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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.
29 = 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 (16x + 60).
Write original equation.
9x – 5 =
4
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
1. 24 – 3m = 5m.
24 – 3m = 5m
24 – 3m + 3m = 5m + 3m
24 = 8m
3=m
Write original equation.
Add 3m to each side.
Simplify each side.
Divide each side by 8.
ANSWER
The solution is 3. Check by substituting 3 for m in the
original equation.
GUIDED PRACTICE
for Examples 1 and 2
CHECK
24 – 3m = 5m
?
24 – 3(3) = 5(3)
?
Write original equation.
Substitute 3 for m.
15 = 5(3)
Simplify left side.
15 = 15 
Simplify right side. Solution checks.
GUIDED PRACTICE
for Examples 1 and 2
2. 20 + c = 4c – 7 .
20 + c = 4c – 7
20 + c – c = 4c – c – 7
Write original equation.
Subtract c from each side.
20 = 3c – 7
Simplify each side.
27 = 3c
Add 7 to each side.
9=c
Divide each side by 3.
ANSWER
The solution is 9. Check by substituting 9 for c in the
original equation.
GUIDED PRACTICE
for Examples 1 and 2
CHECK
20 + c = 4c – 7
?
20 + 9 = 4(9) – 7
?
Write original equation.
Substitute 9 for c.
29 = 4(9) – 7
Simplify left side.
29 = 29 
Simplify right side. Solution checks.
GUIDED PRACTICE
for Examples 1 and 2
3. 9 – 3k = 17k – 2k .
9 – 3k = 17k – 2k
9 – 3k + 3k = 17k – 2k + 3k
9 = 17k + k
–8=k
Write original equation.
Add 3k to each side.
Simplify each side.
Subtract 17 from each side.
ANSWER
The solution is – 8. Check by substituting – 8 for k in
the original equation.
GUIDED PRACTICE
for Examples 1 and 2
CHECK
9 – 3k = 17 – 2k
Write original equation.
?
9 –3(– 8) = 17 – (– 8)2 Substitute – 8 for k.
?
33 = 17 – (– 8)2 Simplify left side.
33 = 33 
Simplify right side. Solution checks.
GUIDED PRACTICE
for Examples 1 and 2
4. 5z – 2 = 2(3z – 4) .
5z – 2 = 2(3z – 4)
Write original equation.
5z – 2 = 6z – 8
Distributive property.
–z–2=–8
Subtract 6z from each side.
z=6
Add z to each side.
ANSWER
The solution is 6. Check by substituting 6 for z in the
original equation.
GUIDED PRACTICE
for Examples 1 and 2
CHECK
5z – 2 = 2(3z – 4)
Write original equation.
?
5(6) – 2 = 2(3(6) – 4) Substitute 6 for z.
?
28 = 2(3(6) – 4) Simplify left side.
28 = 28 
Simplify right side. Solution checks.
GUIDED PRACTICE
for Examples 1 and 2
5. 3 – 4a = 5(a – 3) .
3 – 4a = 5(a – 3)
Write original equation.
3 – 4a = 5a – 15
Distributive property.
3 – 9a = – 15
Subtract 5a from each side.
– 9a = – 18
a=2
Subtract 3 from each side.
Divide each side by – 9.
ANSWER
The solution is 2. Check by substituting 2 for a in the
original equation.
GUIDED PRACTICE
for Examples 1 and 2
CHECK
3 – 4a = 5(a – 3)
?
3 – 4(2) = 5(2 – 3)
?
Write original equation.
Substitute 2 for a.
– 5 = 5(2 – 3)
Simplify left side.
–5=–5
Simplify right side. Solution checks.
GUIDED PRACTICE
6.
for Examples 1 and 2
2
8y – 6 = 3 (6y + 15).
2
8y – 6 = 3 (6y + 15).
8y – 6 = 4y + 10
4y – 6 = 10
4y = 16
y=4
Write original equation.
Distributive property
Subtract 4y from each side.
Add 6 to each side.
Divide each side by 4.
ANSWER
The solution is 4. Check by substituting 4 for y in the
original equation.
GUIDED PRACTICE
for Examples 1 and 2
CHECK
2
8y – 6 = 3 (6y + 15). Write original equation.
? 2
8(4) – 6 = (6(4) + 15) Substitute 4 for y.
3
?
26 = 2 (6(4) + 15) Simplify left side.
3
26 = 26 
Simplify right side. Solution checks.
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
Solve a real-world problem
CHECK
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? 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?
GUIDED PRACTICE
for Example 3
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.
50
+
6x
=2(
67
+
(–4x)
)
GUIDED PRACTICE
for Example 3
50 + 6x = 2(67 +(– 4x))
Write equation.
50 + 6x = 134 – 8x
Distributive property
50 + 14x = 134
14x = 84
x= 6
Add 8x to each side.
Subtract 50 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 6 years.
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 4
1
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
8.
for Example 4
9z + 12 = 9(z + 3)
SOLUTION
9z + 12 = 9(z + 3)
Original equation
9z + 12 = 9z + 27
Distributive property
The equation 9z + 12 = 9z + 27 is not true because the
number 9z + 12 cannot be equal to 27 more than itself.
So, the equation has no solution. This can be
demonstrated by continuing to solve the equation.
GUIDED PRACTICE
for Example 4
9z – 9z + 12 = 9z – 9z + 27
12 = 27
Subtract 9z from each side.
Simplify.
ANSWER
The statement 12 = 27 is not true, so the equation has
no solution.
GUIDED PRACTICE
9.
for Example 4
7w + 1 = 8w + 1
SOLUTION
–w+1=1
–w=0
ANSWER
w=0
Subtract 8w from each side.
Subtract 1 from each side.
GUIDED PRACTICE
10.
for Example 4
3(2a + 2) = 2(3a + 3)
SOLUTION
3(2a + 2) = 2(3a + 3) Original equation
6a + 6 = 6a + 6
Distributive property
ANSWER
The statement 6a + 6 = 6a + 6 is true for all values of a.
So, the equation is an identity, and the solution is all
real numbers.