Transcript Solve Equations With Variables on Both Sides

Solve Equations With
Variables on Both Sides
Section 3.4
Steps to Solve Equations
with Variables on Both Sides
1) Simplify each side

Get rid of double Negatives

Distribute

Combine Like Terms
 2) Move variables to same side

“Smaller to the bigger”
 3) Solve by using INVERSE Operations

bkevil
Use Steps to Solve Equation:
2(x + 7) + 3 = 5x - 1
2x + 14 + 3 = 5x – 1
2x + 17 = 5x - 1
-2x
-2x
17 = 3x - 1
+1
+1
18 = 3x
3
3
x= 6
distribute
Combine like terms
Get variables on same side –
use inverse operation
Solve 2 step equation
Check - replace “x” with solution
2(x + 7) + 3 = 5x - 1
2(6 + 7) + 3 =5(6) - 1
2(13) + 3 =5(6) - 1
26 + 3 = 30 - 1
29 = 29

X=6
Replace X = 6
Follow order of
operations on both
sides of equation
Checks
is the solution
to the equation
Use Steps to Solve Equation:
-3x + 4 = 5x – 8
+3x
+3x
4 = 8x - 8
+8
+8
12 = 8x
8
8
x = 3/2
Get variables on same side
of equation – use inverse
operation (add 3x)
Solve 2 step equation
Use Steps to Solve Equation:
4(1 – 2x) = 4 – 6x
4 – 8x = 4 - 6x
+8x
+ 8x
4 = 4 + 2x
-4 -4
0 = 2x
2
2
x= 0
Get rid of ( ) -- distribute
Get variables on same side
– use inverse operation
(add 6x)
Solve 2 step equation
Undo by using inverse
 -2
 -2 undo 2nd
+4
-4
undo 1st
Use Steps to Solve Equation:
9 + 5x = 5x + 9
-5x -5x
9=9
Infinite
Solutions
Get variables on same side
of equation – use inverse
operation (subtract 5x)
When solving, if you get a
TRUE STATEMENT, then
that means that any real
number works.
Use Steps to Solve Equation:
6x – 1 = 6x – 8
-6x
-6x
-1 = - 8
x = no solutions
The solution is no real
numbers or empty set
Get variables on same side
of equation – use inverse
operation (subtract 6x)
The variables zeroed out
and remaining is a false
statement where a
number is equal to a
different number, so there
will be no number that will
work in the equation.
Review Steps to Solve Equations
with Variables on Both Sides
 1)
Simplify each side

Get rid of double Negatives

Distribute

Combine Like Terms
 2) Move variables to same side

“Smaller to the bigger”
 3) Solve by using INVERSE Operations
bkevil
Solve the equation.
1. 2m – 6 + 4m = 12
ANSWER
3
2. 6a – 5(a – 1) = 11
ANSWER
6
Solve the equation.
3. A charter bus company charges $11.25 per ticket
plus a handling charge of $.50 per ticket, and a $15
fee for booking the bus. If a group pays $297 to
charter a bus, how many tickets did they buy?
ANSWER
24 tickets
Solve the equation.
1.
8g – 2 + g = 16
ANSWER
2.
2
3b + 2(b – 4) = 47
ANSWER
11
3. –6 + 4(2c + 1) = –34
ANSWER
–4
4.
2 (x – 6) = 12
3
ANSWER
24
5. Joe drove 405 miles in 7 hours. He drove at a rate of 55
miles per hour during the first part of the trip and 60
miles per hour during the second part. How many hours
did he drive at a rate of 55 miles per hour?
ANSWER
3h
EXAMPLE 1
Solve an equation with variables on both sides
Solve 7 – 8x = 4x – 17.
7 – 8x = 4x – 17
7 – 8x + 8x = 4x – 17 + 8x
7 = 12x – 17
24 = 12x
2=x
Write original equation.
Add 8x to each side.
Simplify each side.
Add 17 to each side.
Divide each side by 12.
ANSWER
The solution is 2. Check by substituting 2 for x in the
original equation.
EXAMPLE 1
Solve an equation with variables on both sides
CHECK
7 – 8x = 4x – 17
?
7 – 8(2) = 4(2) – 17
?
Write original equation.
Substitute 2 for x.
–9 = 4(2) – 17
Simplify left side.
–9 = –9
Simplify right side. Solution checks.
EXAMPLE 2
Solve an equation with grouping symbols
1
Solve 9x – 5 = 4 (16x + 60).
1
9x – 5 = (16x + 60)
4
Write original equation.
9x – 5 = 4x + 15
Distributive property
5x – 5 = 15
Subtract 4x from each side.
5x = 20
x=4
Add 5 to each side.
Divide each side by 5.
GUIDED PRACTICE
for Examples 1 and 2
Solve the equation. Check your solution.
1.
24 – 3m = 5m
ANSWER
3
GUIDED PRACTICE
for Examples 1 and 2
Solve the equation. Check your solution.
2. 20 + c = 4c – 7
ANSWER
9
GUIDED PRACTICE
for Examples 1 and 2
Solve the equation. Check your solution.
3. 9 – 3k = 17k – 2k
ANSWER
–8
GUIDED PRACTICE
for Examples 1 and 2
Solve the equation. Check your solution.
4. 5z – 2 = 2(3z – 4)
ANSWER
6
GUIDED PRACTICE
for Examples 1 and 2
Solve the equation. Check your solution.
5. 3 – 4a = 5(a – 3)
ANSWER
2
GUIDED PRACTICE
for Examples 1 and 2
Solve the equation. Check your solution.
6.
2
8y – 6 = 3 (6y + 15)
ANSWER
4
EXAMPLE 3
Solve a real-world problem
CAR SALES
A car dealership sold 78 new cars and 67 used cars this
year. The number of new cars sold by the dealership has
been increasing by 6 cars each year. The number of used
cars sold by the dealership has been decreasing by 4 cars
each year. If these trends continue, in how many years will
the number of new cars sold be twice the number of used
cars sold?
EXAMPLE 3
Solve a real-world problem
SOLUTION
Let x represent the number of years from now. So, 6x
represents the increase in the number of new cars sold over
x years and –4x represents the decrease in the number of
used cars sold over x years. Write a verbal model.
78
+
6x
=2(
67
+
(– 4 x)
)
EXAMPLE 3
Solve a real-world problem
78 + 6x = 2(67 – 4x)
Write equation.
78 + 6x = 134 – 8x
Distributive property
78 + 14x = 134
14x = 56
x= 4
Add 8x to each side.
Subtract 78 from each side.
Divide each side by 14.
ANSWER
The number of new cars sold will be twice the number of
used cars sold in 4 years.
EXAMPLE 3
CHECK
Solve a real-world problem
You can use a table to check your answer.
YEAR
Used car sold
0
67
1
63
2
59
3
55
4
51
New car sold
78
84
90
96
102
GUIDED PRACTICE
7.
for Example 3
WHAT IF? In Example 3, suppose the car
dealership sold 50 new cars this year instead of 78.
In how many years will the number of new cars
sold be twice the number of used cars sold?
ANSWER
6 yr
EXAMPLE 4
Identify the number of solutions of an equation
Solve the equation, if possible.
a.
3x = 3(x + 4)
b. 2x + 10 = 2(x + 5)
SOLUTION
a.
3x = 3(x + 4)
Original equation
3x = 3x + 12
Distributive property
The equation 3x = 3x + 12 is not true because the number 3x
cannot be equal to 12 more than itself. So, the equation has
no solution. This can be demonstrated by continuing to solve
the equation.
EXAMPLE 4
Identify the number of solutions of an equation
3x – 3x = 3x + 12 – 3x
0 = 12
Subtract 3x from each side.
Simplify.
ANSWER
The statement 0 = 12 is not true, so the equation has
no solution.
EXAMPLE 1
4
b.
Identify the number of solutions of an equation
2x + 10 = 2(x + 5)
Original equation
2x + 10 = 2x + 10
Distributive property
ANSWER
Notice that the statement 2x + 10 = 2x + 10 is true for all
values of x. So, the equation is an identity, and the solution is
all real numbers.
GUIDED PRACTICE
for Example 4
Solve the equation, if possible.
8.
9z + 12 = 9(z + 3)
ANSWER
no solution
GUIDED PRACTICE
for Example 4
Solve the equation, if possible.
9. 7w + 1 = 8w + 1
ANSWER
0
GUIDED PRACTICE
for Example 4
Solve the equation, if possible.
10. 3(2a + 2) = 2(3a + 3)
ANSWER
identity