to find the values of x and 2x, which represent the width and length of the garden, respectively.
Question 4: A car rental company charges $30 per day to rent a car, plus an additional $0.25 per mile driven. If a customer rented a car for 5 days and drove a total of 200 miles, how much would they need to pay?
Answer 4: Let’s assume that the customer rented the car for x days and drove y miles. From the given information, we can set up an equation: 30x + 0.25y = total amount to pay. We can substitute x = 5 and y = 200 into the equation to find the total amount the customer would need to pay.
Question 5: A small bakery sells cupcakes for $2 each and muffins for $3 each. Last week, the bakery made a total of $40 from selling cupcakes and muffins. If they sold a total of 16 cupcakes and muffins combined, how many of each did they sell?
Answer 5: Let’s assume that the bakery sold x cupcakes and y muffins. From the given information, we can set up two equations: 2x + 3y = 40 (equation 1) and x + y = 16 (equation 2). We can solve this system of linear equations to find the values of x and y, which represent the number of cupcakes and muffins sold, respectively.
Question 1: A toy store sells teddy bears for $20 each and stuffed dogs for $15 each. If the store sells a total of 50 toys and makes $830 in sales, how many teddy bears did they sell?
Answer 1: Let’s denote the number of teddy bears as x and the number of stuffed dogs as y. We can set up a system of linear equations: x + y = 50 (total number of toys) and 20x + 15y = 830 (total sales). We can solve this system of equations to find that the store sold 30 teddy bears.
Question 2: A car rental agency charges $30 per day for renting a car and an additional $0.25 per mile. If a customer rented a car for 5 days and drove a total of 200 miles, how much will they be charged?
Answer 2: Let’s denote the number of days as x and the number of miles driven as y. The total cost can be represented by the equation: 30x + 0.25y = total cost. Substituting x = 5 and y = 200 into the equation, we can find that the customer will be charged $130.