Download Section 2.3 Solving Linear Equations

Section 2.3 Solving Linear Equations
Variable:
 Defined as “A symbol (or letter) that is used to represent an unknown numbers”
 Examples: a, b, c, x, y, z, s, t, m, n,⋯
Constant:
 Defined as “A single number”
 Examples: 1, 2, 3, –6, –1, 𝜋, e, −𝜋, 1.6, 5/8, ⋯
Coefficient:
 Defined as “A number written next to the variable”
 3y, 3 is the coefficient of y
 –5x2, –5 is the coefficient of x2
Identify each term of the algebraic expressions below
Expressions
Variable(s)
Coefficient(s)
x+3
1
x
4x – 7y +5
4, -7
x, y
4x2 – 5
4
x2
–2m – 6
-2
m
11x – 50t – 5s
11, -50, -5
x, t, s
m–a
1, -1
m, a
0.5a – 3.7c
0.5, -3.7
a, c
–v + 8
-1
v
Constant
3
5
-5
-6
None
None
None
8
Algebraic Expression:
 Defined as “A combination of variables and numbers using any of the operations of ‘+, –,
×, ÷’ and exponents”
3
8
 Examples: x + 1, 4 x 2 − 7 xy − 5, − x 2 + ,
3 xy + y 3
,
x2 − 7
Like Terms:
 Terms that are constants or terms that contain the same variables raised to the same
exponents
 We can combine and simplify only like terms.
 Like Terms as constant: –4, 1.54, 374, –0.37, 𝜋, ⋯
 Like Terms as variable term: –5t, 20t, 3.4t, 𝜋𝑡
 Note that 3st2, –4st, 3s3t, and –7s3t2 are not like terms since variables are all different
Steps for simplifying expressions:
 Step 1: Identify each like term
 Step 2: Rearrange the expression by like terms
(Note: only one type of like term in expression, then skip this step)
 Step 3: Combine like terms by the distributive property
 Step 4: Add (or subtract) the coefficient and keep the common variable expression
Cheon-Sig Lee
www.coastalbend.edu/lee
Page 1
Section 2.3 Solving Linear Equations
Simplify Algebraic Expressions:
Steps
3𝑥 + 5𝑥
Step 1
One type
(Identifying like terms)
Step 2
(Rearrangement)
Step 3
=(3 + 5)𝑥
(Distribution Law)
Step 4
=8𝑥
(Simplifying)
−2𝑥 + 2 − 7𝑥𝑦 − 5𝑥 + 𝑥 2 𝑦 − 8𝑥𝑦
3𝑥 − 7𝑦 − 5𝑥 + 1
Three Types
Three Types
=3𝑥 − 5𝑥 − 7𝑦 + 1
=−2𝑥 − 5𝑥 + 2 − 7𝑥𝑦 − 8𝑥𝑦 + 𝑥 2 𝑦
=−2𝑥 − 7𝑦 + 1
=−7𝑥 + 2 − 15𝑥𝑦 + 𝑥 2 𝑦
=(3 − 5)𝑥 − 7𝑦 + 1
=(−2 − 5)𝑥 + 2 + (−7 − 8)𝑥𝑦 + 𝑥 2 𝑦
Steps for Evaluating Expressions
 Step 1: Simplify a given expression, if possible
 Step 2: Substitute the values given for any variable
 Step 3: Evaluate the resulting expression
Evaluate Algebraic Expressions bellow for 𝒙 = 𝟐, 𝒚 = 𝟏
Steps
3x + 4
Step 1
(Simplifying Expression)
3x – 7y – 5 x + 1
–2x + 2 – 7xy –5x + x2y – 8xy
= –2x – 7y + 1
= –7x – 15xy + x2y + 2
Step 2
(Substituting)
=3(2) + 4
= –2(2) – 7(-1) + 1
= –7(2) –15(2)(-1) + (2)2(-1) + 2
Step 3
(Evaluating)
=6 + 4
=10
= –4 +7 + 1
=4
= –14 + 30 – 4 + 2
=14
Equation:
A statement that two algebraic expressions are equal
Solution:
Any number that gives a true statement when substituted for the variable
Solution Set:
The solutions to an equation form
Solving Equations
 Step 1: Simplify the algebraic expression on each side
 Step 2: Collect all the variable terms on one side and all the constant terms on the other
side
 Step 3: Isolate the variable and solve.
 Step 4: Check the proposed solution in the original equation
Cheon-Sig Lee
www.coastalbend.edu/lee
Page 2
Section 2.3 Solving Linear Equations
Solve the equations:
Steps
Step 1
Step 2
𝒙 + 𝟒= 𝟔
𝑥 + 4= 6
−4 −4
𝑥= 2
𝟑𝒙 − 𝟒 − 𝟐𝒙 + 𝟐= −𝟔
3𝑥 − 2𝑥 − 4 + 2= −6
𝑥 − 2= −6
𝑥 − 2= −6
+2 +2
𝑥= 4
Step 3
Step 4
Optional
Optional
−𝒙−𝟔
−𝑥 −6
=
−1 −1
𝑥= 6
Optional
Types of Linear Equations:
Types
Number of solutions
Conditional
Finite Number of Solutions
Identity
Infinite Number of Solutions
Contradiction
No Solution
𝟒𝒙 − 𝟒 − 𝟕𝒙 + 𝟐= 𝟑𝒙 − 𝟖
4𝑥 − 7𝑥 − 4 + 2= 3𝑥 − 8
−3𝑥 − 2= 3𝑥 − 8
−3𝑥 − 2= 3𝑥 − 8
+3𝑥
+3𝑥
−2= 6𝑥 − 8
+8
+8
6= 6𝑥
6 6𝑥
=
6 6
1= 𝑥
Optional
Other terms
One solution
Infinitely many solutions
No solution
Exercises
(Solution 1)
5𝑥 + 8𝑥 − 7𝑥 = 20 + 4
13𝑥 − 7𝑥 = 24
6𝑥 = 24
6𝑥 24
=
6
6
𝑥= 4
(Solution 2)
5(2𝑥 − 1)= 20
5(2𝑥 − 1) 20
=
5
5
2𝑥 − 1= 4
+1 +1
2𝑥 = 5
2𝑥 5
=
2 2
5
𝑥=
2
Cheon-Sig Lee
www.coastalbend.edu/lee
Page 3
Section 2.3 Solving Linear Equations
(Solution 3)
16𝑥 − (9𝑥 − 5)= 40
16𝑥 − 9𝑥 + 5= 40
7𝑥 + 5= 40
−5 −5
7𝑥 = 35
7𝑥 35
=
7
7
𝑥= 5
(Solution 4)
13(17 − 𝑥)= 15(6𝑥 + 1)
13 ∙ 17 + 13(−𝑥)= 15 ∙ 6𝑥 + 15 ∙ 1
221 − 13𝑥 = 90𝑥 + 15
+13𝑥 +13𝑥
221= 103𝑥 + 15
−15
−15
206= 103𝑥
206 103𝑥
=
103 103
2= 𝑥
(Solution 5)
4(𝑥 + 1)= 7(𝑥 − 1) − 7
4 ∙ 𝑥 + 4 ∙ 1= 7 ∙ 𝑥 + 7(−1) − 7
4𝑥 + 4= 7𝑥 − 7 − 7
4𝑥 + 4= 7𝑥 − 14
−4𝑥
−4𝑥
4= 3𝑥 − 14
+14
+14
18= 3𝑥
18 3𝑥
=
3
3
6= 𝑥
Cheon-Sig Lee
www.coastalbend.edu/lee
Page 4
Section 2.3 Solving Linear Equations
(Solution 6)
37= −2(23 − 𝑦) + 3(𝑦 − 4)
37= (−2)(23) + (−2)(−𝑦) + 3 ∙ 𝑦 + 3(−4)
37= −46 + 2𝑦 + 3𝑦 − 12
37= 2𝑦 + 3𝑦 − 46 − 12
37= 5𝑦 − 58
+58
+58
95= 5𝑦
95 5𝑦
=
5
5
19= 𝑦
(Solution 7)
Multiply LCD on both sides; LCD = 3
𝑥
− 2= −7
3
𝑥
3 ∙ − 2 ∙ 3= −7 ∙ 3
3
𝑥 − 6= −21
+6 +6
𝑥 = −15
(Solution 8) LCD of 4, 5, and 20 is 20
Multiply LCD on both sides; LCD = 20
𝑥 𝑥 9
+ =
4 5 20
𝑥
𝑥
9
20 ∙ + 20 ∙ = 20 ∙
4
5
20
5 ∙ 𝑥 + 4 ∙ 𝑥= 9
9𝑥 = 9
9𝑥 9
=
9 9
𝑥= 1
Cheon-Sig Lee
www.coastalbend.edu/lee
Page 5
Section 2.3 Solving Linear Equations
(Solution 9)
LCD of 4 and 9 is 36
Multiply LCD on both sides; LCD = 36
36 ∙
(Solution 10)
𝑦
5 𝑦
5
+ 36 ∙ = ∙ 36 − ∙ 36
4
9 9
9
9𝑦 + 4 ∙ 5= 4𝑦 − 5 ∙ 4
9𝑦 + 20= 4𝑦 − 20
−4𝑦
−4𝑦
5𝑦 + 20= −20
−20 −20
5𝑦= −40
5𝑦 −40
=
5
5
𝑦= −8
LCD of 2 and 3 is 6
Multiply LCD on both sides; LCD = 6
𝑥−1
𝑥−1
− 2=
2
3
𝑥−1
𝑥−1
6∙
− 6 ∙ 2= 6 ∙
2
3
3(𝑥 − 1) − 12= 2(𝑥 − 1)
3 ∙ 𝑥 + 3 ∙ (−1) − 12= 2 ∙ 𝑥 + 2(−1)
3𝑥 − 3 − 12= 2 ∙ 𝑥 − 2
3𝑥 − 15= 2𝑥 − 2
−2𝑥
−2𝑥
𝑥 − 15= −2
+15 +15
𝑥 = 13
(Solution 10)
3.6𝑥 = 2.4𝑥 + 8.4
−2.4𝑥 −2.4𝑥
1.2𝑥 = 8.4
1.2𝑥 8.4
=
1.2 1.2
𝑥= 7
Cheon-Sig Lee
www.coastalbend.edu/lee
Page 6