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 = 4x – y = 8

An ordered pair (x, y) is a solution to linear equationif the values for x and y satisfy both equations.

The standard form isax + by = cdx + ey = k

Where a, b, c, d, e, k are constants.

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 uniquesolution. Consistent systemThe equations are called independent equation

2. If the two lines are parallel it is an inconsistent system and nosolution

3. If two lines are identical and every point on the line representssolution and give infinitely many solutions. The equations are calleddependent equations

..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.

dependentInconsistent Unique

y y y

x x x

The Substitution Method Consider the following system of equations.

 2x + y = 53x – 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).

Elimination MethodThe elimination method is the second way to solve linear systemssymbolically. This method is based on the property that ‘ equalsadded to equals are equal.’ That is, if

a = b and c = d

Then a + c = b + dNote that adding the two equations eliminates the variable y

Example 2x – y = 4x + y = 13x = 5

or x = 5/3 and solve for xSubstituting x = 5/3 into the second equation gives 5/3 + y = 1 or y = - 2/3The solution is (5/3, - 2/3)

Solve the system of equations using Graphing Calculator

4.1 Pg 226

No solution Infinitely many solutions

Burning caloriesEx – 4.2, No. 88 pg 254

During strenuous exercise an athelete can burn on Rowing machine Stair climber

10 calories per minute 11.5 calories per minuteIn 60 minute an athelete burns 633 calories by using both exercisemachinesLet x minute in rowing machine, y minute in stair climber The equations are x + y =6010x + 11.5 y = 633Find x and y-10x - 10 y = -600 (Multiply by -10) 10x + 11.5 y = 633Add 1.5y = 33, y = 33/1.5= 22 minute in stair climber and x = 38 minute in rowing machine

Mixing acids No.90 ( Pg 254 )

x represents the amount of 10% solution of Sulphuric acidy represents the amount of 25 % solution of Sulphuric acidAccording 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 = 20Subtract 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

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

4.3 Solving Linear inequalities in Two variables

- 2 -1 1 2 3 - 2 -1 1 2 -2 -1 0 1 2 3 4

x < 1 x > 1

y < 2x - 1-1

x – 2y < 4

-1

- 2

x –intercept = 4 (4, 0)y –intercept = -2 (0, -2)

x - 2y < 40 - 2(0) < 4 0 < 4 which is true statementShade containing (0, 0)

Choose a test pointLet x = 0, y = 0

Solving System of Linear Inequalities ( Pg 259)x + y < 4

y > x

-4 -3 -2 -1 0 1 2 3 4

4

3

2

1(1, 2)

Testing point

x + y < 4 x = 1, y = 2 1 + 2 < 4 ( True )

Shaded region

y > x

(1, 2)

Shaded region

Testing point 2 > 1 ( True)

( 1, 2)Shaded region To solve inequalities

Modeling target heart rates (Ex 4 Pg 260 )For Aerobic Fitness

200

175

150

125

100

75

50

25

0 20 30 40 50 60 70 80

Age in years

(30, 150) is a solution

Heart Rate

Beats Per minute

T = - 0.8A + 196 ( Upper Line )

T = - 0.7 A + 154 (Lower Line )

A person’s Maximum heart rate ( MHR) = 220 - A

150 < - 0.8 (30) + 196 = 172 True150 > - 0.7 (30) + 196 = 133 True

-0.8 (40) + 196 <= 165 When A = 40 yrs

- 0.7 (40) + 196> = 125

Solving a system of linear inequalities with technology

Ex 5 ( Pg 261 )

Shade a solution set for the system of inequalities, using graphing calculator

- 2x + y > 1 or y> 2x + 1

[ - 15, 15, 5 ] by [ - 10, 10, 5 ]

2x + y < 5 or y < 5 – 2x

Hit 2nd and Draw then go to Shade Hit Graph

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 Hit Graph