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Some concept of Intermediate Algebra are Factoring Strategies, Factoring Strategies, Factoring Strategies, Introduction, Inverse_Fcns, Lines_By_Slp-Inter, Log_Change_Base, Multiply Polynomials, Multiply Polynomials. Main points of this lecture are: Linear Systems, Point-Slope Eqn, Modeling, System of Equations, Group, Solution, System of Equations, Equations, Same Time, Corresponding Value
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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)
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
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
Solving by Graphing Procedure
y = 3 x + 1→
Thus (−1, −2) Chks as a Soln
SOLUTION: graph Both Equations
y x
y x = −
= − 6
2 (4, 2)
y = x − 2
y = 6 − x
( ) 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
The ordered pair (2, 1) appears to be the solution
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.