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Linear Functions Problems with Solutions
Linear functions are highly used throughout mathematics and are therefore important to understand. A set of problems involving linear functions , along with detailed solutions, are presented. The problems are designed with emphasis on the meaning of the slope and the y intercept.
Problem 1: f is a linear function. Values of x and f(x) are given in the table below; complete the table.
Problem 2: A family of linear functions is given by
Solution to Problem 2: a)
Problem 3: A high school had 1200 students enrolled in 2003 and 1500 students in 2006. If the student population P ; grows as a linear function of time t, where t is the number of years after 2003. a) How many students will be enrolled in the school in 2010? b) Find a linear function that relates the student population to the time t. Solution to Problem 3: a) The given information may be written as ordered pairs (t , P). The year 2003 correspond to t = 0 and the year 2006 corresponds to t = 3, hence the 2 ordered pairs (0, 1200) and (3, 1500) Since the population grows linearly with the time t, we use the two ordered pairs to find the slope m of the graph of P as follows m = (1500 - 1200) / (6 - 3) = 100 students / year The slope m = 100 means that the students population grows by 100 students every year. From 2003 to 2010 there are 7 years and the students population in 2010 will be P(2010) = P(2003) + 7 * 100 = 1200 + 700 = 1900 students. b) We know the slope and two points, we may use the point slope form to find an equation for the population P as a function of t as follows P - P1 = m (t - t1) P - 1200 = 100 (t - 0) P = 100 t + 1200
Problem 4: The graph shown below is that of the linear function that relates the value V (in $) of a car to its age t, where t is the number of years after 2000.
Problem 5: The cost of producing x tools by a company is given by
Problem 6: A 500-liter tank full of oil is being drained at the constant rate of 20 liters par minute. a) Write a linear function V for the number of liters in the tank after t minutes (assuming that the drainage started at t = 0). b) Find the V and the t intercepts and interpret them. e) How many liters are in the tank after 11 minutes and 45 seconds? Solution to Problem 6: After each minute the amount of oil in the tank deceases by 20 liters. After t minutes, the amount of oil in the tank decreases by 20*t liters. Hence if at the start there 500 liters, after t minute the amount V of oil left in the tank is given by V = 500 - 20 t b) To find the V intercept, set t = 0 in the equation V = 500 - 20 t. V = 500 liters : it is the amount of oil at the start of the drainage. To find the t intercept, set V = 0 in the equation V = 500 - 20 t and solve for t. 0 = 500 - 20 t t = 500 / 20 = 25 minutes : it is the total time it takes to drain the 500 liters of oil. c) Convert 11 minutes 45 seconds in decimal form. t = 11 minutes 45 seconds = 11.75 minutes Calculate V at t = 11.75 minutes. V(11.75) = 500 - 20*11.75 = 265 liters are in the tank after 11 minutes 45 seconds of drainage.
Problem 7: A 50-meter by 70-meter rectangular garden is surrounded by a walkway of constant width x meters.
Problem 8: A driver starts a journey with 25 gallons in the tank of his car. The car burns 5 gallons for every 100 miles. Assuming that the amount of gasoline in the tank decreases linearly, a) write a linear function that relates the number of gallons G left in the tank after a journey of x miles. b) What is the value and meaning of the slope of the graph of G? c) What is the value and meaning of the x intercept? Solution to Problem 8: a) If 5 gallons are burnt for 100 miles then (5 / 100) gallons are burnt for 1 mile. Hence for x miles, x * (5 / 100) gallons are burnt. G is then equal to the initial amount of gasoline decreased by the amount gasoline burnt by the car. Hence G = 25 - (5 / 100) x b) The slope of G is equal to 5 / 1000 and it represent the amount of gasoline burnt for a distance of 1 mile. c) To find the x intercept, we set G = 0 and solve for x. 25 - (5 / 100) x = 0 x = 500 miles : it is the distance x for which all 25 gallons of gasoline will be burnt.
Problem 9: A rectangular wire frame has one of its dimensions moving at the rate of 0.5 cm / second. Its width is constant and equal to 4 cm. If at t = 0 the length of the rectangle is 10 cm,
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Linear Functions Worksheets
This collection of linear functions worksheets is a complete package and leaves no stone unturned. Eighth grade and high school students gain practice in identifying and distinguishing between a linear and a nonlinear function presented as equations, graphs and tables. Identify the function rule, complete tables, evaluate, graph, compare and transform linear functions are some topics dealt with here. Start off your functions practice with our free worksheets!
Identify Linear and Nonlinear Functions from Equation
Try this set of linear vs nonlinear functions worksheet pdfs to determine whether a function is linear or not. If the equation can be written in the slope-intercept form, y=mx+b then it is linear.
- Download the set
Is the Function Linear or Nonlinear? | Graph
8th grade students learn to distinguish between linear and nonlinear functions by observing the graphs. The graph of a linear function is a straight line, while the graph of a nonlinear function is a curve.
Is the Function Linear or Nonlinear? | Table
Examine the input(x) and output(y) values of the table inthese linear function worksheets for grade 8. If the rate of change for y with respect to x remains constant, then the table represents a linear function.
Linear Function | Level 1
Substitute the x-values in the linear expression to find the y-values in each function table worksheet. The easy level deals with integers, while the moderate level focuses on fractions and decimals.
Linear Function | Level 2
Completing the function table in this batch of pdf worksheets comes with a twist. Students are expected to plug in either the input or output values in the function rule to determine the missing value.
Finding the Function Rule
Deciphering the pattern that converts the input into an output value in each table and writing it as an expression is all one is expected to do in this set of printable function rule worksheets for high school students.
Graphing Linear Functions Worksheets
Plotting coordinates and graphing functions are the two major learning outcomes of this section of graphing linear equations worksheet printables. The slopes here are expressed as integers and fractions.
(30 Worksheets)
Comparing Linear Functions
Compare two functions presented as tables, graphs and equations in these printable worksheets. Find the slope, x and y-intercepts to complete the table and answer questions based on comparisons.
Transformation of Linear Functions Worksheets
Draw a transformed graph, write a transformed function using horizontal and vertical shifts, stretches, compressions, and reflections too with these transformation of linear functions worksheets.
(25 Worksheets)
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Linear Equations Word Problems Worksheet with Solutions
The first equation family that many students learn about is linear equations. Linear equations have many applications in the real-world, which can make for a really great set of word problems!
As a student studying algebra, you will encounter many linear equations word problems. That’s why I have put together this linear equations word problems worksheet with solutions!
My hope is that this linear equation word problems worksheet and answer key help you deepen your understanding of linear equations and linear systems!
What Are Linear Equations?
A linear equation is an algebraic equation where the highest power on the variable is one. When graphed, a linear equation will produce a straight line.
There are a few ways that we can write linear equations, with two of the most common being slope-intercept form and standard form .
Slope-intercept form is best way to identify the slope of the line and the y-intercept of the line. In general, the equation of a line in slope-intercept form is written as:
In this form, a represents the slope of the line and b represents the y-intercept of the line.
Equations of lines in standard form are easy to recognize because it is a uniformly recognized form of a line. Standard form allows for easy comparison of coefficients. When two linear equations are in standard form, you can quickly compare the coefficients of x and y.
In general, the standard form of a line is written as:
$$Ax+By=C$$
Note that A and B do not represent the slope of the line or the y-intercept in this form. Instead, A and B are simply constants.
Solving Linear Equations
Any set of word problems relating to linear equations will ask you to solve an equation of some sort. However, there are many different types of solving equations problems that you will encounter as you explore linear equations word problems.
Let’s take a look at a few different types to make sure you know what to expect when you check out the linear equation word problems worksheet with solutions below.
Solving Two-Step Equations and Multi-Step Equations
One of the simplest equation problems that you can solve is a two-step equation. A two-step equation requires you to perform just two steps in order to determine the unique solution to the linear equation.
The first step is to identify the side of the equation with the unknown variable. Your goal will be to isolate this variable (or get it by itself). Consider the following example:
In this example, the variable is on the right-hand side of the equation. To isolate x, we “undo” the operations on the right-hand side of the equation using inverse operations. This just means doing whatever the opposite operation is.
We can apply order of operations in reverse to start with the subtraction and then deal with the multiplication. Adding 5 to both sides and then dividing by 2 will result in:
$$\begin{split} 9+5&=2x-5+5 \\ \\ 14 &= 2x \\ \\ \frac{14}{2} &= \frac{2x}{2} \\ \\ 7 &= x \end{split} $$
This shows that the value of the unknown variable here is x = 7. This is a unique solution that will satisfy the equation.
If you want to learn more about finding the solution of linear equations and explore multi-step equations that use the distributive property, check out these equation solving worksheets !
Systems of Linear Equations
Systems of linear equations are another type of problem that you will see on the linear equations word problems worksheet linked below. A system of linear equations involves two (or more!) linear equations that intersect in some way.
There are a few different ways that linear equations can intersect :
- Once at a single point of intersection
- Never as a result of the lines being parallel
- Always as a result of the lines being on top of one another
When solving a system of linear equations, your goal is to determine both unknown variables. If the lines intersect, the solution to the system will be the point of intersection for the lines.
When given a linear equation, we can find the point of intersection between it and a second equation using a few different methods. I made a video on the substitution method and a video on the elimination method to help you understand these strategies for solving systems before you apply them to word problems involving systems of equations.
What Are Linear Equation Word Problems?
A linear equation word problem involves a real-world situation or scenario that can be solved by setting up and solving linear equations. The equations that are used model the relationships between different quantities in the real-world scenario.
The topics of these problems vary, ranging from applications in science and physics (ie. calculate the speed of the boat) to business applications (ie. how many sales are required to break even?). The problems that you encounter will also vary in depth and difficulty.
In my teaching experience, students tend to struggle with word problems because it isn’t always immediately clear what is being asked. I have seen many students feel very confident in their equation solving skills, yet they still struggle when it comes to solving linear equation word problems.
One reason for this is that you aren’t always given equations from the start while solving word problems.
Tips for Solving Linear Equation Word Problems
During my time as a high school math teacher, I have come across a few tips that I think will help you solve linear equation word problems successfully.
To begin, the first step should always be to define two variables. Read the question carefully and think about the quantities involved. Use variables to represent these quantities.
The second step should be writing an equation that models the scenario. Depending on the problem, you may need to write a second equation as well.
Lastly, think about what the problem is asking you to find.
For example, are you looking for the values of two unknown variables? If so, you are likely going to be setting up and solving a system of linear equations.
If you are being asked for the value of a single variable, the chances are you will be solving a single linear equation.
Now that you have a basic understanding of the concepts involved in solving linear equations word problems, it’s time to try a few!
My goal here is to provide you with a worksheet that you can use to practice and feel confident that you understand linear equations word problems!
While I was writing this worksheet, I made sure to include a wide variety of problems that range in difficulty. You will see a few simpler problems involving a two-step equation or multi-step equations, but you will also see a few problems that involve systems of linear equations.
After solving each word problem, be sure to check the answer key to verify that you fully understand the process used to set up the problem and solve it. Reflecting on your understanding is an important part of developing comfort with any given math concept!
Click below to download the linear equations word problems worksheet with solutions!
Using This Linear Equation Word Problems Worksheet
Being able to read a real-world algebra problem and set up a linear equation (or a system of linear equations) to solve it is a very challenging skill. In my experience as a math teacher, many students struggle with this concept, even if they fully understand the mathematics that the problem requires.
This is the main reason that I put together this linear equation word problems worksheet with solutions. My goal is to provide you with a set of word problems that you can use to check your understanding of solving linear equations in the real-world.
I hope you found this practice worksheet helpful as you continue your studies of algebra and linear equations!
If you are looking for more linear equations math worksheets in PDF formats, check out my collection of solving linear inequalities worksheets and this linear inequality word problems worksheet .
Did you find this linear equation word problems worksheet with solutions helpful? Share this post and subscribe to Math By The Pixel on YouTube for more helpful mathematics content!
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Linear Equations Word Problems Worksheets
Linear equations are equations that have two variables and are a straight line when graphed, based on their slope and y-intercept . Hence,linear equations word problems worksheets have a variety of word problems that help students practice key concepts and build a rock-solid foundation of the concepts.
Benefits of Linear Equations Word Problems Worksheets
Linear equations word problems worksheets are a great resource for students to practice a large variety of word type questions. These worksheets are supported by visuals which help students get a crystal clear understanding of the linear equations word type topic. The variety of problems that these worksheets offer helps students approach these concepts in an engaging and fun manner.
Linear equations word problems worksheets come with visual simulation for students to see the problems in action, an answer key that provides a detailed step-by-step solution for students to understand the process better, and a worksheet with detailed solutions.
Download Linear Equations Word Problems Worksheet PDFs
These math worksheets should be practiced regularly and are free to download in PDF formats.
☛ Check Grade wise Linear Equations Word Problems Worksheets
- 8th Grade Linear Equations Worksheets
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1.20: Word Problems for Linear Equations
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Word problems are important applications of linear equations. We start with examples of translating an English sentence or phrase into an algebraic expression.
Example 18.1
Translate the phrase into an algebraic expression:
a) Twice a variable is added to 4
Solution: We call the variable \(x .\) Twice the variable is \(2 x .\) Adding \(2 x\) to 4 gives:
\[4 + 2x\nonumber\]
b) Three times a number is subtracted from 7.
Solution: Three times a number is \(3 x .\) We need to subtract \(3 x\) from 7. This means:\
\[7-3 x\nonumber\]
c) 8 less than a number.
Solution: The number is denoted by \(x .8\) less than \(x\) mean, that we need to subtract 8 from it. We get:
\[x-8\nonumber\]
For example, 8 less than 10 is \(10-8=2\).
d) Subtract \(5 p^{2}-7 p+2\) from \(3 p^{2}+4 p\) and simplify.
Solution: We need to calculate \(3 p^{2}+4 p\) minus \(5 p^{2}-7 p+2:\)
\[\left(3 p^{2}+4 p\right)-\left(5 p^{2}-7 p+2\right)\nonumber\]
Simplifying this expression gives:
\[\left(3 p^{2}+4 p\right)-\left(5 p^{2}-7 p+2\right)=3 p^{2}+4 p-5 p^{2}+7 p-2 =-2 p^{2}+11 p-2\nonumber\]
e) The amount of money given by \(x\) dimes and \(y\) quarters.
Solution: Each dime is worth 10 cents, so that this gives a total of \(10 x\) cents. Each quarter is worth 25 cents, so that this gives a total of \(25 y\) cents. Adding the two amounts gives a total of
\[10 x+25 y \text{ cents or } .10x + .25y \text{ dollars}\nonumber\]
Now we deal with word problems that directly describe an equation involving one variable, which we can then solve.
Example 18.2
Solve the following word problems:
a) Five times an unknown number is equal to 60. Find the number.
Solution: We translate the problem to algebra:
\[5x = 60\nonumber\]
We solve this for \(x\) :
\[x=\frac{60}{5}=12\nonumber\]
b) If 5 is subtracted from twice an unknown number, the difference is \(13 .\) Find the number.
Solution: Translating the problem into an algebraic equation gives:
\[2x − 5 = 13\nonumber\]
We solve this for \(x\). First, add 5 to both sides.
\[2x = 13 + 5, \text{ so that } 2x = 18\nonumber\]
Dividing by 2 gives \(x=\frac{18}{2}=9\).
c) A number subtracted from 9 is equal to 2 times the number. Find the number.
Solution: We translate the problem to algebra.
\[9 − x = 2x\nonumber\]
We solve this as follows. First, add \(x\) :
\[9 = 2x + x \text{ so that } 9 = 3x\nonumber\]
Then the answer is \(x=\frac{9}{3}=3\)
d) Multiply an unknown number by five is equal to adding twelve to the unknown number. Find the number.
Solution: We have the equation:
\[5x = x + 12.\nonumber\]
Subtracting \(x\) gives
\[4x = 12.\nonumber\]
Dividing both sides by 4 gives the answer: \(x=3\).
e) Adding nine to a number gives the same result as subtracting seven from three times the number. Find the number.
Solution: Adding 9 to a number is written as \(x+9,\) while subtracting 7 from three times the number is written as \(3 x-7\). We therefore get the equation:
\[x + 9 = 3x − 7.\nonumber\]
We solve for \(x\) by adding 7 on both sides of the equation:
\[x + 16 = 3x.\nonumber\]
Then we subtract \(x:\)
\[16 = 2x.\nonumber\]
After dividing by \(2,\) we obtain the answer \(x=8\)
The following word problems consider real world applications. They require to model a given situation in the form of an equation.
Example 18.3
a) Due to inflation, the price of a loaf of bread has increased by \(5 \%\). How much does the loaf of bread cost now, when its price was \(\$ 2.40\) last year?
Solution: We calculate the price increase as \(5 \% \cdot \$ 2.40 .\) We have
\[5 \% \cdot 2.40=0.05 \cdot 2.40=0.1200=0.12\nonumber\]
We must add the price increase to the old price.
\[2.40+0.12=2.52\nonumber\]
The new price is therefore \(\$ 2.52\).
b) To complete a job, three workers get paid at a rate of \(\$ 12\) per hour. If the total pay for the job was \(\$ 180,\) then how many hours did the three workers spend on the job?
Solution: We denote the number of hours by \(x\). Then the total price is calculated as the price per hour \((\$ 12)\) times the number of workers times the number of hours \((3) .\) We obtain the equation
\[12 \cdot 3 \cdot x=180\nonumber\]
Simplifying this yields
\[36 x=180\nonumber\]
Dividing by 36 gives
\[x=\frac{180}{36}=5\nonumber\]
Therefore, the three workers needed 5 hours for the job.
c) A farmer cuts a 300 foot fence into two pieces of different sizes. The longer piece should be four times as long as the shorter piece. How long are the two pieces?
\[x+4 x=300\nonumber\]
Combining the like terms on the left, we get
\[5 x=300\nonumber\]
Dividing by 5, we obtain that
\[x=\frac{300}{5}=60\nonumber\]
Therefore, the shorter piece has a length of 60 feet, while the longer piece has four times this length, that is \(4 \times 60\) feet \(=240\) feet.
d) If 4 blocks weigh 28 ounces, how many blocks weigh 70 ounces?
Solution: We denote the weight of a block by \(x .\) If 4 blocks weigh \(28,\) then a block weighs \(x=\frac{28}{4}=7\)
How many blocks weigh \(70 ?\) Well, we only need to find \(\frac{70}{7}=10 .\) So, the answer is \(10 .\)
Note You can solve this problem by setting up and solving the fractional equation \(\frac{28}{4}=\frac{70}{x}\). Solving such equations is addressed in chapter 24.
e) If a rectangle has a length that is three more than twice the width and the perimeter is 20 in, what are the dimensions of the rectangle?
Solution: We denote the width by \(x\). Then the length is \(2 x+3\). The perimeter is 20 in on one hand and \(2(\)length\()+2(\)width\()\) on the other. So we have
\[20=2 x+2(2 x+3)\nonumber\]
Distributing and collecting like terms give
\[20=6 x+6\nonumber\]
Subtracting 6 from both sides of the equation and then dividing both sides of the resulting equation by 6 gives:
\[20-6=6 x \Longrightarrow 14=6 x \Longrightarrow x=\frac{14}{6} \text { in }=\frac{7}{3} \text { in }=2 \frac{1}{3} \text { in. }\nonumber\]
f) If a circle has circumference 4in, what is its radius?
Solution: We know that \(C=2 \pi r\) where \(C\) is the circumference and \(r\) is the radius. So in this case
\[4=2 \pi r\nonumber\]
Dividing both sides by \(2 \pi\) gives
\[r=\frac{4}{2 \pi}=\frac{2}{\pi} \text { in } \approx 0.63 \mathrm{in}\nonumber\]
g) The perimeter of an equilateral triangle is 60 meters. How long is each side?
Solution: Let \(x\) equal the side of the triangle. Then the perimeter is, on the one hand, \(60,\) and on other hand \(3 x .\) So \(3 x=60\) and dividing both sides of the equation by 3 gives \(x=20\) meters.
h) If a gardener has \(\$ 600\) to spend on a fence which costs \(\$ 10\) per linear foot and the area to be fenced in is rectangular and should be twice as long as it is wide, what are the dimensions of the largest fenced in area?
Solution: The perimeter of a rectangle is \(P=2 L+2 W\). Let \(x\) be the width of the rectangle. Then the length is \(2 x .\) The perimeter is \(P=2(2 x)+2 x=6 x\). The largest perimeter is \(\$ 600 /(\$ 10 / f t)=60\) ft. So \(60=6 x\) and dividing both sides by 6 gives \(x=60 / 6=10\). So the dimensions are 10 feet by 20 feet.
i) A trapezoid has an area of 20.2 square inches with one base measuring 3.2 in and the height of 4 in. Find the length of the other base.
Solution: Let \(b\) be the length of the unknown base. The area of the trapezoid is on the one hand 20.2 square inches. On the other hand it is \(\frac{1}{2}(3.2+b) \cdot 4=\) \(6.4+2 b .\) So
\[20.2=6.4+2 b\nonumber\]
Multiplying both sides by 10 gives
\[202=64+20 b\nonumber\]
Subtracting 64 from both sides gives
\[b=\frac{138}{20}=\frac{69}{10}=6.9 \text { in }\nonumber\]
and dividing by 20 gives
Exit Problem
Write an equation and solve: A car uses 12 gallons of gas to travel 100 miles. How many gallons would be needed to travel 450 miles?
Free Printable Linear Equations worksheets
Math Linear Equations: Discover an extensive collection of free printable worksheets to help students master solving linear equations, enhancing their mathematical skills and confidence.
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Linear Equations worksheets are an essential tool for teachers to help their students master the fundamental concepts of math and algebra. These worksheets provide a variety of problems that challenge students to apply their knowledge and skills in solving linear equations. Teachers can use these worksheets to supplement their lesson plans, provide additional practice for students who may be struggling with the material, or even as a form of assessment to gauge student understanding. With a wide range of topics covered, from simple one-step equations to more complex systems of linear equations, these worksheets cater to students of all grade levels and abilities. By incorporating Linear Equations worksheets into their teaching strategies, educators can ensure that their students are well-equipped to tackle the challenges of math and algebra with confidence.
Quizizz offers a comprehensive collection of Linear Equations worksheets, as well as other math and algebra resources, to support teachers in their quest to provide the best possible education for their students. With Quizizz, teachers can easily create engaging and interactive quizzes that can be used in conjunction with worksheets to reinforce learning and assess student progress. In addition to Linear Equations worksheets, Quizizz also offers a wide range of other math and algebra resources, including lessons, activities, and games, all designed to make learning fun and engaging for students of all grade levels. By utilizing Quizizz in their classrooms, teachers can create a dynamic and interactive learning environment that not only helps students master the concepts of linear equations but also fosters a lifelong love of learning.
Word Problems on Linear Equations
Worked-out word problems on linear equations with solutions explained step-by-step in different types of examples.
There are several problems which involve relations among known and unknown numbers and can be put in the form of equations. The equations are generally stated in words and it is for this reason we refer to these problems as word problems. With the help of equations in one variable, we have already practiced equations to solve some real life problems.
Steps involved in solving a linear equation word problem: ● Read the problem carefully and note what is given and what is required and what is given. ● Denote the unknown by the variables as x, y, ……. ● Translate the problem to the language of mathematics or mathematical statements. ● Form the linear equation in one variable using the conditions given in the problems. ● Solve the equation for the unknown. ● Verify to be sure whether the answer satisfies the conditions of the problem.
Step-by-step application of linear equations to solve practical word problems:
1. The sum of two numbers is 25. One of the numbers exceeds the other by 9. Find the numbers.
Solution: Then the other number = x + 9 Let the number be x. Sum of two numbers = 25 According to question, x + x + 9 = 25 ⇒ 2x + 9 = 25 ⇒ 2x = 25 - 9 (transposing 9 to the R.H.S changes to -9) ⇒ 2x = 16 ⇒ 2x/2 = 16/2 (divide by 2 on both the sides) ⇒ x = 8 Therefore, x + 9 = 8 + 9 = 17 Therefore, the two numbers are 8 and 17.
2.The difference between the two numbers is 48. The ratio of the two numbers is 7:3. What are the two numbers? Solution: Let the common ratio be x. Let the common ratio be x. Their difference = 48 According to the question, 7x - 3x = 48 ⇒ 4x = 48 ⇒ x = 48/4 ⇒ x = 12 Therefore, 7x = 7 × 12 = 84 3x = 3 × 12 = 36 Therefore, the two numbers are 84 and 36.
3. The length of a rectangle is twice its breadth. If the perimeter is 72 metre, find the length and breadth of the rectangle. Solution: Let the breadth of the rectangle be x, Then the length of the rectangle = 2x Perimeter of the rectangle = 72 Therefore, according to the question 2(x + 2x) = 72 ⇒ 2 × 3x = 72 ⇒ 6x = 72 ⇒ x = 72/6 ⇒ x = 12 We know, length of the rectangle = 2x = 2 × 12 = 24 Therefore, length of the rectangle is 24 m and breadth of the rectangle is 12 m.
4. Aaron is 5 years younger than Ron. Four years later, Ron will be twice as old as Aaron. Find their present ages.
Solution: Let Ron’s present age be x. Then Aaron’s present age = x - 5 After 4 years Ron’s age = x + 4, Aaron’s age x - 5 + 4. According to the question; Ron will be twice as old as Aaron. Therefore, x + 4 = 2(x - 5 + 4) ⇒ x + 4 = 2(x - 1) ⇒ x + 4 = 2x - 2 ⇒ x + 4 = 2x - 2 ⇒ x - 2x = -2 - 4 ⇒ -x = -6 ⇒ x = 6 Therefore, Aaron’s present age = x - 5 = 6 - 5 = 1 Therefore, present age of Ron = 6 years and present age of Aaron = 1 year.
5. A number is divided into two parts, such that one part is 10 more than the other. If the two parts are in the ratio 5 : 3, find the number and the two parts. Solution: Let one part of the number be x Then the other part of the number = x + 10 The ratio of the two numbers is 5 : 3 Therefore, (x + 10)/x = 5/3 ⇒ 3(x + 10) = 5x ⇒ 3x + 30 = 5x ⇒ 30 = 5x - 3x ⇒ 30 = 2x ⇒ x = 30/2 ⇒ x = 15 Therefore, x + 10 = 15 + 10 = 25 Therefore, the number = 25 + 15 = 40 The two parts are 15 and 25.
More solved examples with detailed explanation on the word problems on linear equations.
6. Robert’s father is 4 times as old as Robert. After 5 years, father will be three times as old as Robert. Find their present ages. Solution: Let Robert’s age be x years. Then Robert’s father’s age = 4x After 5 years, Robert’s age = x + 5 Father’s age = 4x + 5 According to the question, 4x + 5 = 3(x + 5) ⇒ 4x + 5 = 3x + 15 ⇒ 4x - 3x = 15 - 5 ⇒ x = 10 ⇒ 4x = 4 × 10 = 40 Robert’s present age is 10 years and that of his father’s age = 40 years.
7. The sum of two consecutive multiples of 5 is 55. Find these multiples. Solution: Let the first multiple of 5 be x. Then the other multiple of 5 will be x + 5 and their sum = 55 Therefore, x + x + 5 = 55 ⇒ 2x + 5 = 55 ⇒ 2x = 55 - 5 ⇒ 2x = 50 ⇒ x = 50/2 ⇒ x = 25 Therefore, the multiples of 5, i.e., x + 5 = 25 + 5 = 30 Therefore, the two consecutive multiples of 5 whose sum is 55 are 25 and 30.
8. The difference in the measures of two complementary angles is 12°. Find the measure of the angles. Solution: Let the angle be x. Complement of x = 90 - x Given their difference = 12° Therefore, (90 - x) - x = 12° ⇒ 90 - 2x = 12 ⇒ -2x = 12 - 90 ⇒ -2x = -78 ⇒ 2x/2 = 78/2 ⇒ x = 39 Therefore, 90 - x = 90 - 39 = 51 Therefore, the two complementary angles are 39° and 51°
9. The cost of two tables and three chairs is $705. If the table costs $40 more than the chair, find the cost of the table and the chair. Solution: The table cost $ 40 more than the chair. Let us assume the cost of the chair to be x. Then the cost of the table = $ 40 + x The cost of 3 chairs = 3 × x = 3x and the cost of 2 tables 2(40 + x) Total cost of 2 tables and 3 chairs = $705 Therefore, 2(40 + x) + 3x = 705 80 + 2x + 3x = 705 80 + 5x = 705 5x = 705 - 80 5x = 625/5 x = 125 and 40 + x = 40 + 125 = 165 Therefore, the cost of each chair is $125 and that of each table is $165.
10. If 3/5 ᵗʰ of a number is 4 more than 1/2 the number, then what is the number? Solution: Let the number be x, then 3/5 ᵗʰ of the number = 3x/5 Also, 1/2 of the number = x/2 According to the question, 3/5 ᵗʰ of the number is 4 more than 1/2 of the number. ⇒ 3x/5 - x/2 = 4 ⇒ (6x - 5x)/10 = 4 ⇒ x/10 = 4 ⇒ x = 40 The required number is 40.
Try to follow the methods of solving word problems on linear equations and then observe the detailed instruction on the application of equations to solve the problems.
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Solution: Adding 9 to a number is written as x + 9, while subtracting 7 from three times the number is written as 3x − 7. We therefore get the equation: x + 9 = 3x − 7. We solve for x by adding 7 on both sides of the equation: x + 16 = 3x. Then we subtract x: 16 = 2x. After dividing by 2, we obtain the answer x = 8.
Linear Equations worksheets are an essential tool for teachers to help their students master the fundamental concepts of math and algebra. These worksheets provide a variety of problems that challenge students to apply their knowledge and skills in solving linear equations. Teachers can use these worksheets to supplement their lesson plans ...
Step-by-step application of linear equations to solve practical word problems: 1. The sum of two numbers is 25. One of the numbers exceeds the other by 9. Find the numbers. Let the number be x. Therefore, the two numbers are 8 and 17. 2.The difference between the two numbers is 48. The ratio of the two numbers is 7:3.
equation, solve. Practice Questions. Previous: Ray Method Practice Questions. Next: Equations involving Fractions Practice Questions. The Corbettmaths Practice Questions on Solving Equations.