ALGEBRA 1 Lesson 6-1 Warm-Up. ALGEBRA 1 “Solving Systems by Graphing” (6-1) What is a “system...
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Transcript of ALGEBRA 1 Lesson 6-1 Warm-Up. ALGEBRA 1 “Solving Systems by Graphing” (6-1) What is a “system...
ALGEBRA 1
“Solving Systems by Graphing” (6-1)
What is a “system of linear equations”?
What is the “solution of the system of linear equations”?
How do you find the solution for a system of linear equations”?
System of Linear Equations: two or more lines (linear equation) that are on the same graph
Solution of the System of Linear Equations: a point that all of the lines in a system of equations have in common (in other words, all of the lines in the system of linear equations cross at that point)
Method 1: Make a Graph – Graph all of the lines and find the point where all of the lines cross (the point of intersection)
Example: Solve y = 2x – 3 and y = x – 1
Graph both equations on the same coordinate plane.
Find the point of intersection.
The lines intersect at (2,1), so (2,1) is the solution to the system.
ALGEBRA 1
Solve by graphing. Check your solution.
y = 2x + 1y = 3x – 1
Graph both equations on the same coordinate plane.
y = 2x + 1 The slope is 2. The y-intercept is 1.y = 3x – 1 The slope is 3. The y-intercept is –1.
Solving Systems by GraphingLESSON 6-1
Additional Examples
ALGEBRA 1
(continued)
Find the point of intersection.
The lines intersect at (2, 5), so (2, 5) is the solution of the system.
Check: See if (2, 5) makes both equations true.
y = 2x + 1 y = 3x – 1
5 2(2) + 1 Substitute (2, 5) for (x, y). 5 3(2) – 1
5 4 + 1 5 6 – 1
5 = 5 5 = 5
Solving Systems by GraphingLESSON 6-1
Additional Examples
ALGEBRA 1
Suppose you plan to start taking an aerobics class. Non-members pay $4 per class while members pay a $10 fee plus an additional $2 per class. Write a system of equations that models these plans.
Define: Let = cost of one class
Let = number of classes.
Let = total cost of the classes.
n
C(n)
Solving Systems by GraphingLESSON 6-1
Additional Examples
C
ALGEBRA 1
The system is C(n) = 4n (or y = 4x) C(n) = 10 + 2n (or y = 2x + 10)
Words: cost is membership plus cost of classes fee attended
nC(n)Equation: member = 10 + 2
C(n) nnon-member = 0 + 4
(continued)
Solving Systems by GraphingLESSON 6-1
Additional Examples
ALGEBRA 1
The system below models the cost of taking an aerobics class as a function of the number of classes. Find the solution of the system by graphing. What does the solution mean in terms of the situation?
C(a) = 2a + 10 The slope is 2. The intercept on the vertical axis is 10.
C(a) = 2a + 10
Graph the equations.
The lines intersect at (5, 20).
Part 1: Find the solutions by graphing.
C(a) = 4a The slope is 4. The intercept on the vertical axis is 0.
C(a) = 4a
Solving Systems by GraphingLESSON 6-1
Additional Examples
ALGEBRA 1
Part 2: Interpret the solution.
(continued)
Solving Systems by GraphingLESSON 6-1
Additional Examples
The lines intersect at (5, 20) so the cost will be $20 after 5 classes.
ALGEBRA 1
“Solving Systems by Graphing” (6-1)
When does a system of linear equations have no solutions?
When does a system of linear equations have an infinite number of solutions?
A system of linear equations has no solutions when the two (or more) lines are parallel to one another (in other words, there are no points of intersection).
Example: Solve by graphing: y = -2x + 1 and y = -2x – 1
Graph both equations on the same coordinate plane.
y = -2x + 1 The slope s -2. The y-intercept is 1
y = -2x - 1 The slope s -2. The y-intercept is -1.
The lines are parallel, so there is no solution.
A system of linear equations has an “infinite many solutions” when the graphs of the equation are the same line (in other words, every point on the line is a solution).
Example: Solve by graphing:2x + 4y = 8 and y = - x + 2
Graph both equations on the same coordinate plane.
2x + 4y = 8 The y-intercept is 2 and the x-intercept is 4.
y = - x + 2 The slope s - . The y-intercept is 2.
The graphs are the same line, so there are an infinite number of solution
12
12
12
ALGEBRA 1
Solve by graphing. y = 3x + 2
y = 3x – 2
Graph both equations on the same coordinate plane.
The lines are parallel. There is no solution.
y = 3x + 2 The slope is 3. The y-intercept is 2.
y = 3x – 2 The slope is 3. The y-intercept is –2.
Solving Systems by GraphingLESSON 6-1
Additional Examples
ALGEBRA 1
Solve by graphing.3x + 4y = 12
y = – x + 3
Graph both equations on the same coordinate plane.
The graphs are the same line. The solutions are
an infinite number of ordered pairs (x, y), such
that y = – x + 3.34
34
3x + 4y = 12 The y-intercept is 3.The x-intercept is 4.
y = – x + 3 The slope is – . The y-intercept is 3.34
34
Solving Systems by GraphingLESSON 6-1
Additional Examples