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Copy of Unit 7 Linear Systems
Transcript of Copy of Unit 7 Linear Systems
Linear equation in two unknowns x and y is an equation of the form
ax + by = c
where a, b, and c are numbers
ex: 4x + 5y = 0
This has a = 4, b = 5, c = 0
When you have two variables and enough information for two equations, we can create a system of equations
A solution of Linear Systems is a pair of values that satisfy both equations
4x + 5y = 40
x-y = 1
This is a system of two linear equations with solution x = 5, y = 4.
We can also write the solution as (5, 4) Part One: Representing Linear Systems The solutions to a single linear equation are the points on its graph, which is a straight line. For a point to represent a solution to two linear equations, it must lie simultaneously on both of the corresponding lines.
To locate solutions to a system of two equations in two unknowns, plot the graphs, and locate the intersection points (if any). Part Two: Solving Graphically Part Three: Solving Systems using Substitution The method of elimination is an algebraic way of obtaining the exact solution(s) of a system of equations in two unknowns by manipulating the equations in such a way as to eliminate of the variables (x or y).
In other words: We can eliminate one variable by subtracting or adding the two equations Part Four: Solving by Elimination Part 5: Properties of Linear Systems The problem with the graphical approach is that it only gives approximate solutions; locating the exact point of intersection of two lines would require perfect accuracy, which is impossible in practice. We can transform a system of two linear equations into a single equation with one variable Ex: y = 3x + 2 7x – 4y = 7 Step One: Notice how one of the variables is isolated. We can use this information and input it into the other equation, instead of the variable.
y = 3x + 2
7x – 4(3x + 2) = 7 Step Two: Solve the resulting equation.
7x – 4(3x + 2) = 7
7x – 12x – 8 = 7
– 5x– 8 = 7
– 5x = 7 + 8
– 5x = 15
– 5 – 5
x = – 3 Step Three: Input the new value into either of the equations and solve for the missing variable.
7 (-3) – 4y = 7
-21 – 4y = 7
- 4y = 7 + 21
- 4y = 28
- 4 -4
y = - 7 y = 3 (-3) + 2
y = -9 +2
y = -7 Step One: Make sure both equations are in standard form (Ax + By = C) x + 2y = 9
– x + 3y = 16 Step Two: Decide on addition or subtraction so that one of the variables (in our case x) will cancel each other out. Step Three: Now solve for the remaining variable
y = 5 Step Four: Using one of the original equations, input the new value for the variable solved in the earlier steps.
- x + 3(5) = 16
-x + 15 = 16
–x = 16 - 15
-x = 1
x = -1 x + 2 (5) = 9
x + 10 = 9
x = 9 – 10
x = -1 So the solution for the system (x, y) = (-1, 5) TRY:
4x – 3y = 25
–3x + 8y = 10 Answer y = 5 x = 10 The solution is (10,5) There are 3 possibilities of solutions when graphing. Single solution: Where the slopes are different but the y-intercept can be the same. These lines can also be perpendicular, which means they intersect; their lines form a 90-degree angle, they are also negative reciprocals of each other. No Solution: They would have the same slope, but different y-intercepts. When they have no solution the lines are parallel, parallel lines continue, literally, forever without touching. Infinite Solutions: Where they have the same slope and y-intercept. Extra Resources The Math Warehouse http://www.mathwarehouse.com/ Tutorials for Finite Math http://people.hofstra.edu/stefan_waner/realworld/tutorialsf1/frames2_1.html YourTeacher - YouTube Solving Systems of Equations - YourTeacher.com - Algebra Help Solving Systems by Substitution - YourTeacher.com - Math Help Solving Systems by Graphing - YourTeacher.com - Algebra Help TRY: Determine which of the 3 categories each question belongs to Answers 1. 2. 3. 1. 2. 3. TRY:
y = 3x – 2
y = -x + 2 Answer Blue: y = 3x – 2
Red: y = -x + 2 Example y = -x + 5
y = 1 x + 2
2 Purple Math http://www.purplemath.com/modules/systlin4.htm TRY:
2x – 3y = –2
4x + y = 24 Answer