Adding, subtracting and finding the least common multiple
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This solution deals with adding, subtracting and finding the least common multiple.
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Step by Step Solution
Rearrange:
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
2/3*t-(t-1/t)=0
Step by step solution :
Step 1 :
1
Simplify —
t
Equation at the end of step 1 :
2 1
(— • t) - (t - —) = 0
3 t
Step 2 :
Rewriting the whole as an Equivalent Fraction :
2.1 Subtracting a fraction from a whole
Rewrite the whole as a fraction using t as the denominator :
t t • t
t = — = —————
1 t
Equivalent fraction : The fraction thus generated looks different but has the same value as the whole
Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator
Adding fractions that have a common denominator :
2.2 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:
t • t - (1) t2 - 1
——————————— = ——————
t t
Equation at the end of step 2 :
2 (t2 - 1)
(— • t) - ———————— = 0
3 t
Step 3 :
2
Simplify —
3
Equation at the end of step 3 :
2 (t2 - 1)
(— • t) - ———————— = 0
3 t
Step 4 :
Trying to factor as a Difference of Squares :
4.1 Factoring: t2-1
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 : 1 is the square of 1
Check : t2 is the square of t1
Factorization is : (t + 1) • (t - 1)
Calculating the Least Common Multiple :
4.2 Find the Least Common Multiple
The left denominator is : 3
The right denominator is : t
Prime Factor | Left Denominator | Right Denominator | L.C.M = Max {Left,Right} |
---|---|---|---|
3 | 1 | 0 | 1 |
Product of all Prime Factors | 3 | 1 | 3 |
Algebraic Factor | Left Denominator | Right Denominator | L.C.M = Max {Left,Right} |
---|---|---|---|
t | 0 | 1 | 1 |
Least Common Multiple:
3t
Calculating Multipliers :
4.3 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 = t
Right_M = L.C.M / R_Deno = 3
Making Equivalent Fractions :
4.4 Rewrite the two fractions into equivalent fractions
Two fractions are called equivalent if they have the same numeric value.
For example : 1/2 and 2/4 are equivalent, y/(y+1)2 and (y2+y)/(y+1)3 are equivalent as well.
To calculate equivalent fraction , multiply the Numerator of each fraction, by its respective Multiplier.
L. Mult. • L. Num. 2t • t —————————————————— = —————— L.C.M 3t R. Mult. • R. Num. (t+1) • (t-1) • 3 —————————————————— = ————————————————— L.C.M 3t
Adding fractions that have a common denominator :
4.5 Adding up the two equivalent fractions
2t • t - ((t+1) • (t-1) • 3) 3 - t2
———————————————————————————— = ——————
3t 3t
Trying to factor as a Difference of Squares :
4.6 Factoring: 3 - t2
Check : 3 is not a square !!
Ruling : Binomial can not be factored as the
difference of two perfect squares
Equation at the end of step 4 :
3 - t2
—————— = 0
3t
Step 5 :
When a fraction equals zero :
5.1 When a fraction equals zero ...
Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.
Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.
Here's how:
3-t2
———— • 3t = 0 • 3t
3t
Now, on the left hand side, the 3t cancels out the denominator, while, on the right hand side, zero times anything is still zero.
The equation now takes the shape :
3-t2 = 0
Solving a Single Variable Equation :
5.2 Solve : -t2+3 = 0
Subtract 3 from both sides of the equation :
-t2 = -3
Multiply both sides of the equation by (-1) : t2 = 3
When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:
t = ± √ 3
The equation has two real solutions
These solutions are t = ± √3 = ± 1.7321