Download Solving Linear Equations and Inequalities Warm Up Lesson

Solving Linear Equations and
2-1
2-1 Inequalities
Solving Linear Equations and Inequalities
Warm Up
Lesson Presentation
Lesson Quiz
Holt
Algebra
Holt
Algebra
22
2-1
Solving Linear Equations and Inequalities
Warm Up
Simplify each expression.
2. –(w – 2) –w + 2
1. 2x + 5 – 3x –x + 5
3. 6(2 – 3g) 12 – 18g
Graph on a number line.
4. t > –2
–4 –3 –2 –1
0
1
2
3
4
5
5. Is 2 a solution of the inequality –2x < –6? Explain.
No; when 2 is substituted for x, the inequality is false:
–4 < –6
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Objectives
Solve linear equations using a variety
of methods.
Solve linear inequalities.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Vocabulary
equation
solution set of an equation
linear equation in one variable
identify
contradiction
inequality
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
An equation is a mathematical statement that two
expressions are equivalent. The solution set of an
equation is the value or values of the variable that
make the equation true. A linear equation in one
variable can be written in the form ax = b, where a
and b are constants and a ≠ 0.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Linear Equations in
One variable
4x = 8
3x –
= –9
2x – 5 = 0.1x +2
Nonlinear
Equations
+ 1 = 32
+ 1 = 41
3 – 2x = –5
Notice that the variable in a linear equation is not
under a radical sign and is not raised to a power other
than 1. The variable is also not an exponent and is not
in a denominator.
Solving a linear equation requires isolating the
variable on one side of the equation by using the
properties of equality.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
To isolate the variable, perform the inverse or
opposite of every operation in the equation on
both sides of the equation. Do inverse
operations in the reverse order of operations.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Example 1: Consumer Application
The local phone company charges
$12.95 a month for the first 200 of air
time, plus $0.07 for each additional
minute. If Nina’s bill for the month was
$14.56, how many additional minutes
did she use?
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Example 1 Continued
Let m represent the number of additional minutes that
Nina used.
Model
monthly
charge plus
12.95
Holt Algebra 2
+
additional
minute
charge
times
0.07
*
number of
additional = total
charge
minutes
m
=
14.56
2-1
Solving Linear Equations and Inequalities
Example 1 Continued
Solve.
12.95 + 0.07m = 14.56
–12.95
–12.95
0.07m =
0.07
1.61
0.07
Subtract 12.95 from both
sides.
Divide both sides by 0.07.
m = 23
Nina used 23 additional minutes.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 1
Stacked cups are to be placed in a
pantry. One cup is 3.25 in. high and
each additional cup raises the stack
0.25 in. How many cups fit between
two shelves 14 in. apart?
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 1 Continued
Let c represent the number of additional cups needed.
Model
additional
cup
one cup plus
height
3.25
Holt Algebra 2
+
0.25
times
*
number of
total
additional = height
cups
c
=
14.00
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 1 Continued
Solve.
3.25 + 0.25c = 14.00
–3.25
–3.25
0.25c = 10.75
0.25
0.25
Subtract 3.25 from both
sides.
Divide both sides by 0.25.
c = 43
44 cups fit between the 14 in. shelves.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Example 2: Solving Equations with the Distributive
Property
Solve 4(m + 12) = –36
Method 1
The quantity (m + 12) is multiplied by 4, so divide
by 4 first.
4(m + 12) = –36
4
4
m + 12 = –9
–12 –12
m = –21
Holt Algebra 2
Divide both sides by 4.
Subtract 12 from both sides.
2-1
Solving Linear Equations and Inequalities
Example 2 Continued
Check
4(m + 12) = –36
4(–21 + 12)
4(–9)
–36
Holt Algebra 2
–36
–36
–36 ü
2-1
Solving Linear Equations and Inequalities
Example 2 Continued
Solve 4(m + 12) = –36
Method 2
Distribute before solving.
4m + 48 = –36
–48 –48
Distribute 4.
Subtract 48 from both sides.
4m = –84
4m –84
=
4
4
m = –21
Holt Algebra 2
Divide both sides by 4.
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 2a
Solve 3(2 –3p) = 42.
Method 1
The quantity (2 – 3p) is multiplied by 3, so divide
by 3 first.
3(2 – 3p) = 42
3
3
2 – 3p = 14
–2
–2
–3p = 12
–3
–3
p = –4
Holt Algebra 2
Divide both sides by 3.
Subtract 2 from both sides.
Divide both sides by –3.
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 2a Continued
Check
Holt Algebra 2
3(2 – 3p) =
3(2 + 12)
6 + 36
42
42
42
42
42 ü
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 2a Continued
Solve 3(2 – 3p) = 42 .
Method 2
Distribute before solving.
6 – 9p = 42
–6
–6
–9p = 36
–9p 36
=
–9 –9
p = –4
Holt Algebra 2
Distribute 3.
Subtract 6 from both sides.
Divide both sides by –9.
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 2b
Solve –3(5 – 4r) = –9.
Method 1
The quantity (5 – 4r) is multiplied by –3, so divide by
–3 first.
–3(5 – 4r) = –9
–3
–3
5 – 4r = 3
–5
–5
–4r = –2
Holt Algebra 2
Divide both sides by –3.
Subtract 5 from both sides.
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 2b Continued
Solve –3(5 – 4r) = –9.
Method 1
–4r –2
–4
–4
r=
Divide both sides by –4.
=
Check
–3(5 –4r) = –9
–3(5 – 4• ) –9
–3(5 – 2)
–3(3)
–9
Holt Algebra 2
–9
–9
–9 ü
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 2b Continued
Solve –3(5 – 4r) = –9.
Method 2
Distribute before solving.
–15 + 12r = –9
+15
+15
12r = 6
12r
6
=
12
12
r=
Holt Algebra 2
Distribute 3.
Add 15 to both sides.
Divide both sides by 12.
2-1
Solving Linear Equations and Inequalities
If there are variables on both sides of the
equation, (1) simplify each side. (2) collect all
variable terms on one side and all constants
terms on the other side. (3) isolate the
variables as you did in the previous problems.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Example 3: Solving Equations with
Variables on Both Sides
Solve 3k– 14k + 25 = 2 – 6k – 12.
Simplify each side by combining
–11k + 25 = –6k – 10
like terms.
+11k
+11k
Collect variables on the right side.
25 = 5k – 10
Add.
+10
+ 10
Collect constants on the left side.
35 = 5k
Isolate the variable.
5
5
7=k
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 3
Solve 3(w + 7) – 5w = w + 12.
Simplify each side by combining
–2w + 21 = w + 12
like terms.
+2w
+2w
Collect variables on the right side.
21 =
–12
9 =
3
3=
Holt Algebra 2
3w + 12
–12
3w
3
w
Add.
Collect constants on the left side.
Isolate the variable.
2-1
Solving Linear Equations and Inequalities
You have solved equations that have a
single solution. Equations may also have
infinitely many solutions or no solution.
An equation that is true for all values of the
variable, such as x = x, is an identity. An
equation that has no solutions, such as
3 = 5, is a contradiction because there
are no values that make it true.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Example 4A: Identifying Identities and Contractions
Solve 3v – 9 – 4v = –(5 + v).
3v – 9 – 4v = –(5 + v)
–9 – v = –5 – v
+v
+v
–9 ≠ –5 x
Simplify.
Contradiction
The equation has no solution. The solution set is
the empty set, which is represented by the
symbol .
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Example 4B: Identifying Identities and Contractions
Solve 2(x – 6) = –5x – 12 + 7x.
2(x – 6) = –5x – 12 + 7x
Simplify.
2x – 12 = 2x – 12
–2x
–2x
–12 = –12 ü
Identity
The solutions set is all real number, or ℜ.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 4a
Solve 5(x – 6) = 3x – 18 + 2x.
5(x – 6) = 3x – 18 + 2x
5x – 30 = 5x – 18
–5x
–5x
–30 ≠ –18 x
Simplify.
Contradiction
The equation has no solution. The solution set is
the empty set, which is represented by the
symbol .
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 4b
Solve 3(2 –3x) = –7x – 2(x –3).
3(2 –3x) = –7x – 2(x –3)
6 – 9x = –9x + 6
Simplify.
+ 9x
+9x
6 = 6 ü
Identity
The solutions set is all real numbers, or ℜ.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
An inequality is a statement that compares two
expressions by using the symbols <, >, ≤, ≥, or ≠.
The graph of an inequality is the solution set, the set
of all points on the number line that satisfy the
inequality.
The properties of equality are true for inequalities,
with one important difference. If you multiply or
divide both sides by a negative number, you must
reverse the inequality symbol.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
These properties also apply to inequalities
expressed with >, ≥, and ≤.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Helpful
Hint
To check an inequality, test
•  the value being compared with x
•  a value less than that, and
•  a value greater than that.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Example 5: Solving Inequalities
Solve and graph 8a –2 ≥ 13a + 8. 8a – 2 ≥ 13a + 8
–13a
–13a
–5a – 2 ≥ 8
+2
+2
–5a ≥ 10
–5a ≤ 10
–5
–5
a ≤ –2
Holt Algebra 2
Subtract 13a from both sides.
Add 2 to both sides.
Divide both sides by –5 and
reverse the inequality.
2-1
Solving Linear Equations and Inequalities
Example 5 Continued
Solve and graph 8a – 2 ≥ 13a + 8. Check Test values in
the original inequality.
Test x = –4
•
–10 –9
Test x = –2
–8 –7 –6 –5 –4
–3 –2 –1
Test x = –1
8(–4) – 2 ≥ 13(–4) + 8 8(–2) – 2 ≥ 13(–2) + 8 8(–1) – 2 ≥ 13(–1) + 8
–34 ≥ –44 ü
–18 ≥ –18 ü
–10 ≥ –5 x
So –4 is a
solution.
So –2 is a
solution.
So –1 is not
a solution.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 5
Solve and graph x + 8 ≥ 4x + 17.
x + 8 ≥ 4x + 17
–x
–x
8 ≥ 3x +17
–17
–17
–9 ≥ 3x
–9 ≥ 3x
3
3
–3 ≥ x or x ≤ –3
Holt Algebra 2
Subtract x from both sides.
Subtract 17 from both sides.
Divide both sides by 3.
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 5 Continued
Solve and graph x + 8 ≥ 4x + 17.
Check Test values in the
original inequality.
Test x = –6
•
–6 –5
Test x = –3
–6 + 8 ≥ 4(–6) + 17 –3 +8 ≥ 4(–3) + 17
2 ≥ –7 ü
5≥5
So –6 is a
solution.
So –3 is a
solution.
Holt Algebra 2
ü
–4 –3 –2 –1
0
1
2
3
Test x = 0
0 +8 ≥ 4(0) + 17
8 ≥ 17 x
So 0 is not
a solution.
2-1
Solving Linear Equations and Inequalities
Lesson Quiz: Part I
1. Alex pays $19.99 for cable service each month.
He also pays $2.50 for each movie he orders
through the cable company’s pay-per-view
service. If his bill last month was $32.49, how
many movies did Alex order?
5 movies
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Lesson Quiz: Part II
Solve.
x=6
2. 2(3x – 1) = 34
3. 4y – 9 – 6y = 2(y + 5) – 3 y = –4
4. r + 8 – 5r = 2(4 – 2r)
all real numbers, or ℜ
5. –4(2m + 7) =
no solution, or
Holt Algebra 2
(6 – 16m)
2-1
Solving Linear Equations and Inequalities
Lesson Quiz: Part III
5. Solve and graph.
12 + 3q > 9q – 18
q<5
°
–2 –1
Holt Algebra 2
0
1
2
3
4
5
6
7