Download Extraneous Solution

Homework Log
Wed
Learning Objective:
To find solutions of equations
9/30
Hw: #201 Pg. 101 #1 – 31 odd
Lesson
2–1
9/30/15 Lesson 2 – 1 Solutions of
Equations Day 1
Advanced Math/Trig
Ch 2 TEST on
Tues 10/27
Learning Objective

To find solutions of
equations

To be aware of extraneous
solutions
Solve a Multi-Step
Equation
1. 3(2x – 1) – 2(3x + 4) = 11x
6x – 3 – 6x – 8 = 11x
- 11 = 11x
11 11
x=–1
{–1}
Solve a Multi-Step
Equation
2. 11 + 3x – 7 = 6x + 5 – 3x
4 + 3x = 3x + 5
- 3x -3x
4=5
NEVER TRUE!!!
No Solution!!
{∅}
Solve a Multi-Step
Equation
3. 6x + 5 – 2x = 4 + 4x + 1
4x + 5 = 4x + 5
-5
-5
4x = 4x
-4x - 4x
0=0
ALWAYS TRUE!!!
Infinite Solution!! {R}
Solve a Rational Equation
mult by common denom
4.
5 y  9 (y-1)(y-2)
(y-1)(y-2)y  4 (y-1)(y-2) y  3


y 1
y2
( y  1)( y  2)
( y  2)( y  4)  ( y  1)( y  3)  5 y  9
y2  2 y  8  y2  4 y  3  5 y  9
2
2
y  2y 8  y  y  6
2
2
y
y
2y 8  y  6
BUT, 2 makes denom = 0
y2
So, NO SOLUTION
Solve a Rational Equation
5.
3(𝑥+1) −2
𝑥+1
1
+3
3(𝑥+1)
1
=
7𝑥
3(𝑥+1)
3(𝑥+1)
1
LCD:
3(x + 1)
–6 + 3(3x + 3) = 7x
–6 + 9x + 9 = 7x
3 + 9x = 7x
3 = – 2x
x = –3/2
3
−
2
Always get the absolute
value by ITSELF before
splitting it up and
solving!!
Multi-Step Absolute Value
Equation
6.
2 𝑥+9 +3=7
–3 –3
2 𝑥+9 =4
2
2
𝑥+9 =2
x+9=2
or
–9 –9
x = – 7 or x = – 11
x+9=–2
–9 –9
{–7, –11}
Absolute Value Equations
7.
𝑥+3 =0
x + 3 = 0 (No – 0!)
–3–3
x=–3
{–3}
8.
𝑥 + 3 = −4
No Solution!!
{∅}
Absolute Value ≠ Negative!
Absolute Value Equations
9.
−2 7𝑥 − 8 + 4 = −8
–4 –4
−2 7𝑥 − 8 = −12
–2
–2
7𝑥 − 8 = 6
7x – 8 = 6
or
7x – 8 = – 6
7x = 14
7x = 2
2
2
x=2
or x =
,2
7
7
 Extraneous
Solution – Extra, but wrong
answer
plug your answers back into
the ORIGINAL equation to CHECK!!
 ALWAYS
Extraneous Solutions
10.
3𝑥 + 2 = 4𝑥 + 5
3x+2 = 4x+5
–2
–2
3x = 4x + 3
– 4x – 4x
–x=3
–1 –1
x=–3
or
3x+2 = – (4x+5)
3x+2 = – 4x – 5
–2
–2
3x = – 4x – 7
+ 4x + 4x
7x = – 7
7
7
or x = – 1
Check for Extraneous
Solutions
10. x = – 3
3(−3) + 2 = 4 −3 + 5
−9 + 2 = −12 + 5
−7 = −7
NO!!
x=–1
3(−1) + 2 = 4(−1) + 5
−3 + 2 = −4 + 5
−1 = 1
YES!!
Extraneous Solutions
10.
3𝑥 + 2 = 4𝑥 + 5
3x+2 = 4x+5
–2
–2
3x = 4x + 3
– 4x – 4x
–x=3
–1 –1
x=–3
or
3x+2 = – (4x+5)
3x+2 = – 4x – 5
–2
–2
3x = – 4x – 7
+ 4x + 4x
7x = – 7
7
7
or x = – 1 {–1}
Ticket Out the Door
2
𝑥−2
3
𝑥+1
6
(𝑥−2)(𝑥+1)

Solve

Explain your process.
−
=
Homework
#201
Pg. 101
1 – 31 odd