Example 5A: Solving Simple Rational Equations

Slides:

Advertisements

Similar presentations

Multiplying and Dividing Rational Expressions

Advertisements

Rational Expressions To add or subtract rational expressions, find the least common denominator, rewrite all terms with the LCD as the new denominator,

Algebraic Fractions and Rational Equations. In this discussion, we will look at examples of simplifying Algebraic Fractions using the 4 rules of fractions.

Solve an equation with variables on both sides

EXAMPLE 1 Solve a quadratic equation having two solutions Solve x 2 – 2x = 3 by graphing. STEP 1 Write the equation in standard form. Write original equation.

EXAMPLE 1 Solve a simple absolute value equation Solve |x – 5| = 7. Graph the solution. SOLUTION | x – 5 | = 7 x – 5 = – 7 or x – 5 = 7 x = 5 – 7 or x.

Ch 8 - Rational & Radical Functions 8.5 – Solving Rational Equations.

EXAMPLE 1 Solve a linear-quadratic system by graphing Solve the system using a graphing calculator. y 2 – 7x + 3 = 0 Equation 1 2x – y = 3 Equation 2 SOLUTION.

EXAMPLE 3 Solve an equation by factoring Solve 2x 2 + 8x = 0. 2x 2 + 8x = 0 2x(x + 4) = 0 2x = 0 x = 0 or x + 4 = 0 or x = – 4 ANSWER The solutions of.

Standardized Test Practice

Solve a radical equation

Solve an equation with an extraneous solution

5.4 Exponential and Logarithmic Equations Essential Questions: How do we solve exponential and logarithmic equations?

and Logarithmic Equations

Substitute 0 for y. Write original equation. To find the x- intercept, substitute 0 for y and solve for x. SOLUTION Find the x- intercept and the y- intercept.

Substitute 0 for y. Write original equation. To find the x- intercept, substitute 0 for y and solve for x. SOLUTION Find the x- intercept and the y- intercept.

Solve an equation with an extraneous solution

( ) EXAMPLE 3 Standardized Test Practice SOLUTION 5 x = – 9 – 9

3-2 Solving Equations by Using Addition and Subtraction Objective: Students will be able to solve equations by using addition and subtraction.

3.6 Solving Absolute Value Equations and Inequalities

Inequalities Symbols and line graphs. Symbols  < is less than  > is greater than  < is less than or equal to  > is greater than or equal to points.

3.7 Absolute value DAY 2. Solve for x----no notes on this slide (just watch). |x| = 5 |x + 2| = 5 x = 5 or x = -5 x + 2 = 5 or x + 2 = -5 x =

Math – Rational Equations 1. A rational equation is an equation that has one or more rational expressions in it. To solve, we start by multiplying.

Solving Systems of Equations: The Elimination Method Solving Systems of Equations: The Elimination Method Solving Systems of Equations: The Elimination.

Solving Rational Equations

4.8 Using matrices to solve systems! 2 variable systems – by hand 3 or more variables – using calculator!

Notes Over 7.6 Solving a Simple Radical Equation Solve the equation. Check your solution.

Warm Up. Solving Differential Equations General and Particular solutions.

Objectives: 1.Be able to solve a system of equations by using the substitution method. 2.Be able to solve a system of equations by using the elimination.

 Cross Multiply – used to solve simple rational equations  Can only be done when each side has a single rational expression.

Holt McDougal Algebra Multiplying and Dividing Rational Expressions Simplify rational expressions. Multiply and divide rational expressions. Objectives.

8.2 M ULTIPLYING AND D IVIDING R ATIONAL E XPRESSIONS Simplify rational expressions. Multiply and divide rational expressions. Objectives rational expression.

Holt McDougal Algebra Exponential and Logarithmic Equations and Inequalities Solve logarithmic equations. Objectives.

x + 5 = 105x = 10  x = (  x ) 2 = ( 5 ) 2 x = 5 x = 2 x = 25 (5) + 5 = 105(2) = 10  25 = 5 10 = = 10 5 = 5.

Holt McDougal Algebra 2 Multiplying and Dividing Rational Expressions Multiplying and Dividing Rational Expressions Holt Algebra 2Holt McDougal Algebra.

A rational expression is a fraction with polynomials for the numerator and denominator. are rational expressions. For example, If x is replaced by a number.

Solving Radical Equations and Inequalities Objective: Solve radical equations and inequalities.

1. Add: 5 x2 – 1 + 2x x2 + 5x – 6 ANSWERS 2x2 +7x + 30

3.3 – Solving Systems of Inequalities by Graphing

( ) EXAMPLE 3 Standardized Test Practice SOLUTION 5 x = – 9 – 9

Solving Systems Using Substitution

Solve an equation by multiplying by a reciprocal

Solve a quadratic equation

Solving Linear Equations

3-2: Solving Systems of Equations using Substitution

Solving Equations by Factoring and Problem Solving

Solving One-Step Equations

Solve a system of linear equation in two variables

Chapter 3 Section 3.

3-2: Solving Systems of Equations using Substitution

Solving Systems of Equations using Substitution

Do Now 1) t + 3 = – 2 2) 18 – 4v = 42.

Equations with Variables on Both Sides Day 2

USING GRAPHS TO SOLVE EQUATIONS

Algebra 1 Section 13.6.

1.4 Solving Absolute-Value Equations

Skill Check over Solving Systems by Graphing after Homework Check

8.5 Solving Rational Equations

8.5 Solving Rational Equations

Solving a Radical Equation

3.5 Polynomial and Rational Inequalities

Solving Equations with Variables on Both Sides

1.4 Solving Absolute-Value Equations

Solving Equations with Variables on Both Sides

A rational expression is a quotient of two polynomials

Solving Equations Containing Rational Expressions § 6.5 Solving Equations Containing Rational Expressions.

Example 2B: Solving Linear Systems by Elimination

Example 1: Solving Rational Equations

11-5 Solving Rational Equations

Presentation transcript:

Example 5A: Solving Simple Rational Equations Solve. Check your solution. x2 – 25 x – 5 = 14 (x + 5)(x – 5) (x – 5) = 14 Note that x ≠ 5. x + 5 = 14 x = 9

Example 5A Continued x2 – 25 x – 5 = 14 Check (9)2 – 25 9 – 5 14 56 4 14 14 14

Example 5B: Solving Simple Rational Equations Solve. Check your solution. x2 + 3x - 10 x – 2 = 7 (x + 5)(x – 2) (x – 2) = 7 Note that x ≠ 2. x + 5 = 7 x = 2 Because the left side of the original equation is undefined when x = 2, there is no solution.

Example 5B Continued Check A graphing calculator shows that 2 is not a solution.

Check It Out! Example 5a Solve. Check your solution. x2 + x – 12 x + 4 = –7 (x – 3)(x + 4) (x + 4) = –7 Note that x ≠ –4. x – 3 = –7 x = –4 Because the left side of the original equation is undefined when x = –4, there is no solution.

Check It Out! Example 5a Continued Check A graphing calculator shows that –4 is not a solution.

Check It Out! Example 5b Solve. Check your solution. 4x2 – 9 2x + 3 = 5 (2x + 3)(2x – 3) (2x + 3) = 5 Note that x ≠ – . 3 2 2x – 3 = 5 x = 4

Check It Out! Example 5b Continued = 5 Check 4(4)2 – 9 2(4) + 3 5 55 11 5 5 5