C = 5d + 2 2c + d = 4 Do Now. Homework Solutions 4)5x – 3y = – 4 15x – 9y = – 12 3x + 2y = 9 – 15x + 10y = 45 – 19y = – 57 y = 3 3x + 2y = 9 3x + 2(3)

Slides:

Advertisements

Similar presentations

Practice Problems Return to MENU Practice Problems

Advertisements

9.3 Systems of Linear Equations and Problem Solving BobsMathClass.Com Copyright © 2010 All Rights Reserved. 1 Applications of Linear Systems of Equations.

7.4 HW Answers (4, -1) (5, 3) (-½, -2) (9, -3) (-10, -5) (19, 16) (5, 6) (-7, -12) (2, 1) (4, 4)

Word Problems Create and solve a system of equations to answer each word problem.

Purpose: To Solve Word Problems using two variables instead of one. Homework: p ALL.

Bell Ringer: Solve each system. 1) 4x – 6y = -4 8x + 2y = 48 2) y = x -2 4x + 2y = 14.

LINEAR SYSTEMS WORD PROBLEMS EQ: HOW ARE REAL-WORLD PROBLEMS SOLVED USING LINEAR SYSTEMS OF EQUATIONS? Please turn cleear everything but a pencil.

5.2 Solving Systems of Linear Equations by Substitution

A system of linear equations allows the relationship between two or more linear equations to be compared and analyzed. 5.1 – Systems of Linear Equations.

Please close your laptops and turn off and put away your cell phones, and get out your note-taking materials. Today’s daily homework quiz will be given.

5.1 Solving Inequalities by Addition and Subtraction

EXAMPLE 1 Writing and Solving an Inequality Season Tickets Individual tickets for a college hockey game cost $8 each plus a one-time transaction fee of.

Money Problems: By Dr. Marcia Tharp and Dr. Julia Arnold.

5-5B Linear Systems and Problems Solving Algebra 1 Glencoe McGraw-HillLinda Stamper.

Lesson #13- Translating Words into Math

Solving Systems of Equations:

Do Now Solve each system using Elimination. 2x + 3y = 8 x – y = 2

3.2 Solving Linear Systems by Substitution

Solving Systems of Equations using Substitution

Warm Up I can solve systems of equations by substitution.

Solving Systems Word Problems. Steps for Solving Word Problems Step 1: Define the 2 variables Step 2: Set up your 2 equations Step 3: Solve the system.

Please close your laptops and turn off and put away your cell phones, and get out your note-taking materials.

3x – 5y = 11 x = 3y + 1 Do Now. Homework Solutions 2)2x – 2y = – 6 y = – 2x 2x – 2(– 2x) = – 6 2x + 4x = – 6 6x = – 6 x = – 1y = – 2x y = – 2(– 1) y =

HW#6: Real-Life Examples Answers 1. a.) 10 t-shirts need to be ordered for both shops to charge an equal amount. b.) If 9 or less t-shirts are ordered,

Word Problems & Systems of Equations

Variables and Expressions 1.3. Vocabulary Algebra: Branch of mathematics that uses symbols such as variables Variable: A letter or symbol used to represent.

ALGEBRA II Jeopardy Linear Systems.

Algebra Chapter 4 Quiz Review.

. Your parents gave you their credit card to go buy a new outfit. You went to the mall and bought a total of 15 items. You only purchased pants and shirts.

Lesson 1-1 Using Variables. Vocabulary A variable is a symbol, usually a letter, that represents on or more numbers. An algebraic expression is a mathematical.

– Problem Solving Objectives:

 Solve by substitution: 1. y = 2x+3 y = 4x+5 2. A tow company charges $2 per mile plus a fee of $20. Another company charges $5 per mile plus a $10 fee.

Do Now Two hamburgers and three cokes cost $ Four hamburgers and four cokes cost $ Find the cost of one hamburger and one coke.

Applications of Cost Example: The admission price to a basketball game at Southridge High School is $4 for children and $7 for adults. If 600 tickets were.

1. Define variables 2. Write as a system of equations 3. Solve showing all steps 4. State your solution (in words!)

Chapter 3 –Systems of Linear Equations

7.3 Elimination Method Using Addition and Subtraction.

Two Step equations. Understand The Problem EX: Chris’s landscaping bill is 380$. The plants cost 212$,and the labor cost 48$ per hour. Total bill = plants.

Linear Applications: Standard Form. Read the problem carefully. Define the variables. Write the equation in standard form. Solve for the x-intercept by.

4.2B Word Problems - Solving Linear System by Substitution.

Problem of the Day Solve the following system any way: 6x + y = -39 3x + 2y = -15.

Section 11.1 Systems of Linear Equations; Substitution and Elimination.

Warm-Up Dave bought 7 tacos and 3 burritos for $36. For the same price he could have bought 2 tacos and 6 burritos. How much do both tacos and burritos.

Solving Addition and Subtraction Equations Lesson 2.3 and 2.4.

The Substitution Method Objectives: To solve a system of equations by substituting for a variable.

Warm-Up 1) Determine whether (-1,7) is a solution of the system. 4 minutes 3x – y = -10 2) Solve for x where 5x + 3(2x – 1) = 5. -x + y = 8.

3-2 Solving Systems Algebraically Hubarth Algebra II.

7.2 Solving Linear Systems by Substitution. Steps: 1. Solve one of the equations for one of the variables. 2.Substitute that expression into the other.

1.6 Translating Problems into Equations Objective: To translate simple word problems into equations. Warm – up: – A season ticket good for 39 basketball.

2-6 solving Two-Step Equations What You’ll Learn To solve two step equations To solve two step equations.

ALGEBRA 1 Lesson 1-1 Warm-Up Are the following answers reasonable? Use estimation to determine why or why not

Warm-up 1. Solve the following system of equations by graphing: 3x – y = -3 y – 3x = Determine the solution type from the following system of equations:

Algebra II day 36 Chapter 3 Systems of Linear Equations.

3x – 5y = 11 x = 3y + 1 Do Now. Homework Solutions 2)2x – 2y = – 6 y = – 2x 2x – 2(– 2x) = – 6 2x + 4x = – 6 6x = – 6 x = – 1y = – 2x y = – 2(– 1) y =

6.3 Solving Systems of Linear Equations by the Addition Method

Thursday Warmup Homework Check: Document Camera.

Equations and Problem Solving

5.2 Solving Systems of Linear Equations by Substitution

Systems of Linear Equations; Substitution and Elimination

5.2 Solve by Substitution.

Review: 8.1a Mini-Quiz Graph the systems of equations. Then determine whether each system has no solution, one solution, or infinitely many solutions.

Main Idea and New Vocabulary

Setting Up Application Problems

Unit 1 – Section 9 “Applications of Systems of Equations”

Algebra 1 Section 7.3.

6-3 Solving Systems Using Elimination (Combination)

Warm-up What would you multiply to use the elimination method?

4 minutes Warm-Up 1) Determine whether (-1,7) is a solution of the system. 3x – y = -10 -x + y = 8 2) Solve for x where 5x + 3(2x – 1) = 5.

Using Variables ALGEBRA 1 LESSON 1-1

Presentation transcript:

c = 5d + 2 2c + d = 4 Do Now

Homework Solutions 4)5x – 3y = – 4 15x – 9y = – 12 3x + 2y = 9 – 15x + 10y = 45 – 19y = – 57 y = 3 3x + 2y = 9 3x + 2(3) = 9 3x + 6 = 9 3x = 3 x = 1 Solution:(1,3)

Homework Solutions 7)3c – 3d = – 12 3c – 3d = – 12 c + 2d = 2 – 3c + 6d = 6 – 9d = – 18 d = 2 c + 2d = 2 c + 2(2) = 2 c + 4 = 2 c = – 2 Solution:(– 2,2)

Homework Solutions 9)4x – 3y = 5 y = 2x + 1 4x – 3(2x + 1) = 5 4x – 6x – 3 = 5 – 2x – 3 = 5 – 2x = 8 x = – 4y = 2x + 1 y = 2(– 4) + 1 y = – 7 Solution:(– 4, – 7)

Homework Solutions 11)5x + 2y = 8 y = 1 – x 5x + 2(1 – x) = 8 5x + 2 – 2x = 8 3x + 2 = 8 3x = 6 x = 2y = 1 – x y = 1 – (2) y = – 1 Solution:(2, – 1)

Word Problems For Systems of Equations CDs The play Pizza

The sum of two numbers is 48 and their difference is 24. What are the numbers?

Three times one number added to another number is 18. Twice the first number minus the other number is 12. Find the numbers.

Three shirts and two neckties cost $69. At the same price, two shirts and three neckties cost $61. What is the cost of one shirt and one necktie?

Nicole bought 3 new compact discs and 3 used compact discs for $ At the same prices, Jon bought 3 new compact discs and 5 used compact discs for $ Find the cost of buying a new and a used compact disc. Let n = the cost of a new disc Let u = the cost of an used disc Set up two equations: 3n + 3u = $69.00 ** 3n + 5u = $85.00 Answer: New disc is $15 and Old disc is $8 Define the variables:

Brandon and Rachel had lunch at the mall. Brandon ordered three slices of pizza and two colas. Rachel ordered two slices of pizza and three colas. Brandon’s bill was $6.00 and Rachel’s bill was $5.25. What was the price of one slice of pizza? What was the price of one cola? Let p = the price of one slice of pizza Let c = the price of one cola 3p + 2c = $6.00 2p + 3c = $5.25 2(3p + 2c = $6.00) 3(2p + 3c = $5.25) Answer: 1 slice = 1.50, 1 cola =.75

The play The total attendance at a school play was The cost of the tickets were $6.00 for students and $7.50 for adults. The school drama club had a revenue total of $ How many of each ticket was sold? Let a = the number of adults Let s = the number of students a + s = 1250 $7.50 a + $6.00s = $ $7.50(a + s = 1250 ) Answer: Adults = 575, Students = 675