Download Notes for Lesson 6-1: Solving Systems by Graphing

Notes for Lesson 6-1: Solving Systems by Graphing
6-1.1 – Identifying Solutions of Systems
Vocabulary:
System of linear equations – a system of equations in which all of the equations are linear
Solution of a system of linear equation – any ordered pair that satisfies all the equations in a system
A system of linear equations are 2 or more linear equation that are related to each other in some way.
A lot of times we want to be able to solve these systems simultaneously or at the same time. When that happens
we need to find a value for x and y that will make both equation true at the same time.
Examples: Tell whether the ordered pair is a solution of the given system
2x  5 y  8
3x  2 y  5
x  2y  6
x y 3
Plug the x and y values into both equation and see if they work.
 1,2 ;
(4)  2(1)  6
2(1)  5(2)  8
4,1 ;
4 1  3
3(1)  2(2)  5
 2  10  8
3 4  5
88
7 5
Both are not true so it is not a solution
42 6
33
66
Both are true so it is a solution
6-1.2 – Solving a system of linear equations by graphing
All of the ordered pairs that make a linear equation true are on its graph. So we need to find a point that is on
both of the line graphs, in other words we need the point of intersection of the two lines.
Examples: Solve each system by graphing
5
4
y  x 3
y  x 1
The lines cross at 1,2  so that is the solution
y
2
-5 -4
-2
2
4 5
2
4 5
x
-2
-4
-5
x y 0
1
y   x 1
2
5
4
The lines cross at  2,2  so that is the solution
2
-5 -4
-2
-2
-4
-5
y
x
6-1.3 – Problem solving application
Bowl-o-Rama charges $2.50 per game plus $2 for shoe rental, and Bowling Plaza charges $2 per game plus $4
for shoe rental. For how many games will the cost to bowl be the same at both places?
Write the equations: Bowl-a-Rama: y  2.50 x  2
Bowling Plaza: y  2 x  4
20
Graph both linear equations: 18
16
14
12
10
8
6
4
2
0
y
x
0 2 4 6 8 10 12 14 16 18 20
The numbers of games would be 4 for a cost of $12
Practice B #’s 1, 3, 4, 5