Holt Algebra 1
2-1Solving Equations by Adding or Subtracting2-1 Solving One Step Equations using
inverse operations.
Holt Algebra 1
Lesson PresentationLesson Presentation
2-2
Holt Algebra 1
2-1Solving Equations by Adding or Subtracting
Solve one-step equations in one variable by using addition or subtraction.
Objective
Solve one-step equations in one variable by using multiplication or division.
Holt Algebra 1
2-1Solving Equations by Adding or Subtracting
An equation is a mathematical statement that two expressions are equal.
A solution of an equation is a value of the variable that makes the equation true.
To find solutions, isolate the variable. A variable is isolated when it appears by itself on one side of an equation, and not at all on the other side.
Holt Algebra 1
2-1Solving Equations by Adding or Subtracting
Inverse Operations
Operation Inverse Operation
Addition Subtraction
Subtraction Addition
Isolate a variable by using inverse operations which "undo" operations on the variable.
An equation is like a balanced scale. To keep the balance, perform the same operation on both sides.
Holt Algebra 1
2-1Solving Equations by Adding or Subtracting
WORDS
Addition Property of EqualityYou can add the same number to both sides of an equation, and the statement will still be true.
NUMBERS
3 = 3 3 + 2 = 3 + 2 5 = 5
ALGEBRA a = b a + c = b + c
Properties of Equality
Holt Algebra 1
2-1Solving Equations by Adding or Subtracting
WORDS
Subtraction Property of EqualityYou can subtract the same number from both sides of an equation, and the statement will still be true.
NUMBERS
7 = 7 7 – 5 = 7 – 5 2 = 2
ALGEBRA a = b a – c = b – c
Properties of Equality
Holt Algebra 1
2-1Solving Equations by Adding or Subtracting
Solve the equation. Check your answer.
Example 1A: Solving Equations by Using Addition
Since 8 is subtracted from y, add 8 to both sides to undo the subtraction.
y – 8 = 24 + 8 + 8
y = 32
Check y – 8 = 24
32 – 8 2424 24
To check your solution, substitute 32 for y in the original equation.
Holt Algebra 1
2-1Solving Equations by Adding or Subtracting
Solve the equation. Check your answer.
Example 1B: Solving Equations by Using Addition
= z34
To check your solution, substitute for z in the original equation.
34
+ 716
+ 716
= z – 716
516
Check= z – 7
16 516
34
516
716
–
516
516
Since is subtracted from z, add to
both sides to undo the subtraction.
716
716
Holt Algebra 1
2-1Solving Equations by Adding or Subtracting
Solve the equation. Check your answer.
Check It Out! Example 1a
Since 3.2 is subtracted from n, add 3.2 to both sides to undo the subtraction.
n – 3.2 = 5.6
+ 3.2 + 3.2
n = 8.8
Check n – 3.2 = 5.6
8.8 – 3.2 5.65.6 5.6
To check your solution, substitute 8.8 for n in the original equation.
Holt Algebra 1
2-1Solving Equations by Adding or Subtracting
Solve the equation. Check your answer.
Check It Out! Example 1c
Since 9 is subtracted from m, add 9 to both sides to undo the subtraction.
16 = m – 9 + 9 + 9
25 = m
Check 16 = m – 9
16 25 – 916 16
To check your solution, substitute 25 for m in the original equation.
Holt Algebra 1
2-1Solving Equations by Adding or Subtracting
Remember that subtracting is the same as adding the opposite. When solving equations, you will sometimes find it easier to add an opposite to both sides instead of subtracting.
Holt Algebra 1
2-1Solving Equations by Adding or Subtracting
Example 3: Solving Equations by Adding the Opposite
p = 311
+ 511
+ 511
Solve – + p = – . 211
511
511
Since – is added to p, add
to both sides.
511
Check + p = – 211
511
–
2 511 11
– – 311
+
211
– 211
–
To check your solution, substitute for p in the original equation.
311
Holt Algebra 1
2-1Solving Equations by Adding or Subtracting
Solve –2.3 + m = 7. Check your answer.
Check It Out! Example 3a
Since –2.3 is added to m, add 2.3 to both sides.
–2.3 + m = 7+2.3 + 2.3
m = 9.3
Check –2.3 + m = 7 –2.3 + 9.3 7
7 7
To check your solution, substitute 9.3 for m in the original equation.
Holt Algebra 1
2-1Solving Equations by Adding or Subtracting
Check It Out! Example 3b
z = 2
+ 3 4
+ 3 4
Solve – + z = . Check your answer. 5 4
3 4
3 4
Since – is added to z, add
to both sides.
3 4
Check + z = 5 4
3 4
–
– 5 3 4 4
+ 2
5 4
5 4
To check your solution, substitute 2 for z in the original equation.
– + z =
5 4
3 4
Holt Algebra 1
2-1Solving Equations by Adding or Subtracting
Over 20 years, the population of a town decreased by 275 people to a population of 850. Write and solve an equation to find the original population.
Example 4: Application
Write an equation to represent the relationship.
+ 275 + 275
p =1125
p – d = c
original population minus
current population
decrease in
populationis
p – 275 = 850 Since 275 is subtracted from p, add 275 to both sides to undo the subtraction.
p – d = c
The original population was 1125 people.
Holt Algebra 1
2-1Solving Equations by Adding or Subtracting
A person's maximum heart rate is the highest rate, in beats per minute, that the person's heart should reach. One method to estimate maximum heart rate states that your age added to your maximum heart rate is 220. Using this method, write and solve an equation to find a person's age if the person's maximum heart rate is 185 beats per minute.
Check It Out! Example 4
Holt Algebra 1
2-1Solving Equations by Adding or Subtracting
Check It Out! Example 4 Continued
a + r = 220
ageadded to
220maximum heart rate is
Write an equation to represent the relationship.
– 185 – 185
a = 35
a + 185 = 220 Substitute 185 for r. Since 185 is added to a, subtract 185 from both sides to undo the addition.
a + r = 220
A person whose maximum heart rate is 185 beats per minute would be 35 years old.
Holt Algebra 1
2-1Solving Equations by Adding or Subtracting
Inverse Operations
Operation Inverse Operation
Multiplication Division
Division Multiplication
Solving an equation that contains multiplication or division is similar to solving an equation that contains addition or subtraction. Use inverse operations to undo the operations on the variable.
Holt Algebra 1
2-1Solving Equations by Adding or Subtracting
WORDS
Multiplication Property of EqualityYou can multiply both sides of an equation by the same number, and the statement will still be true.
NUMBERS
6 = 6 6(3) = 6(3) 18 = 18
ALGEBRA a = b ac = bc
Properties of Equality
Holt Algebra 1
2-1Solving Equations by Adding or Subtracting
Properties of EqualityDivision Property of EqualityYou can divide both sides of an equation by the same nonzero number, and the statement will still be true.
WORDS
a = b (c ≠ 0)
8 = 8
2 = 2
ALGEBRA
NUMBERS 84
84
=
ac
ac=
Holt Algebra 1
2-1Solving Equations by Adding or Subtracting
Solve the equation.
Example 1A: Solving Equations by Using Multiplication
Since j is divided by 3, multiply both sides by 3 to undo the division.–24 = j
–8 –8
To check your solution, substitute –24 for j in the original equation.
–8 =j3
–8 –243
Check –8 =j3
Holt Algebra 1
2-1Solving Equations by Adding or Subtracting
Solve the equation.
Example 1B: Solving Equations by Using Multiplication
Since n is divided by 6, multiply both sides by 6 to undo the division.n = 16.8
2.8 2.8
To check your solution, substitute 16.8 for n in the original equation.
= 2.8n6
2.8 16.86
Check = 2.8n6
Holt Algebra 1
2-1Solving Equations by Adding or Subtracting
Solve the equation. Check your answer.
Example 2B: Solving Equations by Using Division
Since v is multiplied by –6, divide both sides by –6 to undo the multiplication.
0.8 = v
–4.8 –4.8
To check your solution, substitute 0.8 for v in the original equation.
–4.8 = –6v
–4.8 –6(0.8)
Check –4.8 = –6v
Holt Algebra 1
2-1Solving Equations by Adding or Subtracting
Solve the equation. Check your answer.
Check It Out! Example 2a
Since c is multiplied by 4, divide both sides by 4 to undo the multiplication.
4 = c
16 16
To check your solution, substitute 4 for c in the original equation.
16 = 4c
16 4(4)
Check 16 = 4c
Holt Algebra 1
2-1Solving Equations by Adding or Subtracting
Solve the equation. Check your answer.
Check It Out! Example 2b
Since y is multiplied by 0.5, divide both sides by 0.5 to undo the multiplication.y = –20
–10 –10
To check your solution, substitute –20 for y in the original equation.
0.5y = –10
0.5(–20) –10
Check 0.5y = –10
Holt Algebra 1
2-1Solving Equations by Adding or Subtracting
Remember that dividing is the same as multiplying by the reciprocal. When solving equations, you will sometimes find it easier to multiply by a reciprocal instead of dividing. This is often true when an equation contains fractions.
Holt Algebra 1
2-1Solving Equations by Adding or Subtracting
Solve the equation.
Example 3A: Solving Equations That Contain Fractions
w = 24
20 20
To check your solution, substitute 24 for w in the original equation.
w = 2056
Check w = 2056
The reciprocal of is . Since w is
multiplied by , multiply both sides
by .
56
65
566
5
20
Holt Algebra 1
2-1Solving Equations by Adding or Subtracting
Solve the equation.
Example 3B: Solving Equations That Contain Fractions
= z3
16
To check your solution,
substitute for z in the
original equation.
32
= z32
18
The reciprocal of is 8. Since z is
multiplied by , multiply both sides
by 8.
181
8
Check18
316
= z
316
316
Holt Algebra 1
2-1Solving Equations by Adding or Subtracting
Solve the equation. Check your answer.Check It Out! Example 3a
– = b14
To check your solution,
substitute – for b in the
original equation.
54
15
The reciprocal of is 5. Since b is
multiplied by , multiply both sides
by 5.
151
5
= b54–
= b5 4Check11–
Holt Algebra 1
2-1Solving Equations by Adding or Subtracting
Example 4: Application
Write an equation to represent the relationship.
Ciro puts of the money he earns from mowing lawns into a college education fund. This year Ciro added $285 to his college education fund. Write and solve an equation to find how much money Ciro earned mowing lawns this year.
14
one-fourth times earnings equals college fund
m = $1140
Substitute 285 for c. Since m is divided by 4, multiply both sides by 4 to undo the division.
Ciro earned $1140 mowing lawns.
Holt Algebra 1
2-1Solving Equations by Adding or Subtracting
Check it Out! Example 4
Write an equation to represent the relationship.
The distance in miles from the airport that a plane should begin descending, divided by 3, equals the plane's height above the ground in thousands of feet. A plane began descending 45 miles from the airport. Use the equation to find how high the plane was flying when the descent began.
Distance divided by 3 equals height in thousands of feet
15 = h
Substitute 45 for d.
The plane was flying at 15,000 ft when the descent began.
Holt Algebra 1
2-1Solving Equations by Adding or Subtracting
Homework
• Problems to try before you come to class.
• Page80 #22 to 48 even• Page87-88 #22-42
even
• Don’t forget to bring your textbook to class.