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8.E: Solving Linear Equations (Exercises)
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8.1 - Solve Equations using the Subtraction and Addition Properties of Equality
In the following exercises, determine whether the given number is a solution to the equation.
- x + 16 = 31, x = 15
- w − 8 = 5, w = 3
- −9n = 45, n = 54
- 4a = 72, a = 18
In the following exercises, solve the equation using the Subtraction Property of Equality.
- y + 2 = −6
- a + \(\dfrac{1}{3} = \dfrac{5}{3}\)
- n + 3.6 = 5.1
In the following exercises, solve the equation using the Addition Property of Equality.
- u − 7 = 10
- x − 9 = −4
- c − \(\dfrac{3}{11} = \dfrac{9}{11}\)
- p − 4.8 = 14
In the following exercises, solve the equation.
- n − 12 = 32
- y + 16 = −9
- f + \(\dfrac{2}{3}\) = 4
- d − 3.9 = 8.2
- y + 8 − 15 = −3
- 7x + 10 − 6x + 3 = 5
- 6(n − 1) − 5n = −14
- 8(3p + 5) − 23(p − 1) = 35
In the following exercises, translate each English sentence into an algebraic equation and then solve it.
- The sum of −6 and m is 25.
- Four less than n is 13.
In the following exercises, translate into an algebraic equation and solve.
- Rochelle’s daughter is 11 years old. Her son is 3 years younger. How old is her son?
- Tan weighs 146 pounds. Minh weighs 15 pounds more than Tan. How much does Minh weigh?
- Peter paid $9.75 to go to the movies, which was $46.25 less than he paid to go to a concert. How much did he pay for the concert?
- Elissa earned $152.84 this week, which was $21.65 more than she earned last week. How much did she earn last week?
8.2 - Solve Equations using the Division and Multiplication Properties of Equality
In the following exercises, solve each equation using the Division Property of Equality.
- 13a = −65
- 0.25p = 5.25
- −y = 4
In the following exercises, solve each equation using the Multiplication Property of Equality.
- \(\dfrac{n}{6}\) = 18
- y −10 = 30
- 36 = \(\dfrac{3}{4}\)x
- \(\dfrac{5}{8} u = \dfrac{15}{16}\)
In the following exercises, solve each equation.
- −18m = −72
- \(\dfrac{c}{9}\) = 36
- 0.45x = 6.75
- \(\dfrac{11}{12} = \dfrac{2}{3} y\)
- 5r − 3r + 9r = 35 − 2
- 24x + 8x − 11x = −7−14
8.3 - Solve Equations with Variables and Constants on Both Sides
In the following exercises, solve the equations with constants on both sides.
- 8p + 7 = 47
- 10w − 5 = 65
- 3x + 19 = −47
- 32 = −4 − 9n
In the following exercises, solve the equations with variables on both sides.
- 7y = 6y − 13
- 5a + 21 = 2a
- k = −6k − 35
- 4x − \(\dfrac{3}{8}\) = 3x
In the following exercises, solve the equations with constants and variables on both sides.
- 12x − 9 = 3x + 45
- 5n − 20 = −7n − 80
- 4u + 16 = −19 − u
- \(\dfrac{5}{8} c\) − 4 = \(\dfrac{3}{8} c\) + 4
In the following exercises, solve each linear equation using the general strategy.
- 6(x + 6) = 24
- 9(2p − 5) = 72
- −(s + 4) = 18
- 8 + 3(n − 9) = 17
- 23 − 3(y − 7) = 8
- \(\dfrac{1}{3}\)(6m + 21) = m − 7
- 8(r − 2) = 6(r + 10)
- 5 + 7(2 − 5x) = 2(9x + 1) − (13x − 57)
- 4(3.5y + 0.25) = 365
- 0.25(q − 8) = 0.1(q + 7)
8.4 - Solve Equations with Fraction or Decimal Coefficients
In the following exercises, solve each equation by clearing the fractions.
- \(\dfrac{2}{5} n − \dfrac{1}{10} = \dfrac{7}{10}\)
- \(\dfrac{1}{3} x + \dfrac{1}{5} x = 8\)
- \(\dfrac{3}{4} a − \dfrac{1}{3} = \dfrac{1}{2} a + \dfrac{5}{6}\)
- \(\dfrac{1}{2}\)(k + 3) = \(\dfrac{1}{3}\)(k + 16)
In the following exercises, solve each equation by clearing the decimals.
- 0.8x − 0.3 = 0.7x + 0.2
- 0.36u + 2.55 = 0.41u + 6.8
- 0.6p − 1.9 = 0.78p + 1.7
- 0.10d + 0.05(d − 4) = 2.05
PRACTICE TEST
- \(\dfrac{23}{5}\)
- n − 18 = 31
- 4y − 8 = 16
- −8x − 15 + 9x − 1 = −21
- −15a = 120
- \(\dfrac{2}{3}\)x = 6
- x + 3.8 = 8.2
- 10y = −5y + 60
- 8n + 2 = 6n + 12
- 9m − 2 − 4m + m = 42 − 8
- −5(2x + 1) = 45
- −(d + 9) = 23
- 2(6x + 5) − 8 = −22
- 8(3a + 5) − 7(4a − 3) = 20 − 3a
- \(\dfrac{1}{4} p + \dfrac{1}{3} = \dfrac{1}{2}\)
- 0.1d + 0.25(d + 8) = 4.1
- Translate and solve: The difference of twice x and 4 is 16.
- Samuel paid $25.82 for gas this week, which was $3.47 less than he paid last week. How much did he pay last week?
Contributors and Attributions
Lynn Marecek (Santa Ana College) and MaryAnne Anthony-Smith (Formerly of Santa Ana College). This content is licensed under Creative Commons Attribution License v4.0 "Download for free at http://cnx.org/contents/[email protected] ."
A zeroing feedback gradient-based neural dynamics model for solving dynamic quadratic programming problems with linear equation constraints in finite time
- Original Article
- Published: 27 May 2024
Cite this article
- Shangfeng Du 1 ,
- Dongyang Fu ORCID: orcid.org/0000-0003-0426-4356 1 ,
- Long Jin 2 ,
- Yang Si 1 &
- Yongze Li 1
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Gradient-based neural dynamics (GND) models are a classical algorithm for solving optimization problems, but it has non-negligible flaws in solving dynamic problems. In this study, a novel GND model, namely the zeroing feedback gradient-based neural dynamics (ZF-GND) models, is proposed based on the original GND model for tracking down the exact solution of dynamic quadratic programming problem (DQP). Further, a nonlinear projection function is designed to accelerate the convergence of the model. An upper bound on the convergence time of the ZF-GND model is rigorously defined through theoretical analysis. The superior effect of the ZF-GND model in terms of convergence is verified through comparison experiments. Finally, an application of robot motion planning is introduced to verify the practicality of the ZF-GND model.
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Acknowledgements
This research was funded in part by the National Key Research and Development Program of China under Grant No. 2022YFC3103101, Key Special Project for Introduced Talents Team of Southern Marine Science and Engineering Guangdong Laboratory under Contract GML2021GD0809, National Natural Science Foundation of China under Contract 42206187, Key projects of the Guangdong Education Department under Grant No. 2023ZDZX4009.
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Du, S., Fu, D., Jin, L. et al. A zeroing feedback gradient-based neural dynamics model for solving dynamic quadratic programming problems with linear equation constraints in finite time. Neural Comput & Applic (2024). https://doi.org/10.1007/s00521-024-09762-3
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DOI : https://doi.org/10.1007/s00521-024-09762-3
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CHAPTER 2 Solving Equations and Inequalities 84 University of Houston Department of Mathematics Additional Example 2: Solution: Additional Example 3: Solution: We first multiply both sides of the equation by 12 to clear the equation of fractions. Then solve as usual.
Equations Understand solving linear equations. • I can solve simple and multi-step equations. • I can describe how to solve equations. • I can analyze the measurements used to solve a problem and judge the level of accuracy appropriate for the solution. • I can apply equation-solving techniques to solve real-life problems. 1.1 Solving ...
Chapter 1: Linear Equations 1.1 Solving Linear Equations - One Step Equations Solving linear equations is an important and fundamental skill in algebra. In algebra, we are often presented with a problem where the answer is known, but part of the problem is missing. The missing part of the problem is what we seek to find.
In this unit we are going to be looking at simple equations in one variable, and the equations will be linear - that means there'll be no x2 terms and no x3's, just x's and numbers. For example, we will see how to solve the equation 3x+15 = x+25. 2. Solving equations by collecting terms Suppose we wish to solve the equation 3x+15 = x+25
Precalculus: Linear Equations Practice Problems 7. 9(x+3)−6 = 24−2x−3+11x 9x+27−6 = 21+9x 9x+21 = 21+9x 9x+21−9x = 21+9x−9x 21 = 21 We have to interpret what we have found. Since 21 is always equal to 21, the equation is true for any value of x that we try. Therefore, there are an infinite number of solutions. 8. y = −2x+1
Use linear equations to solve real-life problems. Solving Linear Equations by Adding or Subtracting An equation is a statement that two expressions are equal. A linear equation in one variable is an equation that can be written in the form ax + b = 0, where a and b are constants and a ≠ 0.
I. Linear Equations a. Definition: A linear equation in one unknown is an equation in which the only exponent on the unknown is 1. b. The General Form of a basic linear equation is: ax b c. c. To Solve: the goal is to write the equation in the form variable = constant. d. The solution to an equation is the set of all values that check in the ...
SOLVING LINEAR EQUATIONS Recall that whatever operation is performed on one side of the equation must also be performed on the other. Remember that when an equation involves fractions you can multiply both sides of the equation by the least common denominator and proceed as usual. Model Problems: Solve: 1. 8x 7 55 2. 2 17 3 1 y y 3.
tem of linear equations. Linear algebra arose from attempts to find systematic methods for solving these systems, so it is natural to begin this book by studying linear equations. If a, b, and c are real numbers, the graph of an equation of the form ax+by =c is a straight line (if a and b are not both zero), so such an equation is called a ...
Steps for Solving a Linear Equation in One Variable: Simplify both sides of the equation. Use the addition or subtraction properties of equality to collect the variable terms on one side of the equation and the constant terms on the other. Use the multiplication or division properties of equality to make the coefficient of the variable term ...
A linear equation is an algebraic equation with a degree of 1. This means that the highest exponent on any variable in the equation is 1. A . linear equation in one variable . can be written in the form . ax + b = c , where . a, b, and . c . are real numbers. General guidelines for solving linear equations in one variable: 1. Simplify anything ...
No solution. {−3} {−7} (1) Divide by 5 first, or (2) Distribute the 5 first. Create your own worksheets like this one with Infinite Algebra 2. Free trial available at KutaSoftware.com.
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 pre-algebra and algebra 1 courses (grades 6-9).
Linear Equations in Two Variables In this chapter, we'll use the geometry of lines to help us solve equations. Linear equations in two variables If a, b,andr are real numbers (and if a and b are not both equal to 0) then ax+by = r is called a linear equation in two variables. (The "two variables" are the x and the y.)
What is most important is that they are e↵ective: they succeed in solving any linear equation. Let's apply these ideas to the equations of parts a through g of example 9. Example 9Solutions. a. 2(x +5) = 3x 1; Simplify the left side: 2x +10 = 3x 1; Subtract 2x from both sides: 10 = x 1; Add 1 to both sides: 11 = x.
Step 1: Notice that the coefi cients of the y-terms are opposites. So, you can add the equations to obtain an equation in one variable, x. 2x 14 Add the equations. Step 2: Solve for x. x 7 Divide each side by 2. Step 3: Substitute 7 for x in one of the original equations and solve for y. 7 3y 2 Substitute 7 for .
Solution: Translating the problem into an algebraic equation gives: 2x − 5 = 13 2 x − 5 = 13. We solve this for x x. First, add 5 to both sides. 2x = 13 + 5, so that 2x = 18 2 x = 13 + 5, so that 2 x = 18. Dividing by 2 gives x = 182 = 9 x = 18 2 = 9. c) A number subtracted from 9 is equal to 2 times the number.
8.1 - Solve Equations using the Subtraction and Addition Properties of Equality. In the following exercises, determine whether the given number is a solution to the equation. x + 16 = 31, x = 15. w − 8 = 5, w = 3. −9n = 45, n = 54. 4a = 72, a = 18. In the following exercises, solve the equation using the Subtraction Property of Equality.
this section we solve systems of linear equations in two variables and use systems to solve problems. Solving a System by Graphing Because the graph of each linear equation is a line, points that satisfy both equations lie on both lines. For some systems these points can be found by graphing. EXAMPLE 1 A system with only one solution
This topic covers: - Intercepts of linear equations/functions - Slope of linear equations/functions - Slope-intercept, point-slope, & standard forms - Graphing linear equations/functions - Writing linear equations/functions - Interpreting linear equations/functions - Linear equations/functions word problems
So for the rectangle of length 8 and width 3 the formula would give, P =2(8) +2(3) = 16 + 6= 22. With problems that we will consider here the formula P =2L +2W will be used. Example 7. 4. The perimeter of a rectangle is 44. The length is 5 less than double the width. Find the dimensions.
14 Chapter 1 Solving Linear Equations Solving Real-Life Problems Modeling with Mathematics Use the table to fi nd the number of miles x you need to bike on Friday so that the mean number of miles biked per day is 5. SOLUTION 1. Understand the Problem You know how many miles you biked Monday through Thursday. You are asked to fi nd the number
A hybrid analytical method for solving linear and nonlinear fractional partial differential equations is presented. The proposed analytical approach is an elegant combination of the Natural Transform Method (NTM) and a well-known method, Homotopy Perturbation Method (HPM).
This paper briefly examines how literature addresses the numerical solution of partial differential equations by the spectral Tau method. It discusses the implementation of such a numerical solution for PDE's presenting the construction of the problem's algebraic representation and exploring solution mechanisms with different orthogonal polynomial bases. It highlights contexts of ...
Study is devoted to solving dynamics quadratic programming (DQP) problems with linear equation constraints, and a zeroing feedback GND model (ZF-GND) is proposed. The ZF-GND model introduces a time-varying term based on the GND model, which can eliminate the time delay problem in the computation of the GND model and enables the model to obtain ...