Like terms equations examples
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The document provides examples of solving linear equations in three steps: 1) combining like terms, 2) using the inverse operation to isolate the variable, and 3) dividing to solve for the variable. In example 1, the equation 2x + 6x = -24 is solved to get x = -3. In example 2, the equation 8a + 3 - 2a = -17 is solved to get a = -10/3.
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The document reviews the basic rules for exponents that were covered in Algebra 1, including:
- When multiplying like variables, you add their exponents
- When dividing like variables, you subtract their exponents
- When an exponent is raised to another exponent, you multiply the exponents
- When a variable has a negative exponent, you change its position in a fraction and the exponent becomes positive
The document then provides examples of applying multiple exponent rules to evaluate multi-step algebraic expressions involving variables with exponents.
Mathematics 8 Systems of Linear Inequalities
This learner's module discusses and help the students about the topic Systems of Linear Inequalities. It includes definition, examples, applications of Systems of Linear Inequalities.
Solving Systems - Elimination NOTES
The document discusses the elimination method for solving systems of equations. It explains that the elimination method involves choosing a variable to eliminate, adding or subtracting the equations to cancel out the variable, solving the resulting equation for the remaining variable, and then back-substituting to solve for the other variable. Examples are provided demonstrating how to use both addition and multiplication when eliminating variables to solve systems of two equations.
Linear Rules
This document discusses linear relationships and rules for determining the relationship between x and y values in a table. There are three main types of linear rules: 1) simple addition or subtraction, 2) simple multiplication or division, and 3) combination rules using y=mx+c. To determine the rule, you first check if it follows addition/subtraction by looking for a consistent difference between y-x values. If not, you check for multiplication/division by looking at y/x values. If neither, it uses a combination rule where you calculate the slope m from the change in y over change in x, use a point to find the y-intercept c, then write the rule as y=mx+c. Examples
3 2 solving systems of equations (elimination method)
The document describes the elimination method for solving systems of equations. The key steps are:
1) Write both equations in standard form Ax + By = C
2) Determine which variable to eliminate using addition or subtraction
3) Solve the resulting equation for one variable
4) Substitute back into the original equation to solve for the other variable
5) Check that the solution satisfies both original equations
It provides examples showing how to set up and solve systems of equations using elimination, including word problems about supplementary angles and finding two numbers based on their sum and difference.
M1 L5 Remediation Notes
This document discusses solving systems of equations and inequalities through three main methods: graphing, substitution, and elimination. It provides examples of each method. For graphing systems, it explains the three possibilities for the graphs: consistent systems with one solution where the lines intersect, inconsistent systems with no solution where the lines are parallel, and dependent systems with infinite solutions where the lines coincide. It then works through examples of using substitution and elimination to solve systems algebraically. [/SUMMARY]
Elimination method Ch 7
The document discusses the elimination method for solving systems of equations. It provides examples of using addition or multiplication to eliminate a variable and obtain a solution. Key steps include adding or multiplying the equations appropriately, substituting values back to solve for the remaining variable, and obtaining the solution point (x,y).
Business Math Chapter 3
The document covers systems of linear equations, including how to solve them using substitution and elimination methods. It provides examples of solving systems of equations with one solution, no solution, and infinitely many solutions. Quadratic equations are also discussed, including how to solve them by factoring, using the quadratic formula, and identifying the nature of solutions based on the discriminant.
.
This document provides an overview of basic algebra concepts including:
1. Variables, expressions, equations, and manipulating equations through addition, subtraction, multiplication and division while maintaining equality.
2. Solving one-variable equations by isolating the variable on one side of the equation.
3. Calculating the slope of a line using the slope-intercept form given two points on the line.
Linear systems with 3 unknows
This document provides examples for solving systems of linear equations in three variables. It begins with an example using elimination to solve the system 5x - 2y - 3z = -7, etc. step-by-step, reducing it to a 2x2 system and solving for x, y, and z. The next example uses substitution to solve a word problem about ticket sales. It shows setting up and solving a 3x3 system. The document concludes with an example of a system having an infinite number of solutions.
Final presentation
This document provides an overview of solving systems of linear equations through three methods: graphing, substitution, and elimination. It defines systems of linear equations as two or more linear equations with the same variables, where the point of intersection is the solution. Examples are worked through for each solving method. Graphing involves putting the equations in y-intercept form and finding the point where the lines intersect on a graph. Substitution involves solving one equation for a variable and substituting it into the other equation. Elimination involves adding or subtracting the equations to eliminate a variable and then solving for the remaining variable.
Elimination Method Mathematics 8 Linear Equation In 2 variables .pptx
The document provides steps for solving systems of linear equations using the elimination method. It begins with an example of a system and shows the steps of choosing a variable to eliminate, multiplying an equation by a constant to create additive inverses, adding the equations to eliminate the variable, solving the resulting equation for the remaining variable, substituting back into the original equation to find the other variable, and checking the solution. It then provides additional examples demonstrating this process.
Solving Systems by Graphing and Substitution
The document discusses solving systems of linear equations by graphing and substitution. It provides examples of solving systems by graphing, including putting equations in slope-intercept form and finding the intersection point. It also gives examples of solving systems by substitution, including steps like isolating a variable, substituting one equation into the other, solving the resulting equation, and checking solutions. Special cases like no solution and infinitely many solutions are also addressed.
Alg2 lesson 3-2
The document provides examples of using substitution and elimination methods to solve systems of equations. It shows setting one equation equal to the other and solving for one variable in terms of the other to use substitution. It also demonstrates setting corresponding terms of equations equal and combining to solve for one variable and back substitute to find the other variable when using elimination.
Maths
This document discusses linear equations in two variables. It defines linear equations in two variables as equations of the form ax + by = c, where a, b, and c are real numbers and a and b are not both zero. It explains that the graph of any linear equation in two variables is a straight line. It also categorizes different types of systems of linear equations based on the relationship between the lines: intersecting lines have a unique solution; coincident lines have an infinite number of solutions; and parallel lines have no solution. Methods for solving systems of linear equations like substitution, elimination, and graphing are also covered.
Elimination of Systems of Linear Equation
The document discusses solving systems of linear equations by elimination. It involves eliminating one variable at a time through addition or subtraction of equations. This leaves an equation with one variable that can be solved for its value, which is then substituted back into the original equations to solve for the other variable. Two examples are provided showing the full process of setting up equations, eliminating variables, solving for values, and checking solutions.
Lecture 11 systems of nonlinear equations
The document discusses solving systems of nonlinear equations in two variables. It provides examples of nonlinear systems that contain equations that are not in the form Ax + By = C, such as x^2 = 2y + 10. Methods for solving nonlinear systems include substitution and addition. The substitution method involves solving one equation for one variable and substituting into the other equation. The addition method involves rewriting the equations and adding them to eliminate variables. Examples demonstrate both methods and finding the solution set that satisfies both equations.
Exponent properties
This document discusses the rules for simplifying expressions with variables that have exponents.
1) When adding or subtracting terms with the same base and exponent, you combine the coefficients and keep the base and exponent the same.
2) When multiplying terms, you add the exponents and multiply the coefficients.
3) When dividing terms, you subtract the bottom exponent from the top exponent and divide the coefficients.
Laws of exponents
Exponents describe how many times to use a number in a multiplication. The laws of exponents are based on three main ideas: 1) the exponent says how many times to multiply the base, 2) a negative exponent means to divide the base, and 3) a fractional exponent means to take a root of the base. Some key exponent rules are: am * an = am+n, am / an = am-n, (ab)n = an * bn, and an/b = (a/b)n. A zero exponent equals 1 for any base.
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3 2 solving systems of equations (elimination method)
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Elimination Method Mathematics 8 Linear Equation In 2 variables .pptx
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Solution 3
This document shows the step-by-step algebraic manipulation of the equation 9x + 7y - 1 = 0. Through adding 1 to both sides and subtracting 9x, the equation is transformed into y = -9x/7 + 1/7, representing the line that satisfies the original equation.
Solution 1
This document contains the step-by-step work to solve the equation 9x + 2y = 18 for y. It begins with the original equation, subtracts 9x from both sides, and then divides both sides by 2 to isolate y, resulting in the solution y = -9x/2 + 9.
Solving For x
The document shows the steps to solve the equation 3x + 6y - 8 = 4 for x. It begins with adding 8 to both sides, then subtracting 6y from both sides. This leaves 3x = -6y + 12, which is then divided by 3 to isolate x as x = -2y + 4.
Inequalities
The rules for solving inequalities are the same as for equations, except when dividing by a negative number, the inequality sign flips. This is demonstrated through examples of solving various types of inequalities, including those with fractions, variables on both sides, and using the distributive property. The key point is checking if the number divided by is positive or negative to determine if the inequality sign remains the same or flips.
Sol4
The document shows the step-by-step work of solving the equation 3 – 5(x + 1) = 21. It distributes the -5, combines like terms, and isolates x to find that the solution is x = 23/5. The key steps are to always distribute first, distribute the negative sign correctly, and show the work clearly at each step of the solution.
Sol3
The document shows the step-by-step work of solving the equation -2(4x + 5) +3 = -8. It begins with distributing the -2, combining like terms, and performing inverse operations until arriving at the solution of x = 1/8.
Sol2
This document shows the step-by-step work of solving the equation (3x - 6) = 24 for x. It begins with distributing the -1, then combining like terms and solving for x by first adding 6 to both sides and then dividing both sides by -3, resulting in the solution of x = 6.
Dp ex3
The document shows the step-by-step working of the expression 4(2x - 7), which equals 8x - 28 when fully simplified. It cautions that a common mistake is forgetting that the minus sign belongs to the 7 term, and not treating 2x - 7 as a single term.
Dp ex2
This document shows the step-by-step working of distributing a negative sign when multiplying a number by a binomial expression. It starts with the expression -3(6x + 1) and through distributing the negative sign, arrives at the equivalent expression -18x - 3 in 3 lines.
Dp ex1
This document provides an example of distributing a term over a parenthesis in an algebraic expression. It shows the steps of distributing the coefficient 7 over the terms in the parenthesis (x + 6), resulting in the equivalent expression of 7x + 42.
Solving 2 step equations examples
The document provides 7 examples of solving linear equations by performing inverse operations to isolate the variable. Each example shows the step-by-step work including adding, subtracting, multiplying, or dividing both sides of the equation by the same number to simplify it until the variable is alone on one side of the equation. The examples demonstrate solving equations for various types of linear expressions involving addition, subtraction, multiplication, and division of the variable.
Like terms slide show
The document provides 3 examples of combining like terms in algebraic expressions. Each example shows identifying like terms, combining their coefficients, and obtaining a final simplified expression. The examples involve adding and combining terms with variables x, a, and b.
Factoring quadratics ex 3.1
This document describes factoring the quadratic expression 1x^2 - 12x + 32. It shows the steps of multiplying the expression by 1, combining like terms, identifying the factors of the constant term 32 as 1 x 32 and 2 x 16, and determining that the factors of the expression are (x - 4)(x - 8) which results in the fully factored form.
Factoring quadratics ex 4
This document provides step-by-step work to factor the expression 1x^2 - 9. It begins with the expression, multiplies -9 to both sides, then factors -9 into 3*3. This leads to factoring the entire expression as (x-3)(x+3), showing the two factors that multiply to the original expression.
Factoring quadratics ex 1
This document shows the step-by-step factorization of the expression 1x^2 + 6x + 8. It begins with the original expression and shows multiplying and adding like terms. The expression is then factored into (x + 2)(x + 4), showing the work and reasoning for combining the factors.
Perp hints
Two lines are perpendicular if their slopes have a product of -1. The slopes of perpendicular lines will always satisfy the property of having a product of -1.
Perp
El documento contiene varias ecuaciones de líneas. Cada sección presenta dos ecuaciones de líneas, una en función de y = mx + b y la otra en función de y = bx + m. Las ecuaciones describen líneas con diferentes pendientes y ordenadas al origen.
Para hints
The document presents several pairs of lines with the same slope (m) but different y-intercepts (b), demonstrating that parallel lines have the same slope but different y-intercepts. It includes 10 pairs of lines in standard y=mx+b form to illustrate that while the m term is the same between parallel lines, the b term can vary between each pair without affecting their parallel nature.
Coin hints
This document contains 10 linear equations in the form of y=mx+b. The equations represent lines with different slopes and y-intercepts, except for the last two lines which are coincident since they have the same slope of -4 and y-intercept of 0.
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Like terms equations examples
- 1. Ex 1 2x + 6x = -24 1. Simplify (combine like terms) first 2. Use inverse to get rid of constant 3. Divide
- 2. Ex 1 2x + 6x = -24 8x = -24 1. Simplify (combine like terms) first 2. Use inverse to get rid of constant 3. Divide
- 3. Ex 1 2x + 6x = -24 8x = -24 1. Simplify (combine like terms) first 2. Use inverse to get rid of constant There are no constants 3. Divide
- 4. Ex 1 2x + 6x = -24 8x = -24 8 8 1. Simplify (combine like terms) first 2. Use inverse to get rid of constant There are no constants 3. Divide
- 5. Ex 1 2x + 6x = -24 8x = -24 8 8 1. Simplify (combine like terms) first 2. Use inverse to get rid of constant There are no constants 3. Divide x= -3
- 6. Ex 2 8a + 3 – 2a = -17 1. Simplify (combine like terms) first 2. Use inverse to get rid of constant 3. Divide
- 7. Ex 2 8a + 3 – 2a = -17 6a + 3= -17 1. Simplify (combine like terms) first 2. Use inverse to get rid of constant 3. Divide
- 8. Ex 2 8a + 3 – 2a = -17 6a + 3= -17 -3 -3 6a = -20 1. Simplify (combine like terms) first 2. Use inverse to get rid of constant 3. Divide
- 9. Ex 2 8a + 3 – 2a = -17 6a + 3= -17 -3 -3 6a = -20 6 6 1. Simplify (combine like terms) first 2. Use inverse to get rid of constant 3. Divide
- 10. Ex 2 8a + 3 – 2a = -17 6a + 3= -17 -3 -3 6a = -20 6 6 a = -10/3 1. Simplify (combine like terms) first 2. Use inverse to get rid of constant 3. Divide