Algebra Honors (Period 6 & 7)

Fractional Equations 7.4

The total resistance R of an electrical circuit with two resistors R₁ and R₂, that are connected in parallel is given by the formula
1/ R₁ +1/ R₂ = 1/R
What do you notice about the difference between this equation and those with fractional coefficients?

This formula is an example of a fractional equation.
An equation with a variable in the denominator of one or more terms is called a fractional equation. To solve a fractional equation, you can multiply BOTH sides of the equation by the LCD or you could use the method of solving a proportion when the equation consists of one fraction equal to another fraction.
3/x -1/4 = 1/12
The LCD of the fractions is 12x
Multiply BOTH sides of the equation by the LCD, 12x
Notice that x ≠ 0 because in this case 3/0 is undefined
12x(3/x – ¼) = (1/12)(12x)
36 -3x = x
36 = 4x
x = 9
{9}

(2-x)/(3-x) = 4/9
There are two different ways to solve this
First by finding the LCD, which is 9(3-x) Notice that x ≠3 Why?
9(3-x)[(2-x)/(3-x) = (4/9)[9(3-x)]
18-9x =12-4x
6 = 5x
6/5 = x
{6/5}
OR solve as proportion
(2-x)/(3-x) = 4/9

(2-x)(9) = (4)(3-x)
18- 9x = 12- 4x
we are at the same spot as with the first method and we arrive at the same solution
6 = 5x
6/5 = x
{6/5}

The following gets a little more complicated to do and to display here...
Solve
(2/b² - b) – 2/(b-1) = 1
Find the LCD by first factoring the denominators first
b² - b = b(b-1) so the LCD of the two fractions in this equation is in fact
b² - b BUT use it in factored form b(b-1) Notice: b ≠ 1 why?
(2/b² - b) – 2/(b-1) = 1
[b(b-1][ (2/b² - b) – 2/(b-1) ]= 1[b(b-1)]
which separates to
[b(b-1) (2/b² - b)] – [b(b-1)2/(b-1)] = 1
[b(b-1) (2/b(b - 1)] – [b(b-1)2/(b-1)] = 1
2 -2b=b(b-1)
or
2 – 2b = b² - b
solve for b now
0 = b² - b + 2b -2
0 = b² + b – 2
0 = (b-1)(b+2)
b = 1 and b = -2
Remember in this case b ≠ 1 because of the ORIGINAL EQUATION
the solution set is only
{-2)

Multiplying both sides of an equation by a variable expression sometimes results in an equation that has an extra root. You must check each root of the transformed equation to see if it satisfies the original equation.

Algebra Honors (Period 6 & 7)

Equations with Fractional Coefficients 7-3
Solving an equation with fractional coefficients can be easily accomplished by using the LCD of all the fractions in the equation. Clearing the equation of all fractions BEFORE attempting to solve the equation is probably the best way

easy examples:
x/3 + x/7 = 10
The LCD of the fractions is 21
so multiply BOTH SIDES by 21
21(x/3 + x/7) = 10 (21)
7x + 3x = 210
10x = 210
x = 21
{21}
3a/5 – a/2 = 1/20
The LCD of the fractions is 20
20(3a/5 – a/2) = 1/20(20)
4(3a) -10a = 1
12a- 10a = 1
2a = 1
a = ½
{1/2}

x/3- (x+2)/5 = 2
The LCD of the fractions is 15
15[x/3- (x+2)/5] = 2(15)
5x –(3)(x+2) = 30
5x -3x-6 = 30
2x = 36
x = 18

2n + n/3 = n/4 + 5
The LCD is 12
12(2n + n/3) = (n/4 +5)(12)
24n + 4n = 3n + 60
28n = 3n + 60
25n = 60
n = 60/25 = 12/5
{12/5}


More complicated:
(1/4)(n + 2) – (1/6)(n – 2) = 3/2
The LCD is 12
12[(1/4)(n + 2) – (1/6)(n – 2)] = (3/2)(12)
3(n+2) – 2(n-2) = 18
3n + 6 -2n + 4 = 18
n = 8
{8}

Solving some word problems:
one eighth of a number is ten less than one third of the number. Find the number.
Let x = the number.
n/8 = n/3 – 10
LCD is 24
(24)(n/8)= (n/3-10)24
3n = 8n – 240
-5n = -240
5n = 240
n = 48
{48}

Math 6 Honors ( Periods 1, 2, & 3)

Solving Equations 11-7

Now that we have learned about negative integers, we can solve an equation such as
x + 7 = 2
We need to subtract 7 from both sides of the equation
x + 7 = 2
- 7 = - 7

to do this use a side bar and use the rules for adding integers
Notice the signs are different so
ask yourself... Who wins? and By How Much?
stack the winner on top and take the difference
so
x + 7 = 2
- 7 = - 7
x = -5



t - -10 = 19
becomes -- with add the opposite---
t+ + 10 = 19
which is just
t + 10 = 19
so subtract 10 from both sides
t + 10 = 19
- 10 = - 10
t = 9

w - - 26 = -44
"Add the Opposite"
w + + 26 = -44
- 26 = - 26
x = -70

Know your integer rules and it becomes easy!!
Side bars are great, if you need them with difference signs!!


y -^- 6 = 4
add the opposite and you get
y + 6 = 4
now you need to subtract 6 from both sides of the equation
y + 6 = 4
- 6 = - 6
Again the signs are different -- ask your self those all important questions
"Who Wins? and "By How Much?"
Use a side bar, stack the winner on top and take the difference. Make sure to use the winner's sign in your answer!!
y = 2

What about -5u = 125?
Whats happening to u?
It is being multiplied by -5... so you must divide by -5

-5u = 125
-5 -5
u = -25

or written easier to read -5u/-5 = 125/-5
u = -25

(1/-9)c = 33
Need to multiply both sides by the reciprocal of (1/-9) which is (-9/1)

(-9/1)(1/-9)c = 33(-9/1)
c = -297



2- STEP EQUATIONS

What about

3u - 1 = -7
You need to do the reverse of PEMDAS... remember unwrapping the present? We did the exact opposite of what we had done to wrap the present!!
so
3u - 1 = -7
+ 1 = + 1
3u = -6
Now divide by 3 on both sides
3u/3 = -6/3
u = -2


3z - ^- 15 = 9
add the opposite first and you get
3x + 15 = 9
In order to solve this 2 step equation
we need to do the reverse of PEMDAS-- as we did with unwrapping the present so many months ago
3x + 15 = 9
subtract 15 from both sides of the equation
3x + 15 = -9
- 15 = - 15

This time the sides are the same-- so just add them and use their sign
3x + 15 = -9
- 15 = - 15
3x = -24
Now divide both sides by 3
3x = -24
3 3

x = -8

Make sure to BOX your answer!!
What about this one
(1/2)(x) + 3 = 0
subtract 3 from both sides
(1/2)x = -3

Multiple by the reciprocal of 1/2 which is 2/1
(2/1)(1/2)x = -3(2/1)
x = -6
Again box your answer.



What about x = -6 + 3x
OH dear... we have variables on BOTH sides of the equations... we need to get the variables on one side all the constants on the other.
We need to isolate the variable!!

x = -6 + 3x
What if we add six to both sides
x = -6 + 3x
+6 = + 6
x + 6 = 3x
now we need to subtract x from both sides
x + 6 = 3x
- x - x
6 = 2x
so now divide both sides by 2
6/2 = 2x/2
3 = x

How about this one
3 - r = -5 + r
- 3 = - 3
-r = -8 + r
if subtract r from both sides, I will get rid of the +r on the right side

-r = -8 + r
- r = -r
-2r = -8
Now divide by -2 on both sides

-2r/-2 = -8/-2
r = 4