Linear Systems - Intermediate Algebra - Lecture Slides, Slides of Algebra

Download Linear Systems - Intermediate Algebra - Lecture Slides and more Slides Algebra in PDF only on Docsity!

§3.1 2-Var

Linear Systems

Review §

 Any QUESTIONS About

• §’s2.4 → Point-Slope Eqn, Modeling

 Any QUESTIONS About HomeWork

• §’s2.4 → HW-

2.4 MTH 55

Checking System Solution

• To verify or check a solution to a system

of equations:

1. Replace each variable in each equation

with its corresponding value.

2. Verify that each equation is true.

Example  Chk System Soln

• Consider The Equation System

7 (Equation 1) 3 2 (Equation 2)

x y y x

^ +^ =  (^) = − 

 Determine whether each ordered pair is a solution to the system of equations. a. (−3, 2) b. (3, 4)

Example  Chk System Soln

• SOLUTION →

Chk True/False

7 (Equation 1) 3 2 (Equation 2)

x y y x

^ +^ =  (^) = − 

 b. (3, 4) → Sub: 3 for x, & 4 for y

x + y = 7 y = 3 x − 2 3 + 4 = 7 4 = 3(3) − 2 7 = 7 4 = 7 True False

Example  Chk System Soln

• SOLUTION →

Chk True/False

7 (Equation 1) 3 2 (Equation 2)

x y y x

^ +^ =  (^) = − 

 Because (−3, 2) does NOT satisfy EITHER equation, it is NOT a solution for the system.

 Because (3, 4) satisfies ONLY ONE equation, it is NOT a solution to the system of equations

Solve Systems of Eqns by Graphing

• Recall that a graph of an equation is a set of points representing its solution set
• Each point on the graph corresponds to an ordered pair that is a solution of the equation
• By graphing two equations using one set of axes, we can identify a solution of both equations by looking for a point of intersection

Solving by Graphing Procedure

1. Write the equations of the lines in slope- intercept form.
2. Use the slope and y-intercept of each line to plot two points for each line on the same graph.
3. Draw in each line on the graph.
4. Determine the point of intersection (the Common Pt) and write this point as an ordered pair for the Solution

Example  Solve w/ Graphing

• Chk (−1, −2) Soln:
  • y = 3 x + 1
  • x − 2 y = 3

 y = 3 x + 1→

• −2 = 3(−1) + 1
• −2 = −3 + 1
• −2 = −2   x − 2 y = 3 →
• (−1) − 2(−2) = 3
• −1+4 = 3
• 3 = 3 

 Thus (−1, −2) Chks as a Soln

Example  Solve By Graphing

• Solve System:

 SOLUTION: graph Both Equations

• As a check note that [4−2] = [6−4] is true. - The solution is (4, 2)
• The graphs intersect at (4, 2), indicating that for the x-value 4 both x−2 and 6−x share the same value (in this case 2).

y x

y x = −

= − 6

2 (4, 2)

y = x − 2

y = 6 − x

Substitution Summarized

• The substitution method involves

isolating either variable in one

equation and substituting the result

for the same variable in the second

equation. The numerical result is then

back-substituted into the first

equation to find the numerical result

for the

second variable

Example  Solve by Subbing

• Solve the System

( ) 2 5 ( ) 2

3 2 4 1 = − +

− = y x

x y

 SOLUTION: The second equation says that y and − 2 x + 5 represent the same value.

 Thus, in the first equation we can substitute − 2 x + 5 for y

Example  Solve by Subbing

• We have found the x-value of the solution. To find the y-value, we return to the original pair of equations. Substituting x= into either equation will give us the y- value. Choose eqn (2): y = − 2 x + 5 Equation (2) y = − 2 ( ) 2 + 5 Substitute: x = 2 y = − 4 + (^5) Simplifying y = 1 When x = 2

 The ordered pair (2, 1) appears to be the solution

Example  Solve by Subbing

• Check Tentative Solution (2,1)

3 x − 2 y = 4 y = − 2 x + 5 3(2) − 2(1) 4 1 −2(2) + 5 6 − 2 4 1 −4 + 5 4 = 4 True 1 = 1 True

 Since (2, 1) checks in BOTH equations, it IS a solution.