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| → → Linear equations Find here an unlimited supply of printable worksheets for solving linear equations, available as both PDF and html files. You can customize the worksheets to include one-step, two-step, or multi-step equations, variable on both sides, parenthesis, and more. The worksheets suit and courses (grades 6-9). You can choose from SEVEN basic types of equations, ranging from simple to complex, explained below (such as one-step equations, variable on both sides, or having to use the distributive property). Customize the worksheets using the below. For some extra tips and practice problems on linear equations, check out IXL's ! Each worksheet is randomly generated and thus unique. The and is placed on the second page of the file. You can generate the worksheets — both are easy to print. To get the PDF worksheet, simply push the button titled " " or " ". To get the worksheet in html format, push the button " " or " ". This has the advantage that you can save the worksheet directly from your browser (choose File → Save) and then in Word or other word processing program. Sometimes the generated worksheet is not exactly what you want. Just try again! To get a different worksheet using the same options:
Worksheets for simplifying expressions Worksheets for evaluating expressions with variables Worksheets for writing expressions with variables from verbal expressions Worksheets for linear inequalities
 How to Write Linear Equations? (+FREE Worksheet!)In this article, you learn how to write the equation of the lines by using their slope and one point or using two points on the line.  Related Topics
Step by step guide to writing linear equations
The Absolute Best Books to Ace Pre-Algebra to Algebra II The Ultimate Algebra Bundle From Pre-Algebra to Algebra IIWriting linear equations – example 1:. What is the equation of the line that passes through \((1, -2)\) and has a slope of \(6\)? The general slope-intercept form of the equation of a line is \(y=mx+b\), where \(m\) is the slope and \(b\) is the \(y\)-intercept. By substitution of the given point and given slope, we have: \(-2=(6)(1)+b → -2=6+b \) So, \(b= -2-6=-8\), and the required equation is \(y=6x-8\). Writing Linear Equations – Example 2:Write the equation of the line through \((1, 1)\) and \((-1, 3)\). Slop \(= \frac{y_{2}- y_{1}}{x_{2} – x_{1} }=\frac{3- 1}{-1- 1}=\frac{2}{-2}=-1 → m=-1\) To find the value of \(b\), you can use either point. The answer will be the same: \(y=-x+b \) \((1,1) →1=-1+b→ 1+1=b → b=2\) \((-1,3)→3=-(-1)+b→3-1=b → b=2\) The equation of the line is: \(y=-x+2\) Writing Linear Equations – Example 3:What is the equation of the line that passes through \((2,–2)\) and has a slope of \(7\)? The general slope-intercept form of the equation of a line is \(y=mx+b\), where \(m\) is the slope and \(b\) is the \(y-\)intercept. By substitution of the given point and given slope, we have: \(-2=(7)(2)+b → -2=14+b \) So, \(b= –2-14=-16\), and the required equation is \(y=7x-16\). The Best Book to Help You Ace Pre-Algebra Pre-Algebra for Beginners The Ultimate Step by Step Guide to Preparing for the Pre-Algebra TestWriting linear equations – example 4:. Write the equation of the line through \((2,1)\) and \((-1,4)\). Slop \(= \frac{y_{2}- y_{1}}{x_{2} – x_{1} }=\frac{4- 1}{-1- 2}=\frac{3}{-3}=-1 → m= -1\) You can use either point to find the value of \(b\). The answer will be the same: \(y= -x+b \) \( (2,1) →1=-2+b→1+2=b → b=3\) \( (-1,4)→4=-(-1)+b→4-1=b → b=3\) The equation of the line is: \(y=-x+3\) Exercises for Writing Linear EquationsWrite the slope–intercept form of the equation of the line through the given points..
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Unit 2 DownloadsLinear EquationsA linear equation is an equation for a straight line These are all linear equations:
Let us look more closely at one example: Example: y = 2x + 1 is a linear equation:The graph of y = 2x+1 is a straight line
Here are some example values:
Check for yourself that those points are part of the line above! Different FormsThere are many ways of writing linear equations, but they usually have constants (like "2" or "c") and must have simple variables (like "x" or "y"). Examples: These are linear equations:
But the variables (like "x" or "y") in Linear Equations do NOT have:
Examples: These are NOT linear equations:
Slope-Intercept FormThe most common form is the slope-intercept equation of a straight line :
Example: y = 2x + 1
Point-Slope FormAnother common one is the Point-Slope Form of the equation of a straight line: Example: y − 3 = (¼)(x − 2)It is in the form y − y 1 = m(x − x 1 ) where: General FormAnd there is also the General Form of the equation of a straight line:
Example: 3x + 2y − 4 = 0It is in the form Ax + By + C = 0 where: There are other, less common forms as well. As a FunctionSometimes a linear equation is written as a function , with f(x) instead of y :
And functions are not always written using f(x):
The Identity FunctionThere is a special linear function called the "Identity Function": And here is its graph: It is called "Identity" because what comes out is identical to what goes in:
Constant FunctionsAnother special type of linear function is the Constant Function ... it is a horizontal line: No matter what value of "x", f(x) is always equal to some constant value. Using Linear EquationsYou may like to read some of the things you can do with lines:
CLICK HERE TO LEARN ABOUT MTM ALL ACCESS MEMBERSHIP FOR GRADES 6-ALGEBRA 1 Maneuvering the Middle Student-Centered Math Lessons Linear Equations Activity IdeasSome of the links in this post are affiliate links that support the content on this site. Read our disclosure statement for more information.  Linear equations require lots of practice as the skill progresses to include more steps and increases in complexity. Any of these activities can be used from the basics of simplifying expressions to solving equations with variables on both sides. Model by Using Algebra TilesAre you surprised I am starting with Algebra tiles? They are foundational for concrete understanding! You can learn more about using Algebra tiles in your classroom by grabbing our free Getting Started with Algebra Tiles guide, which can be found by checking out our post on Solving Equations . This Modeling Equations with Variables on Both Sides activity is a great way to practice solving equations with Algebra tiles. You just need some Algebra tiles (or your students can draw them!).  Student TeacherPut students into pairs and show an equation on the board. Have one student instruct the other on how to solve as the student listening writes each step and solution. Then, show a new equation and have students switch roles. This gives students a chance to teach and reinforce what they remember about linear equations. I love this activity because it is simple and it makes every student explain their thinking. You, as the teacher, can circulate listening to each pair. Round TableGive students individual white boards and have them work in teams of 2-4. With one equation written on the board, the first person will solve step one. The second person will complete the second step in solving and the third will complete the next step. Keep rotating until the problem is solved and the last person checks the solution. Have groups hold up their boards when they are finished. If you want something like this, we have this Solving Equations with Variables on Both Sides Round Table available!  Board RacesAfter students have had time to practice, implement “Board Races.” Two students will come up to the board and race to solve an equation shown on the board. The person who solves it correctly first stays up at the board for the next equation with a new competitor. I like to have the students who aren’t at the board working the equations on notebook paper to help check the solutions. An element of competition makes repetitive practice more fun! For race type activities, I like to have teams compete (boys v. girls or one side of the room v. the other side of the room). Digital ActivitiesI love our digital activities! This linear equations set of digital activities includes simplifying expressions, solving one and two-step equations, solving multi-step equations, and equations with variables on both sides, making it the perfect review before a linear equations test.  What are some fun ways that students practice solving linear equations in your class? You can check out how to turn any worksheet into an activity which can easily be used for linear equations too!  Getting Started with Algebra TilesCheck out these related products from my shop.  2.2 Linear Equations in One VariableLearning objectives. In this section, you will:
Caroline is a full-time college student planning a spring break vacation. To earn enough money for the trip, she has taken a part-time job at the local bank that pays $15.00/hr, and she opened a savings account with an initial deposit of $400 on January 15. She arranged for direct deposit of her payroll checks. If spring break begins March 20 and the trip will cost approximately $2,500, how many hours will she have to work to earn enough to pay for her vacation? If she can only work 4 hours per day, how many days per week will she have to work? How many weeks will it take? In this section, we will investigate problems like this and others, which generate graphs like the line in Figure 1 . Solving Linear Equations in One VariableA linear equation is an equation of a straight line, written in one variable. The only power of the variable is 1. Linear equations in one variable may take the form a x + b = 0 a x + b = 0 and are solved using basic algebraic operations. We begin by classifying linear equations in one variable as one of three types: identity, conditional, or inconsistent. An identity equation is true for all values of the variable. Here is an example of an identity equation. The solution set consists of all values that make the equation true. For this equation, the solution set is all real numbers because any real number substituted for x x will make the equation true. A conditional equation is true for only some values of the variable. For example, if we are to solve the equation 5 x + 2 = 3 x − 6 , 5 x + 2 = 3 x − 6 , we have the following: The solution set consists of one number: { − 4 } . { − 4 } . It is the only solution and, therefore, we have solved a conditional equation. An inconsistent equation results in a false statement. For example, if we are to solve 5 x − 15 = 5 ( x − 4 ) , 5 x − 15 = 5 ( x − 4 ) , we have the following: Indeed, −15 ≠ −20. −15 ≠ −20. There is no solution because this is an inconsistent equation. Solving linear equations in one variable involves the fundamental properties of equality and basic algebraic operations. A brief review of those operations follows. Linear Equation in One VariableA linear equation in one variable can be written in the form where a and b are real numbers, a ≠ 0. a ≠ 0. Given a linear equation in one variable, use algebra to solve it. The following steps are used to manipulate an equation and isolate the unknown variable, so that the last line reads x = _________, x = _________, if x is the unknown. There is no set order, as the steps used depend on what is given:
Solving an Equation in One VariableSolve the following equation: 2 x + 7 = 19. 2 x + 7 = 19. This equation can be written in the form a x + b = 0 a x + b = 0 by subtracting 19 19 from both sides. However, we may proceed to solve the equation in its original form by performing algebraic operations. The solution is 6. Solve the linear equation in one variable: 2 x + 1 = −9. 2 x + 1 = −9. Solving an Equation Algebraically When the Variable Appears on Both SidesSolve the following equation: 4 ( x −3 ) + 12 = 15 −5 ( x + 6 ) . 4 ( x −3 ) + 12 = 15 −5 ( x + 6 ) . Apply standard algebraic properties. This problem requires the distributive property to be applied twice, and then the properties of algebra are used to reach the final line, x = − 5 3 . x = − 5 3 . Solve the equation in one variable: −2 ( 3 x − 1 ) + x = 14 − x . −2 ( 3 x − 1 ) + x = 14 − x . Solving a Rational EquationIn this section, we look at rational equations that, after some manipulation, result in a linear equation. If an equation contains at least one rational expression, it is a considered a rational equation . Recall that a rational number is the ratio of two numbers, such as 2 3 2 3 or 7 2 . 7 2 . A rational expression is the ratio, or quotient, of two polynomials. Here are three examples. Rational equations have a variable in the denominator in at least one of the terms. Our goal is to perform algebraic operations so that the variables appear in the numerator. In fact, we will eliminate all denominators by multiplying both sides of the equation by the least common denominator (LCD). Finding the LCD is identifying an expression that contains the highest power of all of the factors in all of the denominators. We do this because when the equation is multiplied by the LCD, the common factors in the LCD and in each denominator will equal one and will cancel out. Solve the rational equation: 7 2 x − 5 3 x = 22 3 . 7 2 x − 5 3 x = 22 3 . We have three denominators; 2 x , 3 x , 2 x , 3 x , and 3. The LCD must contain 2 x , 3 x , 2 x , 3 x , and 3. An LCD of 6 x 6 x contains all three denominators. In other words, each denominator can be divided evenly into the LCD. Next, multiply both sides of the equation by the LCD 6 x . 6 x . A common mistake made when solving rational equations involves finding the LCD when one of the denominators is a binomial—two terms added or subtracted—such as ( x + 1 ) . ( x + 1 ) . Always consider a binomial as an individual factor—the terms cannot be separated. For example, suppose a problem has three terms and the denominators are x , x , x − 1 , x − 1 , and 3 x − 3. 3 x − 3. First, factor all denominators. We then have x , x , ( x − 1 ) , ( x − 1 ) , and 3 ( x − 1 ) 3 ( x − 1 ) as the denominators. (Note the parentheses placed around the second denominator.) Only the last two denominators have a common factor of ( x − 1 ) . ( x − 1 ) . The x x in the first denominator is separate from the x x in the ( x − 1 ) ( x − 1 ) denominators. An effective way to remember this is to write factored and binomial denominators in parentheses, and consider each parentheses as a separate unit or a separate factor. The LCD in this instance is found by multiplying together the x , x , one factor of ( x − 1 ) , ( x − 1 ) , and the 3. Thus, the LCD is the following: So, both sides of the equation would be multiplied by 3 x ( x − 1 ) . 3 x ( x − 1 ) . Leave the LCD in factored form, as this makes it easier to see how each denominator in the problem cancels out. Another example is a problem with two denominators, such as x x and x 2 + 2 x . x 2 + 2 x . Once the second denominator is factored as x 2 + 2 x = x ( x + 2 ) , x 2 + 2 x = x ( x + 2 ) , there is a common factor of x in both denominators and the LCD is x ( x + 2 ) . x ( x + 2 ) . Sometimes we have a rational equation in the form of a proportion; that is, when one fraction equals another fraction and there are no other terms in the equation. We can use another method of solving the equation without finding the LCD: cross-multiplication. We multiply terms by crossing over the equal sign. Multiply a ( d ) a ( d ) and b ( c ) , b ( c ) , which results in a d = b c . a d = b c . Any solution that makes a denominator in the original expression equal zero must be excluded from the possibilities. Rational EquationsA rational equation contains at least one rational expression where the variable appears in at least one of the denominators. Given a rational equation, solve it.
Solving a Rational Equation without FactoringSolve the following rational equation: We have three denominators: x , x , 2 , 2 , and 2 x . 2 x . No factoring is required. The product of the first two denominators is equal to the third denominator, so, the LCD is 2 x . 2 x . Only one value is excluded from a solution set, 0. Next, multiply the whole equation (both sides of the equal sign) by 2 x . 2 x . The proposed solution is −1, which is not an excluded value, so the solution set contains one number, −1 , −1 , or { −1 } { −1 } written in set notation. Solve the rational equation: 2 3 x = 1 4 − 1 6 x . 2 3 x = 1 4 − 1 6 x . Solving a Rational Equation by Factoring the DenominatorSolve the following rational equation: 1 x = 1 10 − 3 4 x . 1 x = 1 10 − 3 4 x . First find the common denominator. The three denominators in factored form are x , 10 = 2 ⋅ 5 , x , 10 = 2 ⋅ 5 , and 4 x = 2 ⋅ 2 ⋅ x . 4 x = 2 ⋅ 2 ⋅ x . The smallest expression that is divisible by each one of the denominators is 20 x . 20 x . Only x = 0 x = 0 is an excluded value. Multiply the whole equation by 20 x . 20 x . The solution is 35 2 . 35 2 . Solve the rational equation: − 5 2 x + 3 4 x = − 7 4 . − 5 2 x + 3 4 x = − 7 4 . Solving Rational Equations with a Binomial in the DenominatorSolve the following rational equations and state the excluded values:
The denominators x x and x − 6 x − 6 have nothing in common. Therefore, the LCD is the product x ( x − 6 ) . x ( x − 6 ) . However, for this problem, we can cross-multiply. The solution is 15. The excluded values are 6 6 and 0. 0. The LCD is 2 ( x − 3 ) . 2 ( x − 3 ) . Multiply both sides of the equation by 2 ( x − 3 ) . 2 ( x − 3 ) . The solution is 13 3 . 13 3 . The excluded value is 3. 3. The least common denominator is 2 ( x − 2 ) . 2 ( x − 2 ) . Multiply both sides of the equation by x ( x − 2 ) . x ( x − 2 ) . The solution is 4. The excluded value is 2. 2. Solve − 3 2 x + 1 = 4 3 x + 1 . − 3 2 x + 1 = 4 3 x + 1 . State the excluded values. Solving a Rational Equation with Factored Denominators and Stating Excluded ValuesSolve the rational equation after factoring the denominators: 2 x + 1 − 1 x − 1 = 2 x x 2 − 1 . 2 x + 1 − 1 x − 1 = 2 x x 2 − 1 . State the excluded values. We must factor the denominator x 2 −1. x 2 −1. We recognize this as the difference of squares, and factor it as ( x − 1 ) ( x + 1 ) . ( x − 1 ) ( x + 1 ) . Thus, the LCD that contains each denominator is ( x − 1 ) ( x + 1 ) . ( x − 1 ) ( x + 1 ) . Multiply the whole equation by the LCD, cancel out the denominators, and solve the remaining equation. The solution is −3. −3. The excluded values are 1 1 and −1. −1. Solve the rational equation: 2 x − 2 + 1 x + 1 = 1 x 2 − x − 2 . 2 x − 2 + 1 x + 1 = 1 x 2 − x − 2 . Finding a Linear EquationPerhaps the most familiar form of a linear equation is the slope-intercept form, written as y = m x + b , y = m x + b , where m = slope m = slope and b = y -intercept . b = y -intercept . Let us begin with the slope. The Slope of a LineThe slope of a line refers to the ratio of the vertical change in y over the horizontal change in x between any two points on a line. It indicates the direction in which a line slants as well as its steepness. Slope is sometimes described as rise over run. If the slope is positive, the line slants to the right. If the slope is negative, the line slants to the left. As the slope increases, the line becomes steeper. Some examples are shown in Figure 2 . The lines indicate the following slopes: m = −3 , m = −3 , m = 2 , m = 2 , and m = 1 3 . m = 1 3 . The slope of a line, m , represents the change in y over the change in x. Given two points, ( x 1 , y 1 ) ( x 1 , y 1 ) and ( x 2 , y 2 ) , ( x 2 , y 2 ) , the following formula determines the slope of a line containing these points: Finding the Slope of a Line Given Two PointsFind the slope of a line that passes through the points ( 2 , −1 ) ( 2 , −1 ) and ( −5 , 3 ) . ( −5 , 3 ) . We substitute the y- values and the x- values into the formula. The slope is − 4 7 . − 4 7 . It does not matter which point is called ( x 1 , y 1 ) ( x 1 , y 1 ) or ( x 2 , y 2 ) . ( x 2 , y 2 ) . As long as we are consistent with the order of the y terms and the order of the x terms in the numerator and denominator, the calculation will yield the same result. Find the slope of the line that passes through the points ( −2 , 6 ) ( −2 , 6 ) and ( 1 , 4 ) . ( 1 , 4 ) . Identifying the Slope and y- intercept of a Line Given an EquationIdentify the slope and y- intercept, given the equation y = − 3 4 x − 4. y = − 3 4 x − 4. As the line is in y = m x + b y = m x + b form, the given line has a slope of m = − 3 4 . m = − 3 4 . The y- intercept is b = −4. b = −4. The y -intercept is the point at which the line crosses the y- axis. On the y- axis, x = 0. x = 0. We can always identify the y- intercept when the line is in slope-intercept form, as it will always equal b. Or, just substitute x = 0 x = 0 and solve for y. The Point-Slope FormulaGiven the slope and one point on a line, we can find the equation of the line using the point-slope formula. This is an important formula, as it will be used in other areas of college algebra and often in calculus to find the equation of a tangent line. We need only one point and the slope of the line to use the formula. After substituting the slope and the coordinates of one point into the formula, we simplify it and write it in slope-intercept form. Given one point and the slope, the point-slope formula will lead to the equation of a line: Finding the Equation of a Line Given the Slope and One PointWrite the equation of the line with slope m = −3 m = −3 and passing through the point ( 4 , 8 ) . ( 4 , 8 ) . Write the final equation in slope-intercept form. Using the point-slope formula, substitute −3 −3 for m and the point ( 4 , 8 ) ( 4 , 8 ) for ( x 1 , y 1 ) . ( x 1 , y 1 ) . Note that any point on the line can be used to find the equation. If done correctly, the same final equation will be obtained. Given m = 4 , m = 4 , find the equation of the line in slope-intercept form passing through the point ( 2 , 5 ) . ( 2 , 5 ) . Finding the Equation of a Line Passing Through Two Given PointsFind the equation of the line passing through the points ( 3 , 4 ) ( 3 , 4 ) and ( 0 , −3 ) . ( 0 , −3 ) . Write the final equation in slope-intercept form. First, we calculate the slope using the slope formula and two points. Next, we use the point-slope formula with the slope of 7 3 , 7 3 , and either point. Let’s pick the point ( 3 , 4 ) ( 3 , 4 ) for ( x 1 , y 1 ) . ( x 1 , y 1 ) . In slope-intercept form, the equation is written as y = 7 3 x − 3. y = 7 3 x − 3. To prove that either point can be used, let us use the second point ( 0 , −3 ) ( 0 , −3 ) and see if we get the same equation. We see that the same line will be obtained using either point. This makes sense because we used both points to calculate the slope. Standard Form of a LineAnother way that we can represent the equation of a line is in standard form . Standard form is given as where A , A , B , B , and C C are integers. The x- and y- terms are on one side of the equal sign and the constant term is on the other side. Finding the Equation of a Line and Writing It in Standard FormFind the equation of the line with m = −6 m = −6 and passing through the point ( 1 4 , −2 ) . ( 1 4 , −2 ) . Write the equation in standard form. We begin using the point-slope formula. From here, we multiply through by 2, as no fractions are permitted in standard form, and then move both variables to the left aside of the equal sign and move the constants to the right. This equation is now written in standard form. Find the equation of the line in standard form with slope m = − 1 3 m = − 1 3 and passing through the point ( 1 , 1 3 ) . ( 1 , 1 3 ) . Vertical and Horizontal LinesThe equations of vertical and horizontal lines do not require any of the preceding formulas, although we can use the formulas to prove that the equations are correct. The equation of a vertical line is given as where c is a constant. The slope of a vertical line is undefined, and regardless of the y- value of any point on the line, the x- coordinate of the point will be c . Suppose that we want to find the equation of a line containing the following points: ( −3 , −5 ) , ( −3 , 1 ) , ( −3 , 3 ) , ( −3 , −5 ) , ( −3 , 1 ) , ( −3 , 3 ) , and ( −3 , 5 ) . ( −3 , 5 ) . First, we will find the slope. Zero in the denominator means that the slope is undefined and, therefore, we cannot use the point-slope formula. However, we can plot the points. Notice that all of the x- coordinates are the same and we find a vertical line through x = −3. x = −3. See Figure 3 . The equation of a horizontal line is given as where c is a constant. The slope of a horizontal line is zero, and for any x- value of a point on the line, the y- coordinate will be c . Suppose we want to find the equation of a line that contains the following set of points: ( −2 , −2 ) , ( 0 , −2 ) , ( 3 , −2 ) , ( −2 , −2 ) , ( 0 , −2 ) , ( 3 , −2 ) , and ( 5 , −2 ) . ( 5 , −2 ) . We can use the point-slope formula. First, we find the slope using any two points on the line. Use any point for ( x 1 , y 1 ) ( x 1 , y 1 ) in the formula, or use the y -intercept. The graph is a horizontal line through y = −2. y = −2. Notice that all of the y- coordinates are the same. See Figure 3 . Finding the Equation of a Line Passing Through the Given PointsFind the equation of the line passing through the given points: ( 1 , −3 ) ( 1 , −3 ) and ( 1 , 4 ) . ( 1 , 4 ) . The x- coordinate of both points is 1. Therefore, we have a vertical line, x = 1. x = 1. Find the equation of the line passing through ( −5 , 2 ) ( −5 , 2 ) and ( 2 , 2 ) . ( 2 , 2 ) . Determining Whether Graphs of Lines are Parallel or PerpendicularParallel lines have the same slope and different y- intercepts. Lines that are parallel to each other will never intersect. For example, Figure 4 shows the graphs of various lines with the same slope, m = 2. m = 2. All of the lines shown in the graph are parallel because they have the same slope and different y- intercepts. Lines that are perpendicular intersect to form a 90° 90° -angle. The slope of one line is the negative reciprocal of the other. We can show that two lines are perpendicular if the product of the two slopes is −1 : m 1 ⋅ m 2 = −1. −1 : m 1 ⋅ m 2 = −1. For example, Figure 5 shows the graph of two perpendicular lines. One line has a slope of 3; the other line has a slope of − 1 3 . − 1 3 . Graphing Two Equations, and Determining Whether the Lines are Parallel, Perpendicular, or NeitherGraph the equations of the given lines, and state whether they are parallel, perpendicular, or neither: 3 y = − 4 x + 3 3 y = − 4 x + 3 and 3 x − 4 y = 8. 3 x − 4 y = 8. The first thing we want to do is rewrite the equations so that both equations are in slope-intercept form. First equation: Second equation: See the graph of both lines in Figure 6 From the graph, we can see that the lines appear perpendicular, but we must compare the slopes. The slopes are negative reciprocals of each other, confirming that the lines are perpendicular. Graph the two lines and determine whether they are parallel, perpendicular, or neither: 2 y − x = 10 2 y − x = 10 and 2 y = x + 4. 2 y = x + 4. Writing the Equations of Lines Parallel or Perpendicular to a Given LineAs we have learned, determining whether two lines are parallel or perpendicular is a matter of finding the slopes. To write the equation of a line parallel or perpendicular to another line, we follow the same principles as we do for finding the equation of any line. After finding the slope, use the point-slope formula to write the equation of the new line. Given an equation for a line, write the equation of a line parallel or perpendicular to it.
Writing the Equation of a Line Parallel to a Given Line Passing Through a Given PointWrite the equation of line parallel to a 5 x + 3 y = 1 5 x + 3 y = 1 and passing through the point ( 3 , 5 ) . ( 3 , 5 ) . First, we will write the equation in slope-intercept form to find the slope. The slope is m = − 5 3 . m = − 5 3 . The y- intercept is 1 3 , 1 3 , but that really does not enter into our problem, as the only thing we need for two lines to be parallel is the same slope. The one exception is that if the y- intercepts are the same, then the two lines are the same line. The next step is to use this slope and the given point with the point-slope formula. The equation of the line is y = − 5 3 x + 10. y = − 5 3 x + 10. See Figure 7 . Find the equation of the line parallel to 5 x = 7 + y 5 x = 7 + y and passing through the point ( −1 , −2 ) . ( −1 , −2 ) . Finding the Equation of a Line Perpendicular to a Given Line Passing Through a Given PointFind the equation of the line perpendicular to 5 x − 3 y + 4 = 0 5 x − 3 y + 4 = 0 and passing through the point ( − 4 , 1 ) . ( − 4 , 1 ) . The first step is to write the equation in slope-intercept form. We see that the slope is m = 5 3 . m = 5 3 . This means that the slope of the line perpendicular to the given line is the negative reciprocal, or − 3 5 . − 3 5 . Next, we use the point-slope formula with this new slope and the given point. Access these online resources for additional instruction and practice with linear equations.
2.2 Section ExercisesWhat does it mean when we say that two lines are parallel? What is the relationship between the slopes of perpendicular lines (assuming neither is horizontal nor vertical)? How do we recognize when an equation, for example y = 4 x + 3 , y = 4 x + 3 , will be a straight line (linear) when graphed? What does it mean when we say that a linear equation is inconsistent? When solving the following equation: 2 x − 5 = 4 x + 1 2 x − 5 = 4 x + 1 explain why we must exclude x = 5 x = 5 and x = −1 x = −1 as possible solutions from the solution set. For the following exercises, solve the equation for x . x . 7 x + 2 = 3 x − 9 7 x + 2 = 3 x − 9 4 x − 3 = 5 4 x − 3 = 5 3 ( x + 2 ) − 12 = 5 ( x + 1 ) 3 ( x + 2 ) − 12 = 5 ( x + 1 ) 12 − 5 ( x + 3 ) = 2 x − 5 12 − 5 ( x + 3 ) = 2 x − 5 1 2 − 1 3 x = 4 3 1 2 − 1 3 x = 4 3 x 3 − 3 4 = 2 x + 3 12 x 3 − 3 4 = 2 x + 3 12 2 3 x + 1 2 = 31 6 2 3 x + 1 2 = 31 6 3 ( 2 x − 1 ) + x = 5 x + 3 3 ( 2 x − 1 ) + x = 5 x + 3 2 x 3 − 3 4 = x 6 + 21 4 2 x 3 − 3 4 = x 6 + 21 4 x + 2 4 − x − 1 3 = 2 x + 2 4 − x − 1 3 = 2 For the following exercises, solve each rational equation for x . x . State all x -values that are excluded from the solution set. 3 x − 1 3 = 1 6 3 x − 1 3 = 1 6 2 − 3 x + 4 = x + 2 x + 4 2 − 3 x + 4 = x + 2 x + 4 3 x − 2 = 1 x − 1 + 7 ( x − 1 ) ( x − 2 ) 3 x − 2 = 1 x − 1 + 7 ( x − 1 ) ( x − 2 ) 3 x x − 1 + 2 = 3 x − 1 3 x x − 1 + 2 = 3 x − 1 5 x + 1 + 1 x − 3 = − 6 x 2 − 2 x − 3 5 x + 1 + 1 x − 3 = − 6 x 2 − 2 x − 3 1 x = 1 5 + 3 2 x 1 x = 1 5 + 3 2 x For the following exercises, find the equation of the line using the point-slope formula. Write all the final equations using the slope-intercept form. ( 0 , 3 ) ( 0 , 3 ) with a slope of 2 3 2 3 ( 1 , 2 ) ( 1 , 2 ) with a slope of − 4 5 − 4 5 x -intercept is 1, and ( −2 , 6 ) ( −2 , 6 ) y -intercept is 2, and ( 4 , −1 ) ( 4 , −1 ) ( −3 , 10 ) ( −3 , 10 ) and ( 5 , −6 ) ( 5 , −6 ) ( 1 , 3 ) and ( 5 , 5 ) ( 1 , 3 ) and ( 5 , 5 ) parallel to y = 2 x + 5 y = 2 x + 5 and passes through the point ( 4 , 3 ) ( 4 , 3 ) perpendicular to 3 y = x − 4 3 y = x − 4 and passes through the point ( −2 , 1 ) ( −2 , 1 ) . For the following exercises, find the equation of the line using the given information. ( − 2 , 0 ) ( − 2 , 0 ) and ( −2 , 5 ) ( −2 , 5 ) ( 1 , 7 ) ( 1 , 7 ) and ( 3 , 7 ) ( 3 , 7 ) The slope is undefined and it passes through the point ( 2 , 3 ) . ( 2 , 3 ) . The slope equals zero and it passes through the point ( 1 , −4 ) . ( 1 , −4 ) . The slope is 3 4 3 4 and it passes through the point ( 1 , 4 ) ( 1 , 4 ) . ( –1 , 3 ) ( –1 , 3 ) and ( 4 , –5 ) ( 4 , –5 ) For the following exercises, graph the pair of equations on the same axes, and state whether they are parallel, perpendicular, or neither. y = 2 x + 7 y = − 1 2 x − 4 y = 2 x + 7 y = − 1 2 x − 4 3 x − 2 y = 5 6 y − 9 x = 6 3 x − 2 y = 5 6 y − 9 x = 6 y = 3 x + 1 4 y = 3 x + 2 y = 3 x + 1 4 y = 3 x + 2 x = 4 y = −3 x = 4 y = −3 For the following exercises, find the slope of the line that passes through the given points. ( 5 , 4 ) ( 5 , 4 ) and ( 7 , 9 ) ( 7 , 9 ) ( −3 , 2 ) ( −3 , 2 ) and ( 4 , −7 ) ( 4 , −7 ) ( −5 , 4 ) ( −5 , 4 ) and ( 2 , 4 ) ( 2 , 4 ) ( −1 , −2 ) ( −1 , −2 ) and ( 3 , 4 ) ( 3 , 4 ) ( 3 , −2 ) ( 3 , −2 ) and ( 3 , −2 ) ( 3 , −2 ) For the following exercises, find the slope of the lines that pass through each pair of points and determine whether the lines are parallel or perpendicular. ( −1 , 3 ) and ( 5 , 1 ) ( −2 , 3 ) and ( 0 , 9 ) ( −1 , 3 ) and ( 5 , 1 ) ( −2 , 3 ) and ( 0 , 9 ) ( 2 , 5 ) and ( 5 , 9 ) ( −1 , −1 ) and ( 2 , 3 ) ( 2 , 5 ) and ( 5 , 9 ) ( −1 , −1 ) and ( 2 , 3 ) For the following exercises, express the equations in slope intercept form (rounding each number to the thousandths place). Enter this into a graphing calculator as Y1, then adjust the ymin and ymax values for your window to include where the y -intercept occurs. State your ymin and ymax values. 0.537 x − 2.19 y = 100 0.537 x − 2.19 y = 100 4,500 x − 200 y = 9,528 4,500 x − 200 y = 9,528 200 − 30 y x = 70 200 − 30 y x = 70 Starting with the point-slope formula y − y 1 = m ( x − x 1 ) , y − y 1 = m ( x − x 1 ) , solve this expression for x x in terms of x 1 , y , y 1 , x 1 , y , y 1 , and m m . Starting with the standard form of an equation A x + B y = C A x + B y = C solve this expression for y y in terms of A , B , C A , B , C and x x . Then put the expression in slope-intercept form. Use the above derived formula to put the following standard equation in slope intercept form: 7 x − 5 y = 25. 7 x − 5 y = 25. Given that the following coordinates are the vertices of a rectangle, prove that this truly is a rectangle by showing the slopes of the sides that meet are perpendicular. ( – 1 , 1 ) , ( 2 , 0 ) , ( 3 , 3 ) ( – 1 , 1 ) , ( 2 , 0 ) , ( 3 , 3 ) and ( 0 , 4 ) ( 0 , 4 ) Find the slopes of the diagonals in the previous exercise. Are they perpendicular? Real-World ApplicationsThe slope for a wheelchair ramp for a home has to be 1 12 . 1 12 . If the vertical distance from the ground to the door bottom is 2.5 ft, find the distance the ramp has to extend from the home in order to comply with the needed slope. If the profit equation for a small business selling x x number of item one and y y number of item two is p = 3 x + 4 y , p = 3 x + 4 y , find the y y value when p = $ 453 and x = 75. p = $ 453 and x = 75. For the following exercises, use this scenario: The cost of renting a car is $45/wk plus $0.25/mi traveled during that week. An equation to represent the cost would be y = 45 + .25 x , y = 45 + .25 x , where x x is the number of miles traveled. What is your cost if you travel 50 mi? If your cost were $ 63.75 , $ 63.75 , how many miles were you charged for traveling? Suppose you have a maximum of $100 to spend for the car rental. What would be the maximum number of miles you could travel? 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This problem has been solved!You'll get a detailed solution from a subject matter expert that helps you learn core concepts. Question: Unit 2: Linear Functions Date: Bell: Homework 3: Writing Linear Equations, Applications, & Linear Regression **This is a 2-page documenti ** Point Slope & Two Points: Write a linear equation in slope-intercept form with the given Information 1. slope = -6; passes through (-4,1) 2. slope = passes through (-5, -6) 3. passes through (-4, 11) and (2,8) 4. passes Slope = − 6 passes through ( − 4 , 1 ) .  Not the question you’re looking for?Post any question and get expert help quickly.  Slope Intercept Form WorksheetStudents will practice working with Slope Intercept Form including writing the equation of line given either A) Slope and Intercept B) Slope and a point or C) two points. Also students will practice writing the Slope Intercept Equation of a Line from its graph.  Example QuestionsFind the equation of a line with the given Slope and Y-intercept . Express your answer in Slope Intercept Form  Find the equation of the line with the given Slope that passes through the given point.  Write the equation of the line pictured in the graph below. Express your answer in Slope Intercept Form  Other DetailsThis is a 4 part worksheet:
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Unit 4: Linear Equations Homework 8: Writing Linear Equations REVIEW Direcüons: Write the linear equation in slope-intercept form given the following: 1. slope = Z; y-intercept = O 5. 35 2. slope = 1; y-intercept = -4 10. (2, -4); slope = -3 2) 12. (-2, -4) and (-3, -3) -3+4 Gina Wison (Al Algebra', LLC). 2012-2017
Homework Answer Keys Final Exam Materials Calculator TIps Answer keys are listed and organized by unit. ... Writing Equations from a Graph: File Size: 459 kb: File Type: pdf: Download File. Homework 8: File Size: ... SOLVING SYSTEMS OF LINEAR EQUATIONS. Homework 1: File Size: 1055 kb: File Type: pdf: Download File. Homework 2: File Size:
Study with Quizlet and memorize flashcards containing terms like What is the equation of the line that is parallel to the given line and passes through the point (−4,−6 )?, What is the equation of the line that is perpendicular to the given line and has an x-intercept of 6?, What is the equation, in point-slope form, of the line that is perpendicular to the given line and passes through ...
These Linear Equations Worksheets are a good resource for students in the 5th Grade through the 8th Grade. Writing Linear Equations Worksheets These Linear Equations Worksheets will produce problems for practicing writing linear equations from graphed lines.
Writing Linear Equations Date_____ Period____ Write the slope-intercept form of the equation of each line. 1) 3 x − 2y = − ... Write the standard form of the equation of the line through the given point with the given slope. 9) through: (1, 2), slope = 7 7x − y = 5
You can choose from SEVEN basic types of equations, ranging from simple to complex, explained below (such as one-step equations, variable on both sides, or having to use the distributive property). Customize the worksheets using the generator below. For some extra tips and practice problems on linear equations, check out IXL's linear equations ...
Write an equation in slope-intercept form for the line that satisfies each set of conditions. Substitute = (—4, 12) and (X2,Y2) = (—2, —4) in the slope formula m = Substitute m = —8 and (XI,YI) = (—4, 12) in the point-slope form y— = — Add 12 to each side. Write an equation in slope-intercept form for the line described.
Step by step guide to writing linear equations The equation of a line in slope intercept form is: \(\color{blue}{y=mx+b}\) Identify the slope.
SECTION 2.1 Linear Equations MATH 1310 College Algebra 81 Chapter 2 Solving Equations and Inequalities Section 2.1: Linear Equations Solving an Equation ... graphically, and then check your answer algebraically. 21. 2x −5 = 3 22. −3x −1= 5
Homework 1 U4: L1 (Notes) Writing Linear Equations in Slope-Intercept Form 5.1 Pg 276-277 # 1-25 ODDS, 28 , 30 2 U4: L1b (Notes) Writing Linear Inequalities Given a Graph in Slope-Intercept Form n/a Supplemental Worksheet 3 U4: L2 (Notes) Writing Linear Equations Given Slope and Point (Also… Parallel Lines) 5.2 Pg 282 # 1-11 ALL
WWriting Linear Equationsriting Linear Equations Given a point on a line and the slope of the line, you can write an equation of the line. Example 1 Write an equation in slope-intercept form of the line that passes through the point (−5, 6) and has a slope of 3— 5. Write the slope-intercept form.y = mx + b 6 = 3— — 5 (−5) + b for ...
Free worksheet(pdf) and answer key on the standard form equation of a line. 31 scaffolded questions that start relatively easy and end with some real challenges. Plus model problems explained step by step ... Find the x - and y - intercepts of the Standard Form Linear Equations below. Write each equation in Standard Form using integer ...
Write a linear equation in slope- intercept form that relates the number of shirts collected and the number of hours since midnight. y 3. MAPS The post office and city hall are marked on a coordinate plane. Write the equation of the line in slope- intercept form that passes through these two points. 4. RIGHT TRIANGLES The line containing the base
2-5 Writing Linear Functions Homework. 2-5 Writing and Interpreting Functions ANSWERS. 2-5 Writing and Interpreting Functions. 2-4 Homework ANSWERS. 2-4 Homework. 2-4 Writing Linear Equations ANSWERS. 2-4 Writing Linear Equations. 2-3 Function Notation Practice 2 ANSWERS. 2-3 Function Notation Practice 2.
A linear equation is an equation for a straight line. These are all linear equations: y = 2x + 1 : 5x = 6 + 3y : y/2 = 3 − x: ... There are many ways of writing linear equations, but they usually have constants (like "2" or "c") and must have simple variables (like "x" or "y"). Examples: These are linear equations:
Put students into pairs and show an equation on the board. Have one student instruct the other on how to solve as the student listening writes each step and solution. Then, show a new equation and have students switch roles. This gives students a chance to teach and reinforce what they remember about linear equations.
A linear equation is an equation of a straight line, written in one variable. The only power of the variable is 1. Linear equations in one variable may take the form a x + b = 0 a x + b = 0 and are solved using basic algebraic operations. We begin by classifying linear equations in one variable as one of three types: identity, conditional, or ...
Math; Algebra; Algebra questions and answers; Unit 2: Linear Functions Date: Bell: Homework 3: Writing Linear Equations, Applications, & Linear Regression **This is a 2-page documenti ** Point Slope & Two Points: Write a linear equation in slope-intercept form with the given Information 1. slope = -6; passes through (-4,1) 2. slope = passes through (-5, -6) 3. passes through (-4, 11) and (2,8 ...
Objective. Students will practice working with Slope Intercept Form including writing the equation of line given either A) Slope and Intercept B) Slope and a point or C) two points. Also students will practice writing the Slope Intercept Equation of a Line from its graph.
is linear, write an equation in point-slope form to represent the temperature y at x hour. y — 35 = — 1) Hour 35 39 47 10. SPEED After 2 hours, a car travels 70 miles. After 2.25 hours in the same trip, the car travels 78.75 miles. Write an equation in point-slope form to represent the distance y of the car after x hours. Y — 70 = 35(x — 2)