Add and Subtract
Algebraic Fractions
Common Denominator is Already There
a c a+c
Simple addition + =
b b b
a c a−c
Simple − =
Subtraction b b b
1
Common Denominator is Already There
Slightly more x + 7 2 x − 5 ( x + 7) + (2 x − 5)
+ =
complicated b b b
addition
Slightly more y − 3 3 y − 7 ( y − 3) − (3 y − 7)
− =
complicated b b b
subtraction
Common Denominator is NOT There
Three steps:
1) Find the L.C.M. of the denominators
2) Create equivalent fractions
3) Rewrite the problem
2
Common Denominator is NOT There
1) Find the L.C.M. of the denominators
Factor every denominator if necessary
Write: LCM= followed by all the factors of the first
denominator.
…multiplied by every factor of the second denominator
not written yet.
Common Denominator is NOT There
2) Create equivalent fractions
Write each fraction multiplied by a blank fraction, equal to the final
fraction with the LCM in the denominator
Determine the multiplier for the denominator
Fill in the multiplier in both the numerator and denominator of the blank
fraction.
Multiply the numerators to get the equivalent fraction
3
Common Denominator is NOT There
3) Rewrite the problem
Replace every fraction with its equivalent
You now have a fraction addition and/or subtraction problem
with common denominators
And don’t forget
about
parentheses
Practice
Slightly more x + 7 2 x − 5 ( x + 7) + (2 x − 5)
+ =
complicated b b b
addition
Slightly more y − 3 3 y − 7 ( y − 3) − (3 y − 7)
− =
complicated b b b
subtraction
4
Practice Common Denominator is NOT There
1) Find the L.C.M. of the denominators
Factor every denominator if necessary
Write: LCM= followed by all the factors of the first denominator.
…multiplied by every factor of the second denominator not written yet.
2)Create equivalent fractions
Write each fraction multiplied by a blank fraction, equal to the final fraction with the
LCM in the denominator
Determine the multiplier for the denominator
Fill in the multiplier in both the numerator and denominator of the blank fraction.
Multiply the numerators to get the equivalent fraction
3)Rewrite the Problem
5