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Chapter 4

Chapter 4. Systems of Linear Equations in Two Variables. 4.1 Systems of Linear Equations in two variables. The system of linear equation with two variables. Each equation contains two variables, x and y. Example x + y = 4 x – y = 8 An ordered pair (x, y) is a solution to linear equation

joseph-rosario
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Chapter 4

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  1. Chapter 4 Systems of Linear Equations in Two Variables

  2. 4.1 Systems of Linear Equations in two variables The system of linear equation with two variables. Each equation contains two variables, x and y. Example x + y = 4 x – y = 8 An ordered pair (x, y) is a solution to linear equation if the values for x and y satisfy both equations. The standard form is ax + by = c dx + ey = k Where a, b, c, d, e, k are constants.

  3. Types of Linear Equations in two variables A system of linear equations in two variables can be represented graphically by two lines in the xy-plane 1.The lines intersect at a single point, which represents a unique solution. Consistent system The equations are called independent equation 2. If the two lines are parallel it is an inconsistent system and no solution 3. If two lines are identical and every point on the line represents solution and give infinitely many solutions. The equations are called dependent equations

  4. ..cont. Example The equations x + y = 1 and 2x + 2y = 2 are equivalent. If we divide the second equation by 2 we obtain the first equation. As a result, their graphs are identical and every point on the line represents a solution. Thus there are infinitely many solutions, and the system of equations is a dependent system. y y y x x x dependent Unique Inconsistent

  5. The Substitution Method Consider the following system of equations.  2x + y = 5 3x – 2y = 4  It is convenient to solve the first equation for y to obtain y = 5 – 2x. Now substitute ( 5 – 2x) for y into the second equation. 3x – 2(y) = 4 Second equation 3x – 2(5-2x) = 4, Substitute  to obtain a linear equation in one variable. 3x – 2(5 – 2x) = 4 3x – 10 + 4x = 4 (Distributive property) 7x – 10 = 4 (Combine like terms) 7x = 14 (Add 10 to both sides) x = 2 (Divide both sides by 7) To determine y we substitute x = 2 into y = 5 – 2x to obtain y = 5 – 2(2) = 1  The solution is (2, 1).

  6. Elimination Method The elimination method is the second way to solve linear systems symbolically. This method is based on the property that ‘ equals added to equals are equal.’ That is, if a = b and c = d Then a + c = b + d Note that adding the two equations eliminates the variable y Example 2x – y = 4 x + y = 1 3x = 5 or x = 5/3 and solve for x Substituting x = 5/3 into the second equation gives 5/3 + y = 1 or y = - 2/3 The solution is (5/3, - 2/3)

  7. Solve the system of equations using Graphing Calculator4.1 Pg 226 No solution Infinitely many solutions

  8. Burning caloriesEx – 4.2, No. 88 pg 254 During strenuous exercise an athelete can burn on Rowing machine Stair climber 10 calories per minute11.5 calories per minute In 60 minute an athelete burns 633 calories byusing both exercise machines Let x minute in rowing machine, y minute in stair climber The equations are x + y =60 10x + 11.5 y = 633 Find x and y -10x - 10 y = -600 (Multiply by -10) 10x + 11.5 y = 633 Add 1.5y = 33, y = 33/1.5= 22 minute in stair climber and x = 38 minute in rowing machine

  9. Mixing acids No.90 ( Pg 254 ) x represents the amount of 10% solution of Sulphuric acid y represents the amount of 25 % solution of Sulphuric acid According to statement x + y = 20 10% of x + 25% of y = 0.18(20) 0.10x + .25 y = 3.6 .1x + .25y = 3.6 Multiply by 10 x + 2.5y = 36 x + y = 20 Subtract 1.5 y = 16 y = 16/1.5 = 10.6 x = 20 – y = 20 – 10.6 = 9.4 Mix 9.4 ml of 10 % acid with 10.6 ml of 25% acid

  10. River current(No. 100, pg 254) x Speed of tugboat y Speed of current Distance = Speed x Time 165 = (x – y) 33 Upstream x – y = 5 165 = (x + y) 15 Downstream x + y = 11 By elimination method 2x = 16, x = 8 y = 3 The tugboat travels at a rate of 8 mph and river flows at a rate of 3 mph

  11. 4.3 Solving Linear inequalities in Two variables x < 1 x > 1 -2 -1 0 1 2 3 4 - 2 -1 1 2 3 - 2 -1 1 2 • 1 • - 2 -1 x – 2y < 4 y < 2x - 1 x –intercept = 4 (4, 0) y –intercept = -2 (0, -2) Choose a test point Let x = 0, y = 0 x - 2y < 4 0 - 2(0) < 4 0 < 4 which is true statement Shade containing (0, 0)

  12. Solving System of Linear Inequalities ( Pg 259)x + y < 4y > x Testing point x + y < 4 x = 1, y = 2 1 + 2 < 4 ( True ) Shaded region Testing point 2 > 1 ( True) 4 3 2 1 (1, 2) (1, 2) y > x -4 -3 -2 -1 0 1 2 3 4 Shaded region Shaded region To solve inequalities ( 1, 2)

  13. Modeling target heart rates (Ex 4 Pg 260 )For Aerobic Fitness A person’s Maximum heart rate ( MHR) = 220 - A 200 175 150 125 100 75 50 25 Heart Rate Beats Per minute 0 20 30 40 50 60 70 80 Age in years (30, 150) is a solution T = - 0.8A + 196 ( Upper Line ) T = - 0.7 A + 154 (Lower Line ) • 0.8 (40) + 196 <= 165 When A = 40 yrs • - 0.7 (40) + 196> = 125 150 < - 0.8 (30) + 196 = 172 True 150 > - 0.7 (30) + 196 = 133 True

  14. Solving a system of linear inequalities with technologyEx 5 ( Pg 261 ) Shade a solution set for the system of inequalities, using graphing calculator - 2x + y > 1 or y> 2x + 1 2x + y < 5 or y < 5 – 2x Locate the cursor to the left of Y1, and press ENTER two or three times to shade either above or below the graph of Y1 and Y2 [ - 15, 15, 5 ] by [ - 10, 10, 5 ] Hit Graph Hit 2nd and Draw then go to Shade Hit Graph

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