Transcript Solving Basic Equations

Section 2.1 Solving Linear Equations
Strategy for Solving Algebraic Equations:
1. Use the distributive property to remove parentheses:
becomes 3x – 9 + 3 = 18 – 5x
2. Combine like terms on either side of the equation.
-9 and 3 can be added to get -6.
3x – 6 = 18 - 5x
3. Use the addition or subtraction properties of equality to get
the variables on one side of the = symbol and the constant
terms on the other.
3x and 5x are like terms. Add 5x to each side to get the
variable terms on the left.
3x + 5x – 6 = 18 -5x + 5x
8x - 6 = 18
4. Continue to combine like terms whenever possible.
6 and 21 are like terms. Since 6 is subtracted from 8x, add 6
to both sides to move it to the other side.
8x - 6 + 6 = 18 + 6
8x = 24
5. Undo the operations of multiplication and division to isolate
the variable.
Divide both sides by 8 to get x by itself.
8x/8 = 24/8
x=3
6. Check the results by substituting your found value for x into
the original equation.
3(x - 3) + 3 = 18 – 5x
3(3-3) + 3 = 18 – 5(3)
3(0) + 3 = 18 – 15
3= 3
Now it’s your turn!
Try:
5 + 3(a+4) = 7a – (9-10a) + 4
Step 1: Use distributive property to remove parentheses
Step 2: Combine like terms on each side.
Step 3: Use addition property of equality to combine like terms
between sides.
Step 4: Continue to combine like terms wherever possible.
Step 5: Once variable terms are are combined and isolated, use
multiplication property of equality (multiply both sides by the
reciprocal of the coefficient) to completely isolate the variable.
State your conclusion: a = ____
Step 6: Check your solution in the original equation.
The Trick with Fractions!
Fractions are messy to deal with. When solving equations with
fractions in them we can take advantage of the multiplication property
of equality to get rid of them while keeping an equivalent equation.
What we do is multiply BOTH SIDES of the equation (that is
everything on each side) by the LCD of all the fractions.
2 1
1
 x
3 4
3
T hedenominators are 3 and 4. T heLCD is 12
12 2  12 1 x   1 12
3
4
3
8  3 x  4
Now we can combinelike terms.
8 and - 4 are like terms.
Combine themby subtracting 8 from both sides.
8  8  3 x  4  8
3 x  12
Divide both sides by 3.
3 x  12

3
3
x  4
CHECK!
2 1
1
  4   
3 4
3
2
1
  1  
Change - 1 so thatit has a commondenominator
3
3
2  3
1


3  3 
3
2  3
1

3
3
1
1
 
Yes!
3
3
The Trick with Decimals!
Decimals are also are messy to deal with. When solving equations
with fractions in them we can take advantage of the multiplication
property of equality to get rid of them while keeping an equivalent
equation.
What we do is multiply BOTH SIDES of the equation (that is
everything on each side) by the power of 10 that corresponds to the
number with the most decimal places.
.05a  .2(a  3)  0.1
The number with the most decimal places is .05 (2 places). This
corresponds to 102, or 100. If we multiply both sides by 100, all the
decimal numbers get changed to whole numbers.
100.05a  .2(a  3)  0.1100
5a  20(a  3)  10 Now solve using your regular steps
5a  20a  60  10
25a 60  10
25a 60  60  10  60
25a  50
25a  50

25
25
a  2
Check :
Solve: 0.08k – 0.2(k + 5) = -.1
Not all equations have a solution. Sometimes it is impossible
to find a value for a variable that would make the equation true.
Example: Solve y = y + 2
Is there any number that you can add 2 to it and still get the
same number? No.
What happens when we try to solve it?
Combine like terms. There’s a variable on each side, so to
eliminate the y on the right, you’d have to subtract y from both
sides.
y = y+2
-y -y
0 =0+2
0 = 2 ?? You see, if you try to solve an equation that is
“unsolvable” you will get a false statement. This means that no
matter what value you have for y, the equation will always be
false.
Some equations have an infinite number of solutions.
Example:
3(x – 1) + 1 = 4x – (x + 2)
Step 1: Use distributive property
3x – 3 + 1 = 4x – x – 2
Step 2: Combine like terms on each side
3x – 2 = 3x – 2
Step 3: Combine like terms between sides using addition property of
equality.
3x – 2 = 3x – 2
+2
+2
3x = 3x
-3x -3x
0 = 0 !!
This statement is true no matter what value of x you choose.
Therefore the solution set is {all real numbers}
Section 2.2 Formulas and Functions
A formula or literal equation is an equation involving two or more variables.
The variable that is isolated is called a “function” of the other variables. That is,
it depends on other variables for its value.
Examples:
A  LW
C
5
F  32
9
Example 1 p. 78
Solve C 
5
F  32 
9
for F.
You want to isolate F. We can multiply both sides of the equaiton by 9/5 so we
don’t have fractions when using the distributive property to remove the ( ).
9
9 5
C   F  32
5
5 9
9
C  F  32
5
Now add 32 to both sides
9
C  32  F
5
If the Celsius temperature is 35 degrees, what is the temperature in Fahrenheit?
Example 2 p. 79
Solve 3a – 2b = 6 for a
Isolate a by getting everything that doesn’t have an a on the
other side.
3a  2b  6
3
2
6
a  b
3
3
3
2
a  b2
3
Example 3 p. 80
Solve for P :
A  P  Pr t
Factorout theP with thedistributive property.
A  P(1 rt)
Now divide both sides by (1 rt)
A
P
(1 rt)
Example4 p.81
Solve for a :
3a  7  5ab  b
Get everythingwith an a on one side
and everythingelse on t heotherside.
3a  5ab  b-7
Now factorout t hea with thedistributive property.
a( 3  5b)  b-7
Now divide both sides by ( 3  5b)
b-7
a
3  5b