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Chapter 7

Chapter 7. Systems of Linear Equations and Inequalities. Section 7.1: Graphing Linear Systems. Two or more linear equations in the same variable make up a L inear System. Example: 3x + 2y = 4 -x + 3y = -5

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Chapter 7

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  1. Chapter 7 Systems of Linear Equations and Inequalities

  2. Section 7.1: Graphing Linear Systems • Two or more linear equations in the same variable make up a Linear System. • Example: • 3x + 2y = 4 • -x + 3y = -5 • The solution of the system is an ordered pair which makes both equations true statements. • We can find this solution by graphing both lines and then finding the point of intersection. 3x + 2y = 4 -x + 3y = -5 (2,-1)

  3. Check the Solution • EQUATION 1 • 3x +2y = 4 • 3(2) + 2(-1) = 4 • 6 - 2 = 4 • 4 = 4✓ • EQUATION 2 • -x + 3y = -5 • -(2) + 3(-1) = -5 • -2 + -3 = -5 • -5 = -5 ✓

  4. Solving a Linear System Using Graph & Check • STEP 1: Write each equation in a form that is easy to graph. • STEP 2: Graph both equations in the same coordinate plane. • STEP 3: Estimate the coordinates of the point of intersection. • STEP 4: Check whether the coordinates give a solution by substituting them into each equation of the linear system.

  5. Another Example: • Use the graph-and-check method to solve the linear system. • Equation 1 • x +y= -2 • Rewrite • Y = -x - 2 • Equation 2 • 2x - 3y = -9 • Rewrite • -3y = -2x - 9 • y = 2/3x + 3 • Graph and find the point of intersection. 2x – 3y = -9 (-3, 1) x + y = -2

  6. Check the Solution • EQUATION 1 • 2x -3y = -9 • 2(-3) + -3(1) = -9 • -6 - 3 = -9 • -9 = -9✓ • EQUATION 2 • x +y = -2 • (-3) + 1 = -2 • -2 = -2 ✓

  7. You Try: • Solve the system of linear equations by graphing: • Equation 1: x + y = 4 • Equation 2: 2x + y = 5

  8. Section 7.2: Solving Linear Systems by Substitution • The substitution method of solving a system of Linear Equations involves solving the first equation for one variable in terms of the other. • You then use that solution in the second equation to solve and get one coordinate of the ordered pair. • Finally, you substitute the value from the second equation, back into the first equation to get the second coordinate of the ordered pair.

  9. Example: Substitution MethodSolve for y first • Solve the linear system: • Equation 1 : -x +y = 1 • Equation 2 : 2x + y = -2 • Solve the first equation for one variable in terms of the other. • y = x + 1 • Use that solution in the second equation to solve • 2x + y = -2 Equation 2 • 2x + (x + 1) = -2 Substitute for y • 3x + 1 = -2 Simplify & subtract 1 from both sides • 3x = -3 Divide both sides by 3 • X = -1 • Substitute the value from the second equation, back into the first equation • y = x + 1 Equation 1 • y = (-1) + 1 Substitute for x • y = 0 Simplify • Solution (-1,0) Check: -(-1) + 0 = -1 & 2(-1) + 0 = -2

  10. Example: Solve for x first • Solve the linear system. • Equation 1 : 2x +2y = 3 • Equation 2: x - 4y = -1 • You can easily solve the 2nd equation for x by adding 4y to both sides • x – 4y + 4y = -1 + 4y • x = -1 + 4y • Substitute this value into equation 1 • Equation 1: 2x + 2y = 3 • Substitute: 2(-1 + 4y) + 2y = 3 • Simplify: -2 + 10y = 3 • Add 2: 10y = 5 • Divide by 10: y = ½ • Substitute the y value into Equation 2: • x = -1 + 4y Equation 1 • x = -1 + 4( ½ ) = -1 + 2 = 1 Solution: (1, ½ ) • Check the solution: 2(1) + 2( ½ ) = 3 2 + 1 = 3 1 – 4( ½ ) = -1 1 – 2 = -1

  11. You Try: • Solve the system of linear equations using the substitution method. • Equation 1: 3x – y = -1 • Equation 2: 2x + 4y = 8

  12. Section 7.3: Solving a System of Linear Equations by Combination • Sometimes it is not easy to isolate one of the variables in a linear system • In this case it is easier to solve the system by linear combinations. • A linear combination of two equations is an equation obtained by • Multiplying one or both equations by a constant • Adding the resulting equations

  13. Linear Combinations Example • Solve the linear system. • Equation 1: 4x +3y =16 • Equation 2: 2x -3y = 8 • Solution: • Since one equation has a 3y and the other a -3y we can add the equations to eliminate that variable. • 4x + 3y = 16 • 2x - 3y = 8 • 6x =24 • X = 4 • Substitute the value of x into the 1st equation and solve for y • 4(4) + 3y = 16 • 16 + 3y = 16 • 3y = 16-16 • 3y = 0 • Y = 0 • Check the solution by substituting into the 2nd equation. • 2(4) – 3(0) = 8 • 8 = 8 • The solution is (4,0)

  14. Another Example: Multiply then Add • Solve the linear system. • Equation 1: 3x + 5y = 6 • Equation 2: -4x + 2y = 5 • In this case we can’t just add because it won’t cause anything to cancel, so we first multiply one or both of the equation by a number so that we can add to cancel a variable. • Multiply by 4: (4)(3x + 5y =6) = 12x + 20y = 24 • Multiply by 3: (3)(-4x + 2y =5) = -12x + 6y = 15 • Add: 26y = 39 • Solve for y: y = 1.5 • Substitute: 3x + 5(1.5) = 6 3x + 7.5 = 6 3x = 6 – 7.5 3x = -1.5 x = -0.5 Check by substituting into Equation 2: -4(-0.5) + 2(1.5) = 5 2 + 3 = 5 5 = 5

  15. You Try: • Solve the linear system. • Equation 1: 3x +2y = 8 • Equation 2: 2y = 12 - 5x

  16. Section 7.4:Linear Systems and Problem Solving • In real life, decisions have to be made about how to make a solution in Chemistry or how many items to manufacture for sale in business. This real-life modeling uses linear equations – as we have seen in earlier chapters. • For every variable used you must have a separate equation – hence for a system of two variable, there will be two linear equations that must be written and solved. • The process for writing and solving a system of linear equations is: • Determine your verbal model • Write an equation to match your model • Choose your method for solving the system • Check the solution

  17. Example – Problem Solving • Problem: • In one week a music store sold 7 violins for a total of $1600. Two different types of violins were sold. One type cost $200 and the other type cost $300. How many of each type of violin did the store sell? • Verbal Model: • Type x + Type y = Total Sold • Price type x * Number Type x + Price type y* number type y = sales • Equations: • x + y = 7 • 200x + 300y = 1600

  18. Solving: • Equations: • x + y = 7 • 200x + 300y = 1600 • The since the first equation is easy to solve, let’s try substitution to solve. • y = -x + 7 Rearrange Equation 1 • 200x + 300(-x + 7) = 1600 Substitute into Equation 2 • 200x + -300x + 2100 = 1600 Distribute • -100x + 2100 = 1600 Simplify • -100x = 1600 – 2100 Subtract 2100 from both sides • -100x = -500 Divide by 100 • x = 5 • 5 + y = 7 Substitute into the 1st equation • y = 7 – 5 Subtract 5 from both sides • y = 2 • Check your solutions: 5 + 2 = 7 & 200(5) + 300(2) = 1600

  19. Another Example: Chemistry – Making a Mixture • You combine 2 solutions to form a mixture that is 40% acid. One solution is 20% acid and the other is 50% acid. If you have 90 milliliters of the mixture, how much of each solution was used to create the mixture? • Verbal Model: • Volume of Solution X + Volume of Solution Y = Volume of mixture • Acid in solution X * Acid in solution Y = acid in mixture • Equations: • x + y = 90 • 0.2x + 0.5y = (0.4)(90) = 36

  20. Solving: • Equations: • x + y = 90 • 0.2x + 0.5y = 36 • The since the first equation is easy to solve, let’s try substitution to solve. • y = -x + 90 Rearrange Equation 1 • 0.2x + 0.5(-x + 90) = 36 Substitute into Equation 2 • 0.2x + -0.5x + 45 = 36 Distribute • -0.3x + 45 = 36 Simplify • -0.3x = 36 - 45 Subtract 45 from both sides • -0.3x = -9 Divide by -0.3 • x = 30 • 30 + y = 90 Substitute into the 1st equation • y = 90 – 30 Subtract 30 from both sides • y = 60 • Check your solutions: 30 + 60 = 90 & 0.2(30) + 0.5(60) = 36

  21. You Try: • Job A offers an annual salary of $30,000 plus a bonus of 1% of sales. Job B offers an annual salary of $24,000 plus a bonus of 2% of sales. How much would you have to sell to earn the same amount in each job • Steps to Solve: • Verbal Model • Equations • Solve

  22. Section 7.5: Special Types of Linear Systems • Some Linear Systems can have either no solution, or infinitely many solutions. In this section we will try to recognize these two types of special systems.

  23. Example: A System with No Solution Show that the linear system has no solution. Equation 1: y =3x + 8 Equation 2: y =3x - 4 • Method 1: Demonstrate by Graphing • Because the lines have the same slope but different y-intercepts, they are parallel. • Parallel lines never intersect, so the system has no solution. • Method 2: Substitution • y = 3x – 4 Original Eq. 2 • 3x – 4 = 3x + 8 Substitute Eq.1 into equation 2 for y • – 4 ≠ 8 Subtract 3x from both sides

  24. Example: A System with Infinitely Many Solutions • Show that the Linear System has infinitely many solutions. • Equation 1: 5x + y = 5 • Equation 2: 10x + 2y = 10 • Method 1: Graphing • Rewrite each equation in slope intercept form. • Equation 1: y = -5x + 5 • Equation 2: y = -5x + 5 • These are the same equation so every point on the line is a solution. 5x + y = 5 10x + 2y = 10

  25. A Second Method: Linear CombinationA System with Infinitely Many Solutions • Show that the Linear System has infinitely many solutions. • Equation 1: 5x + y = 5 • Equation 2: 10x + 2y = 10 • Method 2: • Linear Combination • Multiply Equation 1 by 2 and you get: • Equation 1: 10x + 2y = 10 • Equation 2: 10x + 2y = 10 • You can see the equations are the same, so all numbers on the line are a solution to both equations.

  26. You Try: • How many solutions does each system have? System 1: Equation 1: 3x +y = -1 Equation 2: -9x -3y =3 System 2: Equation 1: x -2y =5 Equation 2: -2x + 4y = 2 System 3: Equation 1: 2x +y = 4 Equation 2: 4x -2y =0

  27. Section 7.5: Systems of Linear Inequalities • Solving a system of linear inequalities by graphing involves: • Putting the inequalities into Slope-Intercept Form, y = mx+b • Graphing the lines using the y-intercept & slope. • If the inequality is < or >, make the lines dotted. • If the inequality is < or >, make the lines solid. • Shading the correct side of each line • The overlap of the shaded areas is the solution.

  28. Solving Systems of Linear Inequalities Example: a: 3x + 4y > - 4 b: x + 2y < 2 Put in Slope-Intercept Form:

  29. a: dotted shade above b: dotted shade below Solving Systems of Linear Inequalities Example, continued: Graph each line, make dotted or solid and shade the correct area.

  30. Solving Systems of Linear Inequalities a: 3x + 4y > - 4

  31. Solving Systems of Linear Inequalities a: 3x + 4y > - 4 b: x + 2y < 2

  32. Solving Systems of Linear Inequalities a: 3x + 4y > - 4 b: x + 2y < 2 The area between the green arrows is the region of overlap and thus the solution.

  33. Example: Write a System of Inequalities • Write a system of linear inequalities which describes the region at right. • Identify the lines & Shading • The solid line passes through points (0,1) & (3,4) so we used these to find the slope, m = 1. The y intercept is also 1, and it is shaded below, so the inequality is: • Inequality 1: y ≤ x + 1 • The dotted line is a vertical line at x = 3 and it is shaded to the left, so the equation is: • Inequality 2: x < 3

  34. You Try: • Graph the system of linear inequalities. • Inequality 1: y < 2 • Inequality 2: x ≥ -1 • Inequality 3 : y > x – 2 • Write the system of linear inequalities which describes the graph below. (hint: 4 inequalities)

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