# 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 :

(26)/(b+5)-(1+((3)/b-2))=0

## Step by step solution :

## Step 1 :

```
3
Simplify —
b
```

#### Equation at the end of step 1 :

```
26 3
——————— - (1 + (— - 2)) = 0
(b + 5) b
```

## Step 2 :

#### Rewriting the whole as an Equivalent Fraction :

2.1 Subtracting a whole from a fraction

Rewrite the whole as a fraction using b as the denominator :

```
2 2 • b
2 = — = —————
1 b
```

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:

```
3 - (2 • b) 3 - 2b
——————————— = ——————
b b
```

#### Equation at the end of step 2 :

```
26 (3 - 2b)
——————— - (1 + ————————) = 0
(b + 5) b
```

## Step 3 :

#### Rewriting the whole as an Equivalent Fraction :

3.1 Adding a fraction to a whole

Rewrite the whole as a fraction using b as the denominator :

```
1 1 • b
1 = — = —————
1 b
```

#### Adding fractions that have a common denominator :

3.2 Adding up the two equivalent fractions

```
b + (3-2b) 3 - b
—————————— = —————
b b
```

#### Equation at the end of step 3 :

```
26 (3 - b)
——————— - ——————— = 0
(b + 5) b
```

## Step 4 :

```
26
Simplify —————
b + 5
```

#### Equation at the end of step 4 :

```
26 (3 - b)
————— - ——————— = 0
b + 5 b
```

## Step 5 :

#### Calculating the Least Common Multiple :

5.1 Find the Least Common Multiple

The left denominator is : b+5

The right denominator is : b

Algebraic Factor | Left Denominator | Right Denominator | L.C.M = Max {Left,Right} |
---|---|---|---|

b | 0 | 1 | 1 |

b+5 | 1 | 0 | 1 |

Least Common Multiple:

b • (b+5)

#### Calculating Multipliers :

5.2 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 = b

Right_M = L.C.M / R_Deno = b+5

#### Making Equivalent Fractions :

5.3 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 (y^{2}+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. 26 • b —————————————————— = ————————— L.C.M b • (b+5) R. Mult. • R. Num. (3-b) • (b+5) —————————————————— = ————————————— L.C.M b • (b+5)

#### Adding fractions that have a common denominator :

5.4 Adding up the two equivalent fractions

` 26 • b - ((3-b) • (b+5)) b`^{2} + 28b - 15
———————————————————————— = —————————————
b • (b+5) b • (b + 5)

#### Trying to factor by splitting the middle term

5.5 Factoring b^{2} + 28b - 15

The first term is, b^{2} its coefficient is 1 .

The middle term is, +28b its coefficient is 28 .

The last term, "the constant", is -15

Step-1 : Multiply the coefficient of the first term by the constant 1 • -15 = -15

Step-2 : Find two factors of -15 whose sum equals the coefficient of the middle term, which is 28 .

-15 | + | 1 | = | -14 | ||

-5 | + | 3 | = | -2 | ||

-3 | + | 5 | = | 2 | ||

-1 | + | 15 | = | 14 |

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

#### Equation at the end of step 5 :

` b`^{2} + 28b - 15
————————————— = 0
b • (b + 5)

## Step 6 :

#### When a fraction equals zero :

` 6.1 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:

` b`^{2}+28b-15
————————— • b•(b+5) = 0 • b•(b+5)
b•(b+5)

Now, on the left hand side, the b • b+5 cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :

b^{2}+28b-15 = 0

#### Parabola, Finding the Vertex :

6.2 Find the Vertex of y = b^{2}+28b-15

Parabolas have a highest or a lowest point called the Vertex . Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) . We know this even before plotting "y" because the coefficient of the first term, 1 , is positive (greater than zero).

Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.

Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.

For any parabola,Ab^{2}+Bb+C,the b -coordinate of the vertex is given by -B/(2A) . In our case the b coordinate is -14.0000

Plugging into the parabola formula -14.0000 for b we can calculate the y -coordinate :

y = 1.0 * -14.00 * -14.00 + 28.0 * -14.00 - 15.0

or y = -211.000

#### Parabola, Graphing Vertex and X-Intercepts :

Root plot for : y = b^{2}+28b-15

Axis of Symmetry (dashed) {b}={-14.00}

Vertex at {b,y} = {-14.00,-211.00}

b -Intercepts (Roots) :

Root 1 at {b,y} = {-28.53, 0.00}

Root 2 at {b,y} = { 0.53, 0.00}

#### Solve Quadratic Equation by Completing The Square

6.3 Solving b^{2}+28b-15 = 0 by Completing The Square .

Add 15 to both side of the equation :

b^{2}+28b = 15

Now the clever bit: Take the coefficient of b , which is 28 , divide by two, giving 14 , and finally square it giving 196

Add 196 to both sides of the equation :

On the right hand side we have :

15 + 196 or, (15/1)+(196/1)

The common denominator of the two fractions is 1 Adding (15/1)+(196/1) gives 211/1

So adding to both sides we finally get :

b^{2}+28b+196 = 211

Adding 196 has completed the left hand side into a perfect square :

b^{2}+28b+196 =

(b+14) • (b+14) =

(b+14)^{2}

Things which are equal to the same thing are also equal to one another. Since

b^{2}+28b+196 = 211 and

b^{2}+28b+196 = (b+14)^{2}

then, according to the law of transitivity,

(b+14)^{2} = 211

We'll refer to this Equation as Eq. #6.3.1

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of

(b+14)^{2} is

(b+14)^{2/2} =

(b+14)^{1} =

b+14

Now, applying the Square Root Principle to Eq. #6.3.1 we get:

b+14 = √ 211

Subtract 14 from both sides to obtain:

b = -14 + √ 211

Since a square root has two values, one positive and the other negative

b^{2} + 28b - 15 = 0

has two solutions:

b = -14 + √ 211

or

b = -14 - √ 211

### Solve Quadratic Equation using the Quadratic Formula

6.4 Solving b^{2}+28b-15 = 0 by the Quadratic Formula .

According to the Quadratic Formula, b , the solution for Ab^{2}+Bb+C = 0 , where A, B and C are numbers, often called coefficients, is given by :

__ __

- B ± √ B^{2}-4AC

b = ————————

2A

In our case, A = 1

B = 28

C = -15

Accordingly, B^{2} - 4AC =

784 - (-60) =

844

Applying the quadratic formula :

-28 ± √ 844

b = ——————

2

Can √ 844 be simplified ?

Yes! The prime factorization of 844 is

2•2•211

To be able to remove something from under the radical, there have to be 2 instances of it (because we are taking a square i.e. second root).

√ 844 = √ 2•2•211 =

± 2 • √ 211

√ 211 , rounded to 4 decimal digits, is 14.5258

So now we are looking at:

b = ( -28 ± 2 • 14.526 ) / 2

Two real solutions:

b =(-28+√844)/2=-14+√ 211 = 0.526

or:

b =(-28-√844)/2=-14-√ 211 = -28.526

## Two solutions were found :

- b =(-28-√844)/2=-14-√ 211 = -28.526
- b =(-28+√844)/2=-14+√ 211 = 0.526