Download Chapter 2.5

Chapter 2.5
Solve Equations with
Variables on Both Sides
Key Concept
• Some equations have variables on both sides.
To solve such equations, you can collect the
variable terms on one side of the equation
and the constant terms on the other side of
the equation.
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
Write original equation.
?
7 – 8(2) = 4(2) – 17
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
Solve
1
(16x + 60).
4
9x – 5 =
9x – 5 =
1
(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
On Your Own
Solve the equation. Check your solution.
1. 24 – 3m = 5m
ANSWER
3
GUIDED PRACTICE
On Your Own
Solve the equation. Check your solution.
2. 20 + c = 4c – 7
ANSWER
9
GUIDED PRACTICE
On Your Own
Solve the equation. Check your solution.
3. 9 – 3k = 17k – 2k
ANSWER
–8
GUIDED PRACTICE
On Your Own
Solve the equation. Check your solution.
4. 5z – 2 = 2(3z – 4)
ANSWER
6
GUIDED PRACTICE
On Your Own
Solve the equation. Check your solution.
5. 3 – 4a = 5(a – 3)
ANSWER
2
GUIDED PRACTICE
On Your Own
Solve the equation. Check your solution.
6.
8y – 6 =
ANSWER
2
(6y + 15)
3
4
NUMBER OF SOLUTIONS
• Equations do not always have one solution. An
equation that is true for all values of the
variable is an identity . So, the solution of an
identity is all real numbers. Some equations
have no solution.
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)
3x = 3x + 12
Original equation
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.
Identify
EXAMPLE
4
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.
1
EXAMPLE
4
Identify
b.
the number of solutions of an equation
2x + 10 = 2(x + 5)
2x + 10 = 2x + 10
Original equation
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
On Your Own
Solve the equation, if possible.
8. 9z + 12 = 9(z + 3)
ANSWER
no solution
GUIDED PRACTICE
On Your Own
Solve the equation, if possible.
9. 7w + 1 = 8w + 1
ANSWER
0
GUIDED PRACTICE
On Your Own
Solve the equation, if possible.
10. 3(2a + 2) = 2(3a + 3)
ANSWER
identity