Download Solve the system of equations.

11-6 Systems of Equations
Warm Up
Problem of the Day
Lesson Presentation
Course 3
11-6 Systems of Equations
Warm Up
Solve for the indicated variable.
1. P = R – C for R
2. V = 1 Ah for A
3
3. R = C – S for C
t
Course 3
R=P+C
3V = A
h
Rt + S = C
11-6 Systems of Equations
Problem of the Day
At an audio store, stereos have 2 speakers
and home-theater systems have 5 speakers.
There are 30 sound systems with a total of
99 speakers. How many systems are stereo
systems and how many are home-theater
systems?
17 stereo systems, 13 home-theater systems
Course 3
11-6 Systems of Equations
Learn to solve systems of equations.
Course 3
11-6 Systems
Insert Lesson
Title Here
of Equations
Vocabulary
system of equations
solution of a system of equations
Course 3
11-6 Systems of Equations
A system of equations is a set of two or
more equations that contain two or more
variables. A solution of a system of
equations is a set of values that are solutions
of all of the equations. If the system has two
variables, the solutions can be written as
ordered pairs.
Course 3
11-6 Systems of Equations
Caution!
When solving systems of equations,
remember to find values for all of the
variables.
Course 3
11-6 Systems of Equations
Additional Example 1A: Solving Systems of
Equations
Solve the system of equations.
y = 4x – 6
y=x+3
The expressions x + 3 and 4x – 6 both equal y.
So by the Transitive Property they equal each
other.
y = 4x – 6
y=x+3
4x – 6 = x + 3
Course 3
11-6 Systems of Equations
Additional Example 1A Continued
Solve the equation to find x.
4x – 6 = x + 3
–x
–x
Subtract x from both sides.
3x - 6 =
3
+6
+6
Add 6 to both sides.
3x
9
Divide both sides by 3.
3
=
3
x
=
3
To find y, substitute 3 for x in one of the original
equations.
y=x+3=3+3=6
The solution is (3, 6).
Course 3
11-6 Systems of Equations
Additional Example 1B: Solving Systems of
Equations
y = 2x + 9
y = -8 + 2x
2x + 9 = -8 + 2x
– 2x
– 2x
9 ≠ -8
Transitive Property
Subtract 2x from both sides.
The system of equations has no solution.
Course 3
11-6 Systems of Equations
Check It Out: Example 1A
Solve the system of equations.
y=x–5
y = 2x – 8
The expressions x – 5 and 2x – 8 both equal y.
So by the Transitive Property they equal each
other.
y=x–5
y = 2x – 8
x – 5 = 2x – 8
Course 3
11-6 Systems of Equations
Check It Out: Example 1A Continued
Solve the equation to find x.
x – 5 = 2x – 8
–x
–x
–5 = x – 8
+8
+8
Subtract x from both sides.
Add 8 to both sides.
3=x
To find y, substitute 3 for x in one of the original
equations.
y = x – 5 = 3 – 5 = –2
The solution is (3, –2).
Course 3
11-6 Systems of Equations
Check It Out: Example 1B
y = 3x + -7
y = 6 + 3x
3x + -7 = 6 + 3x
– 3x
– 3x
-7 ≠ 6
Transitive Property
Subtract 3x from both sides.
The system of equations has no solution.
Course 3
11-6 Systems of Equations
To solve a general system of two
equations with two variables, you can
solve both equations for x or both for y.
Course 3
11-6 Systems of Equations
Additional Example 2A: Solving Systems of
Equations by Solving for a Variable
Solve the system of equations.
5x + y = 7
5x + y = 7
-y
-y
5x
x – 3y = 11
Solve both
x – 3y = 11
equations
+ 3y
+ 3y
for x.
x
= 11 + 3y
=7 -y
5(11 + 3y)= 7 - y
55 + 15y = 7 – y
- 15y
- 15y
55
Course 3
Subtract 15y
from both
= 7 – 16y sides.
11-6 Systems of Equations
Additional Example 2A Continued
55
=
–7
48
–16
=
-3
=
7 – 16y
–7
-16y
-16
y
Subtract 7 from
both sides.
Divide both sides
by –16.
x = 11 + 3y
= 11 + 3(-3) Substitute –3 for y.
= 11 + –9 = 2
The solution is (2, –3).
Course 3
11-6 Systems of Equations
Helpful Hint
You can solve for either variable. It is usually
easiest to solve for a variable that has a
coefficient of 1.
Course 3
11-6 Systems of Equations
Additional Example 2B: Solving Systems of
Equations by Solving for a Variable
Solve the system of equations.
–2x + 10y = –8
Solve both x – 5y = 4
–2x + 10y = –8
x – 5y = 4
equations
–10y
–10y for x.
+5y
+5y
–2x
= –8 – 10y
x
= 4 + 5y
–2x = –8 – 10y
–2
–2 –2
x = 4 + 5y
4 + 5y = 4 + 5y Subtract 5y
- 5y
- 5y from both
sides.
4=4
Since 4 = 4 is always true, the system of
equations has an infinite number of solutions.
Course 3
11-6 Systems of Equations
Check It Out: Example 2A
Solve the system of equations.
x+y=5
x+y=5
–x
–x
Solve both
equations
for y.
y=5–x
5 – x = –1 – 3x
+x
+ x
5
Course 3
= –1 – 2x
3x + y = –1
3x + y = –1
– 3x
– 3x
y = –1 – 3x
Add x to
both sides.
11-6 Systems of Equations
Check It Out: Example 2A Continued
5
+1
6
= –1 – 2x
+1
=
–2x
–3 = x
Add 1 to both sides.
Divide both sides
by –2.
y=5–x
= 5 – (–3)
Substitute –3 for x.
=5+3=8
The solution is (–3, 8).
Course 3
11-6 Systems of Equations
Check It Out: Example 2B
Solve the system
x + y = –2
x + y = –2
–x
–x
y = –2 – x
of equations.
Solve both
equations
for y.
–2 – x = 2 + 3x
Course 3
–3x + y = 2
–3x + y = 2
+ 3x
+ 3x
y = 2 + 3x
11-6 Systems of Equations
Check It Out: Example 2B Continued
–2 – x = 2 + 3x
+x
+x
–2
–2
= 2 + 4x
–2
–4
=
4x
–1 = x
y = 2 + 3x
= 2 + 3(–1) = –1
The solution is (–1, –1).
Course 3
Add x to both
sides.
Subtract 2 from
both sides.
Divide both sides
by 4.
Substitute –1
for x.
11-6 Systems
Insert Lesson
of Equations
Title Here
Lesson Quiz
Solve each system of equations.
1. y = 5x + 10
no solution
y = –7 + 5x
2. y = 2x + 1
y = 4x
1
(2 , 2)
3. 6x – y = –15 (–2,3)
2x + 3y = 5
4. Two numbers have a sum of 23 and a
difference of 7. Find the two numbers.
15 and 8
Course 3