Transcript Solving Equations With Variables on Both Sides
Solving Equations With Variables on Both Sides Section 2-4
Goals
Goal
• To solve equations with variables on both sides.
• To identify equations that are identities or have no solution.
Rubric
Level 1 – Know the goals.
Level 2 – Fully understand the goals.
Level 3 – Use the goals to solve simple problems.
Level 4 – Use the goals to solve more advanced problems.
Level 5 – Adapts and applies the goals to different and more complex problems.
• Identity Vocabulary
Equations With Variables on Both Sides
To solve an equation with variables on both sides, use inverse operations to
"collect"
variable terms on one side of the equation.
Helpful Hint Equations are often easier to solve when the variable has a positive coefficient. Keep this in mind when deciding on which side to "collect" variable terms.
Example: Solving Equations with Variables on Both Sides
Solve 7
n
– 2 = 5
n
+ 6.
7
n
–5
n
– 2 = 5
n
–5
n
+ 6 2
n
– 2 = 6 + 2 + 2 2
n
= 8
To collect the variable terms on one side, subtract 5n from both sides.
Since n is multiplied by 2, divide both sides by 2 to undo the multiplication.
n
= 4
Your Turn:
Solve 4
b
+ 2 = 3
b
.
4
b
–3
b
+ 2 = 3
b
–3
b b
+ 2 = 0 – 2 – 2
b
= –2
To collect the variable terms on one side, subtract 3b from both sides.
Your Turn:
Solve 0.5 + 0.3
y
= 0.7
y
– 0.3. 0.5 + 0.3
–0.3
y y
= 0.7
0.5 = 0.4
y
–0.3
y y
– 0.3
– 0.3
+0.3 + 0.3
0.8 = 0.4
y To collect the variable terms on one side, subtract 0.3y from both sides.
Since 0.3 is subtracted from 0.4y, add 0.3 to both sides to undo the subtraction.
2 =
y Since y is multiplied by 0.4, divide both sides by 0.4 to undo the multiplication.
Using the Distributive Property
To solve more complicated equations, you may need to first simplify by using the Distributive Property or combining like terms.
Example: Simplifying Both Sides
Solve 4 – 6
a
+ 4
a
= –1 – 5(7 – 2
a
). 4 – 6
a
+ 4
a
4 – 6
a
+ 4
a
= –1 –5 (7 – 2
a
) = –1 –5 (7) –5 (–2
a
)
Distribute –5 to the expression in parentheses.
4 – 6
a
+ 4
a
= –1 – 35 + 10
a
4 – 2
a
= –36 + 10 +36 +36
a Combine like terms.
Since –36 is added to 10a, add 36 to both sides.
40 – 2
a
= 10
a
+ 2
a
+2
a
40 = 12
a To collect the variable terms on one side, add 2a to both sides.
Example: Continued
40 = 12
a Since a is multiplied by 12, divide both sides by 12.
Your Turn:
Solve .
1 Distribute to the expression in 2 parentheses.
3 =
b
– 1 + 1 + 1 4 =
b To collect the variable terms on one side, subtract b from both sides.
2 Since 1 is subtracted from b, add 1 to both sides.
Your Turn:
Solve 3
x
+ 15 – 9 = 2(
x
+ 2). 3
x
3
x
+ 15 – 9 = + 15 – 9 = 2 2 ( (
x x
+ 2) ) +
Distribute 2 to the expression in parentheses.
2 (2) 3
x
+ 15 – 9 = 2
x
+ 4
x
3
x x
+ 6 = 2 = –2
x
–2x –2x + 4 + 6 = 4 – 6 – 6
Combine like terms.
To collect the variable terms on one side, subtract 2x from both sides.
Since 6 is added to x, subtract 6 from both sides to undo the addition.
Summary
Solving linear equations in one variable
1) Clear the equation of fractions by multiplying both sides of the equation by the LCD of all denominators in the equation.
2) 3) 4) 5) 6) Use the distributive property to remove grouping symbols such as parentheses.
Combine like terms on each side of the equation.
Use the addition property of equality to rewrite the equation as an equivalent equation with variable terms on one side and numbers on the other side.
Use the multiplication property of equality to isolate the variable.
Check the proposed solution in the original equation.
Conditional Equations & Contradictions
A
conditional equation
is an equation that is true for some values of the variable and false for other values of the variable. Normal equations with a finite number of solutions.
A
contradiction
is an equation that is false for every replacement value of the variable.
Example
: Solve the equation 4
x
– 8 = 4
x
– 1. 4
x
– 8 = 4
x
– 1 – 8 = – 1 Subtract 4
x
from both sides.
This is a
false statement
so the equation is a contradiction .
The solution set is the empty set, written { } or
.
Contradictions
WORDS Contradiction
When solving an equation, if you get a false equation, the original equation is a contradiction, and it has no solutions.
NUMBERS
1 = 1 + 2 1 = 3
ALGEBRA
x
=
x
+ 3
–
x
–
x
0 = 3
Identities
An
identity
is an equation that is satisfied for all values of the variable for which both sides of the equation are defined.
Example
: Solve the equation 5
x
+ 3 = 2
x
+ 3(
x
+ 1).
5
x
+ 3 = 2
x
+ 3(
x
+ 1).
5
x
+ 3 = 2
x
+ 3
x
+ 3 5
x
+ 3 = 5
x
+ 3 3 = 3 Distribute to remove parentheses.
Combine like terms.
Subtract 5
x
from both sides.
This is a
true statement
for all real numbers
x
.
The solution set is all real numbers.
Identities
WORDS NUMBERS ALGEBRA Identity
When solving an equation, if you get an equation that is always true, the original equation is an identity, and it has infinitely many solutions.
2 + 1 = 2 + 1 3 = 3
2 + x = 2 + x –x 2 = 2
–x
Summary
1) A
conditional equation
is an equation in one variable that has a finite (normally one solution) number of solutions.
2) An
identity
is an equation that is true for all values of the variable (ie. the variable is eliminated and results in a true statement). An equation that is an identity has infinitely many solutions. 3) A
contradiction
is an equation that is false for any value of the variable (ie. the variable is eliminated and results in a false statement). It has no solutions.
Example: Identity
Solve 10 – 5
x
+ 1 = 7
x
+ 11 – 12
x
. 10 – 5
x
+ 1 = 7
x
+ 11 – 12
x
10 – 5
x
+ 1 = 7
x
+ 11 – 12
x
11 – 5
x
= 11 – 5
x
+ 5
x
11 = 11 + 5
x
Identify like terms.
Combine like terms on the left and the right.
Add 5x to both sides.
True statement.
The equation 10 – 5
x
+ 1 = 7
x
+ 11 – 12
x
is an
identity
. All values of
x
will make the equation true. All real numbers are solutions.
Example: Contradiction
Solve 12
x
– 3 +
x
= 5
x
– 4 + 8
x
. 12
x
– 3 +
x
= 5
x
– 4 + 8
x
12
x
– 3 +
x
= 5
x
– 4 + 8
x
13
x
–13
x
– 3 = 13
x
–13
x
– 4 –3 = –4
Identify like terms.
Combine like terms on the left and the right.
Subtract 13x from both sides.
False statement.
The equation 12
x
– 3 +
x
= 5
x
– 4 + 8
x
is a
contradiction
. There is no value of
x
that will make the equation true. There are no solutions.
Your Turn:
Solve 4
y
+ 7 –
y
= 10 + 3
y
. 4
y
+ 7 –
y
= 10 + 3
y
4
y
+ 7 –
y
= 10 + 3
y
3
y
–3
y
+ 7 = 3
y
–3
y
+ 10 7 = 10
Identify like terms.
Combine like terms on the left and the right.
Subtract 3y from both sides.
False statement.
The equation 4
solutions
.
y
+ 7 –
y
= 10 + 3
y
is a
contradiction
. There is no value of
y
that will make the equation true. There are
no
Your Turn:
Solve 2
c
+ 7 +
c
= –14
+
3
c
+ 21. 2
c
+ 7 +
c
= –14
+
3
c
+ 21 2
c
+ 7 +
c
= –14
+
3
c
+ 21 3
c
–3
c
+ 7 = 3
c
–3
c
+ 7 7 = 7
Identify like terms.
Combine like terms on the left and the right.
Subtract 3c both sides.
True statement.
The equation 2
c
+ 7 +
c
= –14
+
3
c
+ 21 is an
identity
. All values of
c
will make the equation true. All real numbers are solutions (
infinite number of solutions
).
Strategy for Problem Solving
1) 2) 3) 4) • • • • UNDERSTAND the problem. During this step, become comfortable with the problem. Some way of doing this are: Read and reread the problem Propose a solution and check. Construct a drawing.
Choose a variable to represent the unknown TRANSLATE the problem into an equation.
SOLVE the equation.
INTERPRET the result.
Check
the proposed solution in stated problem and
state
your conclusion.
Finding an Unknown Number
Example:
The product of twice a number and three is the same as the difference of five times the number and ¾. Find the number.
1.) Understand
Read and reread the problem. If we let
x
= the unknown number, then “twice a number” translates to 2
x
, “the product of twice a number and three” translates to 2
x
· 3, “five times the number” translates to 5
x
, and “the difference of five times the number and ¾” translates to 5
x –
¾.
Continued
Example continued: 2.) Translate
The product of twice a number and 3 is the same as the difference of 5 times the number and ¾ 2
x
·
3 = 5
x
– ¾ Continued
Example continued: 3.) Solve
2
x
· 3 = 5
x
– ¾ 6
x
= 5
x
– ¾ 6
x
+ (– 5
x
) = 5
x
+ (– 5
x
) – ¾
x
= – ¾ (Simplify left side) (Add –5
x
to both sides) (Simplify both sides)
4.) Interpret Check:
Replace “number” in the original statement of the problem with – ¾. The product of twice – ¾ and 3 is 2(– ¾)(3) = – 4.5. The difference of five times – ¾ and ¾ is 5(– ¾) – ¾ = – 4.5. We get the same results for both portions.
State:
The number is – ¾.
Your Turn:
Example:
A car rental agency advertised renting a Buick Century for $24.95 per day and $0.29 per mile. If you rent this car for 2 days, how many whole miles can you drive on a $100 budget?
1.) Understand
Read and reread the problem. Let’s propose that we drive a total of 100 miles over the 2 days. Then we need to take twice the daily rate and add the fee for mileage to get 2(24.95) + 0.29(100) = 49.90 + 29 = 78.90. This gives us an idea of how the cost is calculated, and also know that the number of miles will be greater than 100. If we let
x
= the number of whole miles driven, then 0.29
x
= the cost for mileage driven Continued
Continued: 2.) Translate
Daily costs plus mileage costs is equal to maximum budget 2(24.95) + 0.29
x
= 100 Continued
Continued: 3.) Solve
2(24.95) + 0.29
x
= 100 49.90 + 0.29
x
= 100 (Simplify left side) 49.90 – 49.90 + 0.29
x
= 100 – 49.90 (Subtract 49.90 from both sides) 0.29
x
= 50.10 0 .
29
x
0 .
29 50 .
10 0 .
29
x
172.75 (Simplify both sides) (Divide both sides by 0.29) (Simplify both sides) Continued
Continued: 4.) Interpret Check:
Recall that the original statement of the problem asked for a “whole number” of miles. If we replace “number of miles” in the problem with 173, then 49.90 + 0.29(173) = 100.07, which is over our budget. However, 49.90 + 0.29(172) = 99.78, which is within the budget.
State:
The maximum number of whole number miles is 172.
Joke Time
• Where do you find a dog with no legs?
• Right where you left him.
• What did E.T.’s Mom say when he got home?
• Where on Earth have you been?
• What happened when the pig pen broke?
• The pigs had to use a pencil.