Solution - Adding, subtracting and finding the least common multiple
Other Ways to Solve
Adding, subtracting and finding the least common multipleStep by Step Solution
Reformatting the input :
Changes made to your input should not affect the solution:
(1): "b2" was replaced by "b^2". 5 more similar replacement(s).
Step 1 :
ab + b2
Simplify ———————
a2
Step 2 :
Pulling out like terms :
2.1 Pull out like factors :
ab + b2 = b • (a + b)
Equation at the end of step 2 :
1 ((b2)-(a2)) b•(a+b)
((—•b)+(———————————•(b3)))+(———————•b2)
a a a2
Step 3 :
Multiplying exponential expressions :
3.1 b1 multiplied by b2 = b(1 + 2) = b3
Equation at the end of step 3 :
1 ((b2)-(a2)) b3•(a+b)
((—•b)+(———————————•(b3)))+————————
a a a2
Step 4 :
b2 - a2
Simplify ———————
a
Trying to factor as a Difference of Squares :
4.1 Factoring: b2 - a2
Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)
Proof : (A+B) • (A-B) =
A2 - AB + BA - B2 =
A2 - AB + AB - B2 =
A2 - B2
Note : AB = BA is the commutative property of multiplication.
Note : - AB + AB equals zero and is therefore eliminated from the expression.
Check : b2 is the square of b1
Check : a2 is the square of a1
Factorization is : (b + a) • (b - a)
Equation at the end of step 4 :
1 (a+b)•(b-a) b3•(a+b)
((—•b)+(———————————•b3))+————————
a a a2
Step 5 :
Equation at the end of step 5 :
1 b3•(a+b)•(b-a) b3•(a+b)
((—•b)+——————————————)+————————
a a a2
Step 6 :
1
Simplify —
a
Equation at the end of step 6 :
1 b3•(a+b)•(b-a) b3•(a+b)
((—•b)+——————————————)+————————
a a a2
Step 7 :
Adding fractions which have a common denominator :
7.1 Adding fractions which have a common denominator
Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:
b + b3 • (a+b) • (b-a) -a2b3 + b5 + b
—————————————————————— = ——————————————
a a
Equation at the end of step 7 :
(-a2b3 + b5 + b) b3 • (a + b)
———————————————— + ————————————
a a2
Step 8 :
Step 9 :
Pulling out like terms :
9.1 Pull out like factors :
-a2b3 + b5 + b = -b • (a2b2 - b4 - 1)
Trying to factor a multi variable polynomial :
9.2 Factoring a2b2 - b4 - 1
Try to factor this multi-variable trinomial using trial and error
Factorization fails
Calculating the Least Common Multiple :
9.3 Find the Least Common Multiple
The left denominator is : a
The right denominator is : a2
| Algebraic Factor | Left Denominator | Right Denominator | L.C.M = Max {Left,Right} |
|---|---|---|---|
| a | 1 | 2 | 2 |
Least Common Multiple:
a2
Calculating Multipliers :
9.4 Calculate multipliers for the two fractions
Denote the Least Common Multiple by L.C.M
Denote the Left Multiplier by Left_M
Denote the Right Multiplier by Right_M
Denote the Left Deniminator by L_Deno
Denote the Right Multiplier by R_Deno
Left_M = L.C.M / L_Deno = a
Right_M = L.C.M / R_Deno = 1
Making Equivalent Fractions :
9.5 Rewrite the two fractions into equivalent fractions
L. Mult. • L. Num. -b • (a2b2-b4-1) • a —————————————————— = ———————————————————— L.C.M a2 R. Mult. • R. Num. b3 • (a+b) —————————————————— = —————————— L.C.M a2
Adding fractions that have a common denominator :
9.6 Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator
Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:
-b • (a2b2-b4-1) • a + b3 • (a+b) -a3b3 + ab5 + ab3 + ab + b4
————————————————————————————————— = ———————————————————————————
a2 a2
Step 10 :
Pulling out like terms :
10.1 Pull out like factors :
-a3b3 + ab5 + ab3 + ab + b4 =
-b • (a3b2 - ab4 - ab2 - a - b3)
Final result :
+b • (a3b2 + ab4 + ab2 + a + b3)
————————————————————————————————
a2
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