Download + 2 +2

Solving Algebraic Equations
Equation:
2+3=5
Solving Algebraic Equations
Equation:
2+3=5
-3 -3
2+0=2
2=2
Solving Algebraic Equations
Equation:
2+3=5
+11 +11
2 + 14 = 16
16 = 16
Solving Algebraic Equations
5 (3) =15
÷3 ÷3
5 (1) = 5
5=5
Solving Algebraic Equations
Solving Algebraic Equations
x + 13 = 20
Solving Algebraic Equations
x + 13 = 20
Solving Algebraic Equations
x + 13 = 20
-13 -13
Solving Algebraic Equations
x + 13 = 20
-13 -13
x + 0 = 17
x = 17
Solving Algebraic Equations
x + 13 = 20
-13 -13
x+0=7
x=7
Check:
7 + 13 = 20
20 = 20
Solving Algebraic Equations
Opposites:
+
x
÷
√
x2
*When you see a number right next to a variable, with no operation, it
means multiply. If you see a number right next to a parentheses (), it
means multiply.
4x is the same as 4xx, but that would be confusing to read.
Solving Algebraic Equations
3t=24
Solving Algebraic Equations
3t=24
Solving Algebraic Equations
3t=24
÷3 ÷3
Solving Algebraic Equations
3t=24
÷3 ÷3
1t = 8
t=8
Solving Algebraic Equations
3t=24
÷3 ÷3
1t = 8
t=8
Check
3(8) = 24
24 = 24
Solving Algebraic Equations
r – 7 = 22
8 + y = 17
Solving Algebraic Equations
r – 7 = 22
8 + y = 17
Solving Algebraic Equations
r – 7 = 22
+7 +7
8 + y = 17
-8
-8
Solving Algebraic Equations
r – 7 = 22
+7 +7
r = 29
Check
29 – 7 = 22
22 = 22
8 + y = 17
-8
-8
y=9
8 + 9 = 17
17 = 17
Solving Algebraic Equations
9s= 27
v/4= 6
Solving Algebraic Equations
9s= 27
v/4= 6
Solving Algebraic Equations
9s= 27
÷ 9 ÷9
×4
v/4= 6
×4
Solving Algebraic Equations
9s= 27
÷ 9 ÷9
v/4= 6
×4
×4
s=3
Check:
9(3) = 27
27 = 27
v = 24
24/4 = 6
6=6
Solving Algebraic Equations
p =9

d2= 64
Solving Algebraic Equations
p =9

d2= 64
Solving Algebraic Equations
d2= 64
p =9
2

2
√
√
Solving Algebraic Equations
d2= 64
p =9
2

2
√
√
Solving Algebraic Equations
d2= 64
p =9
2

Check:
81  9
99
√
2
p
= 92
d=
p
= 81
d=8
√
64

82 = 64
64 = 64

When you have an equation that has multiplication or
division and addition or subtraction, you have to move the
addition or subtraction first.
say you have 4 x 2 - 2 = 6
if you get rid of the multiplication first by dividing both sides
by 2, what happens?
4-2=3
2=3??!?!
It just doesn’t work.
Solving Algebraic Equations
4x2–2=6
get rid of the subtraction first by adding 2 to both
sides,
4x2–2=6
+ 2 +2
4x2 =8
8=8
That’s more like it.
Algebraic Equations
Multiplication with Subtraction
2x – 4 = 8
+4 +4
2x = 12
2
2
x = 6
Algebraic Equations
Multiplication with Addition
5x + 10 = 80
-10 -10
5x = 70
5
5
x = 14
Algebraic Equations
Multiplication with Subtraction
-3x – 4 = -82
+4 +4
-3x = -78
-3
-3
x = 26
Algebraic Equations
Division with Addition
x/5 + 2 = 8
-2 -2
x
(5)
5
= 6 (5)
x = 30
Algebraic Equations
C.L.T.
(4x +6x )+ 20 = 80
10x + 20 = 80
-20 -20
10x = 60
10
10
x=6
Algebraic Equations
C.L.T.
(3x + 2x) +(20 - 8)= 92
5x + 12 = 92
-12 -12
5x = 80
5
5
x = 16
Common Formulas
SIMPLE INTEREST:
Interest = Principle x Rate x Time
DISTANCE:
Distance = Rate x Time
TOTAL COST:
Total Cost = (number of units)x (price per unit)
Substituting in Formulas
If Sarah drove 390 miles at an average speed
of 60 miles per hour, how long did it take
her to complete her trip?
Substituting in Formulas
If Sarah drove 390 miles at an average speed
of 60 miles per hour, how long did it take
her to complete her trip?
1) Write the formula you need:
Distance = Rate x Time
D = RT
Substituting in Formulas
If Sarah drove 390 miles at an average speed
of 60 miles per hour, how long did it take
her to complete her trip?
D = RT
2) Substitute numbers for variables where
possible:
390 = 60T
Substituting in Formulas
Write the formula you need:
D = RT
2) Substitute numbers for variables where possible:
390 = 60T
3) Solve the equation using your algebra skills
390 = 60T
÷60 ÷60
6.5 = T
1)
It took Sarah 6.5 hours to complete her trip.
Substituting in Formulas
1)
If the total cost of a shipment of pet food cans was $350 and there
were 875 cans, how much did each can cost?
Write the formula you need:
Cost = # of Units x Price per unit
C = UP
2)
Substitute numbers for variables where possible:
350 = 875P
3) Solve the equation using your algebra skills
350 = 875P
÷875 ÷875
.4 = P
2)
Each can costs .4 dollars – or 40 cents.
Solving Algebraic Equations

Try the practice problems on pages 144 and
145 and check your answers online.

Try pages 24 and 25 in the GED Math
Practice booklet and enter your answers on
Blackboard.