Adding, Subtracting And Finding The Least Common Multiple

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Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation : (20/c^2-2*c)+5-(10/c-2)=0

Step by step solution :

Step 1 :

10 Simplify —— c

Equation at the end of step 1 :

20 10 ((————-2c)+5)-(——-2) = 0 (c2) c

Step 2 :

Rewriting the whole as an Equivalent Fraction :

2.1 Subtracting a whole from a fraction Rewrite the whole as a fraction using c as the denominator :

2 2 • c 2 = — = ————— 1 c

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 denominatorCombine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

10 - (2 • c) 10 - 2c ———————————— = ——————— c c

Equation at the end of step 2 :

20 (10 - 2c) ((———— - 2c) + 5) - ————————— = 0 (c2) c

Step 3 :

20 Simplify —— c2

Equation at the end of step 3 :

20 (10 - 2c) ((—— - 2c) + 5) - ————————— = 0 c2 c

Step 4 :

Rewriting the whole as an Equivalent Fraction :

4.1 Subtracting a whole from a fraction Rewrite the whole as a fraction using c2 as the denominator :

2c 2c • c2 2c = —— = ——————— 1 c2

Adding fractions that have a common denominator :

4.2 Adding up the two equivalent fractions

20 - (2c • c2) 20 - 2c3 —————————————— = ———————— c2 c2

Equation at the end of step 4 :

(20 - 2c3) (10 - 2c) (—————————— + 5) - ————————— = 0 c2 c

Step 5 :

Rewriting the whole as an Equivalent Fraction :

5.1 Adding a whole to a fraction Rewrite the whole as a fraction using c2 as the denominator :

5 5 • c2 5 = — = —————— 1 c2

Step 6 :

Pulling out like terms :

6.1 Pull out like factors : 20 - 2c3 = -2 • (c3 - 10)

Trying to factor as a Difference of Cubes:

6.2 Factoring: c3 - 10 Theory : A difference of two perfect cubes, a3 - b3 can be factored into (a-b) • (a2 +ab +b2)Proof : (a-b)•(a2+ab+b2) = a3+a2b+ab2-ba2-b2a-b3 = a3+(a2b-ba2)+(ab2-b2a)-b3 = a3+0+0-b3 = a3-b3Check : 10 is not a cube !! Ruling : Binomial can not be factored as the difference of two perfect cubes

Polynomial Roots Calculator :

6.3 Find roots (zeroes) of : F(c) = c3 - 10Polynomial Roots Calculator is a set of methods aimed at finding values of c for which F(c)=0 Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers c which can be expressed as the quotient of two integersThe Rational Root Theorem states that if a polynomial zeroes for a rational number P/Q then P is a factor of the Trailing Constant and Q is a factor of the Leading CoefficientIn this case, the Leading Coefficient is 1 and the Trailing Constant is -10. The factor(s) are: of the Leading Coefficient : 1 of the Trailing Constant : 1 ,2 ,5 ,10 Let us test ....

PQP/QF(P/Q)Divisor
-1 1 -1.00 -11.00
-2 1 -2.00 -18.00
-5 1 -5.00 -135.00
-10 1 -10.00 -1010.00
1 1 1.00 -9.00
2 1 2.00 -2.00
5 1 5.00 115.00
10 1 10.00 990.00

Polynomial Roots Calculator found no rational roots

Adding fractions that have a common denominator :

6.4 Adding up the two equivalent fractions

-2 • (c3-10) + 5 • c2 -2c3 + 5c2 + 20 ————————————————————— = ——————————————— c2 c2

Equation at the end of step 6 :

(-2c3 + 5c2 + 20) (10 - 2c) ————————————————— - ————————— = 0 c2 c

Step 7 :

Polynomial Roots Calculator :

7.1 Find roots (zeroes) of : F(c) = -2c3+5c2+20 See theory in step 6.3 In this case, the Leading Coefficient is -2 and the Trailing Constant is 20. The factor(s) are: of the Leading Coefficient : 1,2 of the Trailing Constant : 1 ,2 ,4 ,5 ,10 ,20 Let us test ....

PQP/QF(P/Q)Divisor
-1 1 -1.00 27.00
-1 2 -0.50 21.50
-2 1 -2.00 56.00
-4 1 -4.00 228.00
-5 1 -5.00 395.00
-5 2 -2.50 82.50
-10 1 -10.00 2520.00
-20 1 -20.00 18020.00
1 1 1.00 23.00
1 2 0.50 21.00
2 1 2.00 24.00
4 1 4.00 -28.00
5 1 5.00 -105.00
5 2 2.50 20.00
10 1 10.00 -1480.00
20 1 20.00 -13980.00

Polynomial Roots Calculator found no rational roots

Step 8 :

Pulling out like terms :

8.1 Pull out like factors : 10 - 2c = -2 • (c - 5)

Calculating the Least Common Multiple :

8.2 Find the Least Common Multiple The left denominator is : c2 The right denominator is : c

Number of times each Algebraic Factor appears in the factorization of:
Algebraic Factor Left Denominator Right Denominator L.C.M = Max {Left,Right}
c 212

Least Common Multiple: c2

Calculating Multipliers :

8.3 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 = 1 Right_M = L.C.M / R_Deno = c

Making Equivalent Fractions :

8.4 Rewrite the two fractions into equivalent fractionsTwo 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. (-2c3+5c2+20) —————————————————— = ————————————— L.C.M c2 R. Mult. • R. Num. -2 • (c-5) • c —————————————————— = —————————————— L.C.M c2

Adding fractions that have a common denominator :

8.5 Adding up the two equivalent fractions

(-2c3+5c2+20) - (-2 • (c-5) • c) -2c3 + 7c2 - 10c + 20 ———————————————————————————————— = ————————————————————— c2 c2

Checking for a perfect cube :

8.6 -2c3 + 7c2 - 10c + 20 is not a perfect cube

Trying to factor by pulling out :

8.7 Factoring: -2c3 + 7c2 - 10c + 20 Thoughtfully split the expression at hand into groups, each group having two terms :Group 1: -10c + 20 Group 2: -2c3 + 7c2 Pull out from each group separately :Group 1: (c - 2) • (-10)Group 2: (2c - 7) • (-c2)Bad news !! Factoring by pulling out fails : The groups have no common factor and can not be added up to form a multiplication.

Polynomial Roots Calculator :

8.8 Find roots (zeroes) of : F(c) = -2c3 + 7c2 - 10c + 20 See theory in step 6.3 In this case, the Leading Coefficient is -2 and the Trailing Constant is 20. The factor(s) are: of the Leading Coefficient : 1,2 of the Trailing Constant : 1 ,2 ,4 ,5 ,10 ,20 Let us test ....

PQP/QF(P/Q)Divisor
-1 1 -1.00 39.00
-1 2 -0.50 27.00
-2 1 -2.00 84.00
-4 1 -4.00 300.00
-5 1 -5.00 495.00
-5 2 -2.50 120.00
-10 1 -10.00 2820.00
-20 1 -20.00 19020.00
1 1 1.00 15.00
1 2 0.50 16.50
2 1 2.00 12.00
4 1 4.00 -36.00
5 1 5.00 -105.00
5 2 2.50 7.50
10 1 10.00 -1380.00
20 1 20.00 -13380.00

Polynomial Roots Calculator found no rational roots

Equation at the end of step 8 :

-2c3 + 7c2 - 10c + 20 ————————————————————— = 0 c2

Step 9 :

When a fraction equals zero :

9.1 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:

-2c3+7c2-10c+20 ——————————————— • c2 = 0 • c2 c2

Now, on the left hand side, the c2 cancels out the denominator, while, on the right hand side, zero times anything is still zero.The equation now takes the shape : -2c3+7c2-10c+20 = 0

Cubic Equations :

9.2 Solve -2c3+7c2-10c+20 = 0Future releases of Tiger-Algebra will solve equations of the third degree directly.Meanwhile we will use the Bisection method to approximate one real solution.

Approximating a root using the Bisection Method :

We now use the Bisection Method to approximate one of the solutions. The Bisection Method is an iterative procedure to approximate a root (Root is another name for a solution of an equation).The function is F(c) = -2c3 + 7c2 - 10c + 20At c= 3.00 F(c) is equal to -1.00 At c= 2.00 F(c) is equal to 12.00 Intuitively we feel, and justly so, that since F(c) is negative on one side of the interval, and positive on the other side then, somewhere inside this interval, F(c) is zero Procedure :(1) Find a point "Left" where F(Left) < 0(2) Find a point 'Right' where F(Right) > 0(3) Compute 'Middle' the middle point of the interval [Left,Right](4) Calculate Value = F(Middle)(5) If Value is close enough to zero goto Step (7) Else : If Value < 0 then : Left <- Middle If Value > 0 then : Right <- Middle(6) Loop back to Step (3)(7) Done!! The approximation found is MiddleFollow Middle movements to understand how it works :

Left Value(Left) Right Value(Right) 3.000000000 -1.000000000 2.000000000 12.000000000 3.000000000 -1.000000000 0.000000000 20.000000000 3.000000000 -1.000000000 1.500000000 14.000000000 3.000000000 -1.000000000 2.250000000 10.156250000 3.000000000 -1.000000000 2.625000000 5.808593750 3.000000000 -1.000000000 2.812500000 2.751464844 3.000000000 -1.000000000 2.906250000 0.967468262 3.000000000 -1.000000000 2.953125000 0.007286072 2.976562500 -0.490391731 2.953125000 0.007286072 2.964843750 -0.240071177 2.953125000 0.007286072 2.958984375 -0.116023347 2.953125000 0.007286072 2.956054688 -0.054276487 2.953125000 0.007286072 2.954589844 -0.023472189 2.953125000 0.007286072 2.953857422 -0.008087306 2.953125000 0.007286072 2.953491211 -0.000399179 2.953125000 0.007286072 2.953491211 -0.000399179 2.953308105 0.003443806 2.953491211 -0.000399179 2.953399658 0.001522403 2.953491211 -0.000399179 2.953445435 0.000561634 2.953491211 -0.000399179 2.953468323 0.000081233 2.953479767 -0.000158972 2.953468323 0.000081233 2.953474045 -0.000038869 2.953468323 0.000081233 2.953474045 -0.000038869 2.953471184 0.000021182 2.953472614 -0.000008843 2.953471184 0.000021182 2.953472614 -0.000008843 2.953471899 0.000006169 2.953472257 -0.000001337 2.953471899 0.000006169

Next Middle will get us close enough to zero: F( 2.953472167 ) is 0.000000540 The desired approximation of the solution is: c ≓ 2.953472167 Note, ≓ is the approximation symbol

One solution was found :

c ≓ 2.953472167

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