Solving Algebraic Equations

Solving Algebraic Equations

Equation:

2 + 3 = 5


Equation:

2 + 3 = 5 -3 -3 2 + 0 = 2

2 = 2


Equation:

2 + 3 = 5 +11 +112 + 14 = 16 16 = 16


5 (3) =15 ÷3 ÷3 5 (1) = 5

5 = 5



x + 13 = 20


x + 13 = 20


x + 13 = 20 -13 -13


x + 13 = 20 -13 -13x + 0 = 17

x = 17


x + 13 = 20 -13 -13x + 0 = 7

x = 7

Check:7 + 13 = 20

20 = 20


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.


3t=24


3t=24


3t=24÷3 ÷3


3t=24÷3 ÷31t = 8t = 8


3t=24÷3 ÷31t = 8t = 8

Check3(8) = 2424 = 24


r – 7 = 22 8 + y = 17


r – 7 = 22 8 + y = 17


r – 7 = 22 8 + y = 17 + 7 + 7 -8 -8


r – 7 = 22 8 + y = 17 + 7 + 7 -8 -8 r = 29 y = 9

Check29 – 7 = 22 8 + 9 = 17

22 = 22 17 = 17


9s= 27 v/4= 6


9s= 27 v/4= 6


9s= 27 v/4= 6 ÷ 9 ÷9 ×4 ×4


9s= 27 v/4= 6 ÷ 9 ÷9 ×4 ×4 s = 3 v = 24

Check:9(3) = 27 24/4 = 6 27 = 27 6 = 6


= 9 d2= 64

€

p


= 9 d2= 64

€

p


= 9 d2= 642 2 √ √

€

p


= 9 d2= 642 2 √ √

€

p


= 9 d2= 642 2 √ √

p = 92 d =

p = 81 d = 8

Check:

82 = 6464 = 64

€

p

€

64

€

81 = 99 = 9

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 = 6if you get rid of the multiplication first by dividing both sides

by 2, what happens? 4-2=32=3??!?! It just doesn’t work.


4 x 2 – 2 = 6get rid of the subtraction first by adding 2 to both sides,

4 x 2 – 2 = 6+ 2 +24 x 2 = 88 = 8That’s more like it.

Algebraic EquationsMultiplication with Subtraction

2x – 4 = 8+4+4

2x = 122 2x = 6

Algebraic EquationsMultiplication with Addition

5x + 10 = 80-10-10

5x = 705 5x = 14

Algebraic EquationsMultiplication with Subtraction

-3x – 4 = -82+4+4

-3x = -78-3 -3x = 26

Algebraic EquationsDivision with Addition

x/5 + 2 = 8-2-2

= 6

x = 30

x5

(5) (5)

Algebraic Equations

C.L.T.

-20-2010x = 6010

4x +6x + 20 = 80( )10x + 20 = 80

10x = 6

Algebraic Equations

C.L.T.

-12-125x = 805

3x + 2x + 20 - 8 = 925x + 12 = 92

5x = 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?



1) Write the formula you need:Distance = Rate x Time

D = RT



D = RT2) Substitute numbers for variables where

possible:390 = 60T


1) 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 ÷606.5 = T

It took Sarah 6.5 hours to complete her trip.


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?

2) Write the formula you need:Cost = # of Units x Price per unitC = UP

2) Substitute numbers for variables where possible:350 = 875P

3) Solve the equation using your algebra skills 350 = 875P

÷875 ÷875.4 = P

Each can costs .4 dollars – or 40 cents.


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.

Documents

Solving Algebraic Equations