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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.