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.