TYPES OF SOLUTIONS OF LINEAR EQUATIONS

Slides:

Advertisements

Similar presentations

SOLUTION EXAMPLE 1 A linear system with no solution Show that the linear system has no solution. 3x + 2y = 10 Equation 1 3x + 2y = 2 Equation 2 Graph the.

Advertisements

Linear Equations with Different Kinds of Solutions

Solve an equation with variables on both sides

SOLVING SYSTEMS USING SUBSTITUTION

Solve an equation using subtraction EXAMPLE 1 Solve x + 7 = 4. x + 7 = 4x + 7 = 4 Write original equation. x + 7 – 7 = 4 – 7 Use subtraction property of.

Solving Equations with variables on both sides of the Equals Chapter 3.5.

Step 1: Simplify Both Sides, if possible Distribute Combine like terms Step 2: Move the variable to one side Add or Subtract Like Term Step 3: Solve for.

3-4 Solving Systems of Linear Equations in 3 Variables

The student will be able to: solve equations with variables on both sides. Equations with Variables on Both Sides Objectives Designed by Skip Tyler, Varina.

Solve Equations with Variables on Both Sides

5-3 Equations as Relations

3.3 Equations w/ Variables on both sides. 3.3 – Eq. w/ Variables on both sides Goals / “I can…”  Solve equations with variables on both sides  Identify.

Martin-Gay, Beginning Algebra, 5ed Using Both Properties Divide both sides by 3. Example: 3z – 1 = 26 3z = 27 Simplify both sides. z = 9 Simplify.

1.4 Solving Multi-Step Equations. To isolate the variable, perform the inverse or opposite of every operation in the equation on both sides of the equation.

1. solve equations with variables on both sides. 2. solve equations containing grouping symbols. Objectives The student will be able to:

Use the substitution method

Solve Linear Systems by Substitution January 28, 2014 Pages

1. solve equations with variables on both sides. 2. solve equations with either infinite solutions or no solution Objectives The student will be able to:

Solve Linear Systems by Substitution Students will solve systems of linear equations by substitution. Students will do assigned homework. Students will.

Multi Step Equations. Algebra Examples 3/8 – 1/4x = 1/2x – 3/4 3/8 – 1/4x = 1/2x – 3/4 8(3/8 – 1/4x) = 8(1/2x – 3/4) (Multiply both sides by 8) 8(3/8.

Notes 3.4 – SOLVING MULTI-STEP INEQUALITIES

Solving Equations with Variable on Both Sides Objective: Students will solve equations with variables on both sides. Section 3.4.

Reasoning with linear equations and inequalities Students will understand that equation is a statement of equality between two expression. Students find.

6-2 Solving Systems Using Substitution Hubarth Algebra.

Martin-Gay, Beginning Algebra, 5ed

Solving Equations with Variables on Both Sides. Review O Suppose you want to solve -4m m = -3 What would you do as your first step? Explain.

3.5 Solving Equations with Variables on Both Sides.

Substitution Method: Solve the linear system. Y = 3x + 2 Equation 1 x + 2y=11 Equation 2.

Rewrite a linear equation

Q1W5: Solving Linear Equations With One Variable

Objectives The student will be able to:

Objectives The student will be able to:

6-3: Solving Equations with variables on both sides of the equal sign

SOLVING ONE-VARIABLE EQUATIONS •. Goal: Find the one value

Parallel and Perpendicular Lines

ALGEBRA 1 CHAPTER 7 LESSON 5 SOLVE SPECIAL TYPES OF LINEAR SYSTEMS.

Example: Solve the equation. Multiply both sides by 5. Simplify both sides. Add –3y to both sides. Simplify both sides. Add –30 to both sides. Simplify.

Solving Equations with the Variable on Both Sides

Equations with variables on both sides Whiteboard practice

Equations with Variables on Both Sides

6-2 Solving Systems Using Substitution

Solving Systems using Substitution

Objectives The student will be able to:

Objectives The student will be able to:

Equations: Multi-Step Examples ..

1.3 Solving Linear Equations

2 Understanding Variables and Solving Equations.

Equations and Inequalities

Equations with Variables on Both Sides

Equations with Variables on Both Sides

Linear Equations and Applications

Solving Multi-Step Equations

Justify your reasoning.

Evaluating expressions and Properties of operations

- Finish Unit 1 test - Solving Equations variables on both sides

Equations with variables on both sides Whiteboard practice

Solving Multi-Step Equations

What is the difference between simplifying and solving?

Section Solving Linear Systems Algebraically

Solving Multi-Step Equations

Objectives The student will be able to:

Objectives The student will be able to:

Objectives The student will be able to:

Warm-Up 2x + 3 = x + 4.

2-3 Equations With Variables on Both Sides

Objectives The student will be able to:

Objectives The student will be able to:

Solving Equations with Fractions

EXAMPLE 4 Solve proportions Solve the proportion. ALGEBRA a x 16

Presentation transcript:

TYPES OF SOLUTIONS OF LINEAR EQUATIONS Algebraic Form Number of Solutions Description a = b None There are no values of the variable for which the equation is true. x = a One The equation is true for exactly one value of the variable. a = a Infinitely many The equation is true for all values of the variable.

1. 2x – 4 = -x - 1 + x + x 3x – 4 = - 1 + 4 + 4 3x = 3 3 = 3 x = 1 Tell whether each equation has one solution, infinitely many solutions, or no solutions. Justify your answer. 1. 2x – 4 = -x - 1 + x + x Add 1x to each side Simplify Add 4 to each side Divide both sides by 3 3x – 4 = - 1 + 4 + 4 3x = 3 3 = 3 x = 1 The result is an equation of the form x = a. This equation is true for exactly one value. So, the equation has one solution.

Tell whether each equation has one solution, infinitely many solutions, or no solutions. Justify your answer. 2. 2x – 4 = 2(x – 2) Distributive Property Subtract 2x from both sides Simplify 2x – 4 = 2x – 4 - 2x - 2x – 4 = - 4 The result is an equation of the form a = a. This equation is true for all values of x. So, the equation has infinitely many solutions.

Tell whether each equation has one solution, infinitely many solutions, or no solutions. Justify your answer. 3. 2x – 4 = 2(x + 1) Distributive Property Subtract 2x from both sides Simplify 2x – 4 = 2x + 2 - 2x - 2x – 4 = 2 The result is an equation of the form a = b. There are no values of x for which the equation is true. So, the equation no solution.

a. 5x + 8 = 5(x + 3) b. 9x = 8 + 5x - 5x - 5x 5x + 8 = 5x + 15 Tell whether each equation has one solution, infinitely many solutions, or no solutions. Justify your answer. a. 5x + 8 = 5(x + 3) b. 9x = 8 + 5x - 5x - 5x 5x + 8 = 5x + 15 - 5x - 5x 4x = 8 8 = 15 4 4 x = 2 no solution form a = b one solution form x = a

c. 6x + 12 = 6(x + 2) d. 7x - 11 = 11 - 7x + 7x + 7x 6x + 12 = 6x + 12 Tell whether each equation has one solution, infinitely many solutions, or no solutions. Justify your answer. c. 6x + 12 = 6(x + 2) d. 7x - 11 = 11 - 7x + 7x + 7x 6x + 12 = 6x + 12 - 6x - 6x 14x - 11 = 11 + 11 + 11 12 = 12 14x = 22 14 14 x = 1 8 14 = 1 4 7 infinitely many solutions form a = a one solution form x = a