Enter a problem...
Algebra Examples
1c-3-1c=3c(c-3)1c−3−1c=3c(c−3)
Step 1
Step 1.1
Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.
c-3,c,c(c-3)c−3,c,c(c−3)
Step 1.2
Since c-3,c,c(c-3)c−3,c,c(c−3) contains both numbers and variables, there are four steps to find the LCM. Find LCM for the numeric, variable, and compound variable parts. Then, multiply them all together.
Steps to find the LCM for c-3,c,c(c-3)c−3,c,c(c−3) are:
1. Find the LCM for the numeric part 1,1,11,1,1.
2. Find the LCM for the variable part c1,c1c1,c1.
3. Find the LCM for the compound variable part c-3,c-3c−3,c−3.
4. Multiply each LCM together.
Step 1.3
The LCM is the smallest positive number that all of the numbers divide into evenly.
1. List the prime factors of each number.
2. Multiply each factor the greatest number of times it occurs in either number.
Step 1.4
The number 11 is not a prime number because it only has one positive factor, which is itself.
Not prime
Step 1.5
The LCM of 1,1,11,1,1 is the result of multiplying all prime factors the greatest number of times they occur in either number.
11
Step 1.6
The factor for c1c1 is cc itself.
c1=cc1=c
cc occurs 11 time.
Step 1.7
The LCM of c1,c1c1,c1 is the result of multiplying all prime factors the greatest number of times they occur in either term.
cc
Step 1.8
The factor for c-3c−3 is c-3c−3 itself.
(c-3)=c-3(c−3)=c−3
(c-3)(c−3) occurs 11 time.
Step 1.9
The LCM of c-3,c-3c−3,c−3 is the result of multiplying all factors the greatest number of times they occur in either term.
c-3c−3
Step 1.10
The Least Common Multiple LCMLCM of some numbers is the smallest number that the numbers are factors of.
c(c-3)c(c−3)
c(c-3)c(c−3)
Step 2
Step 2.1
Multiply each term in 1c-3-1c=3c(c-3)1c−3−1c=3c(c−3) by c(c-3)c(c−3).
1c-3(c(c-3))-1c(c(c-3))=3c(c-3)(c(c-3))1c−3(c(c−3))−1c(c(c−3))=3c(c−3)(c(c−3))
Step 2.2
Simplify the left side.
Step 2.2.1
Simplify each term.
Step 2.2.1.1
Cancel the common factor of c-3c−3.
Step 2.2.1.1.1
Factor c-3c−3 out of c(c-3)c(c−3).
1c-3((c-3)c)-1c(c(c-3))=3c(c-3)(c(c-3))1c−3((c−3)c)−1c(c(c−3))=3c(c−3)(c(c−3))
Step 2.2.1.1.2
Cancel the common factor.
1c-3((c-3)c)-1c(c(c-3))=3c(c-3)(c(c-3))1c−3((c−3)c)−1c(c(c−3))=3c(c−3)(c(c−3))
Step 2.2.1.1.3
Rewrite the expression.
c-1c(c(c-3))=3c(c-3)(c(c-3))c−1c(c(c−3))=3c(c−3)(c(c−3))
c-1c(c(c-3))=3c(c-3)(c(c-3))c−1c(c(c−3))=3c(c−3)(c(c−3))
Step 2.2.1.2
Cancel the common factor of cc.
Step 2.2.1.2.1
Move the leading negative in -1c−1c into the numerator.
c+-1c(c(c-3))=3c(c-3)(c(c-3))c+−1c(c(c−3))=3c(c−3)(c(c−3))
Step 2.2.1.2.2
Cancel the common factor.
c+-1c(c(c-3))=3c(c-3)(c(c-3))c+−1c(c(c−3))=3c(c−3)(c(c−3))
Step 2.2.1.2.3
Rewrite the expression.
c-(c-3)=3c(c-3)(c(c-3))c−(c−3)=3c(c−3)(c(c−3))
c-(c-3)=3c(c-3)(c(c-3))c−(c−3)=3c(c−3)(c(c−3))
Step 2.2.1.3
Apply the distributive property.
c-c--3=3c(c-3)(c(c-3))c−c−−3=3c(c−3)(c(c−3))
Step 2.2.1.4
Multiply -1−1 by -3−3.
c-c+3=3c(c-3)(c(c-3))c−c+3=3c(c−3)(c(c−3))
c-c+3=3c(c-3)(c(c-3))c−c+3=3c(c−3)(c(c−3))
Step 2.2.2
Simplify by adding terms.
Step 2.2.2.1
Subtract cc from cc.
0+3=3c(c-3)(c(c-3))0+3=3c(c−3)(c(c−3))
Step 2.2.2.2
Add 00 and 33.
3=3c(c-3)(c(c-3))3=3c(c−3)(c(c−3))
3=3c(c-3)(c(c-3))3=3c(c−3)(c(c−3))
3=3c(c-3)(c(c-3))3=3c(c−3)(c(c−3))
Step 2.3
Simplify the right side.
Step 2.3.1
Cancel the common factor of c(c-3)c(c−3).
Step 2.3.1.1
Cancel the common factor.
3=3c(c-3)(c(c-3))3=3c(c−3)(c(c−3))
Step 2.3.1.2
Rewrite the expression.
3=33=3
3=33=3
3=33=3
3=33=3
Step 3
Since 3=33=3, the equation will always be true for any value of cc.
All real numbers
Step 4
The result can be shown in multiple forms.
All real numbers
Interval Notation:
(-∞,∞)(−∞,∞)