Question

Jordan is a manager of a car dealership. He has 3 professional car washers to clean the entire lot of cars, Jennifer, Arianna, and Matthew. Jennifer can wash all the cars in 14 hours. Arianna can wash all the cars in 11 hours. Matthew is new to the car dealership, so no one knows how long it will take him.

Jordan assigns all of them to wash the cars together. Explain to Jordan how this task can tell him how long it would take Matthew to complete the task if he worked by himself. Use complete sentences?

Answer 1

@ganeshie8 I have this so far:

Answer 2

Step 1: Set up the equation by adding all the workers together on one side and the total time on the other:

1/x=1/14+1/11+1/M. x equals the number of hours it will take them to do this all together. M equals Matthews hours that we don't know.

Step 2: Multiply all the fractions by the lcm

154xm(1/11+1/14+1/m=1/x)→154xm/11+154xm/14+154xm/m=154/x→ 14mx+11mx+m=x

Answer 3

actually there is no solution

Answer 4

are you sure? because this is a question on an assignment and its an essay haha so do I just say it has no solution?

Answer 5

@Muzzack

Answer 6

well i calculated it already:
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

\(\large\ 1/x-(1/14+1/11+1/m)=0\)

1 1
Simplify —— + ——
14 11

Find the Least Common Multiple (L.C.M)

The left denominator is : 14

The right denominator is: 11

Factor the left and right denominators

Least Common Multiple: \(\color{green}{\large\cal\ 154~}\)

Calculate multipliers for the two fractions

L.C.M
L. Multiplier = —————————————— = 11
L. Denominator


L.C.M
R. Multiplier = —————————————— = 14
R. Denominator

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. = 11
—————————————————— = ———
L.C.M 154

R. Mult. • R. Num. 14
—————————————————— = ———
L.C.M 154

Adding fractions that have a common denominator :
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:

11 + 14 25
——————— = ———
154 154

Equation at the end of step 1 :

1 25 1
— - (——— + —) = 0
x 154 m

Step 2 :

25 1
Simplify ——— + —
154 m

Calculating the Least Common Multiplier :

2.1 Find the Least Common Multiple (L.C.M)

The left denominator is : 154

The right denominator is : m

|———————————————|—————————————————————————————————————|
| | Number of times each prime factor |
| | appears the factorization of: |
| | |
| Prime | Left Right L.C.M = Max |
| Factors |Denominator Denominator {Left,Right}|
| —————————— |——————————— ——————————— ————————————|
| 2 | 1 0 1 |
| 7 | 1 0 1 |
| 11 | 1 0 1 |
|———————————————|—————————————————————————————————————|
|Product of all | |
| Prime Factors | 154 1 154 |
|———————————————|—————————————————————————————————————|
| | # of times each Algebraic Factor |
| | appears the factorization of: |
| | |
| Algebraic | Left Right L.C.M=Max |
| Factors |Denominator Denominator {Left,Right}|
| —————————— |——————————— ——————————— ————————————|
| m | 0 1 1 |
|———————————————|—————————————————————————————————————|

Least Common Multiple: 154m

Calculating Multipliers :

2.2 Calculate multipliers for the two fractions

L.C.M
L. Multiplier = —————————————— = m
L. Denominator


L.C.M
R. Multiplier = —————————————— = 154
R. Denominator

Making Equivalent Fractions :

2.3 Rewrite the two fractions into equivalent fractions

L. Mult. • L. Num. 25 • m
—————————————————— = ——————
L.C.M 154m

R. Mult. • R. Num. 154
—————————————————— = ————
L.C.M 154m

Adding fractions that have a common denominator :

2.4 Adding up the two equivalent fractions

25 • m + 154 25m + 154
———————————— = —————————
154m 154m

Equation at the end of step 2 :

1 (25m + 154)
— - ——————————— = 0
x 154m

Step 3 :

1 25m+154
Simplify — - ———————
x 154m

Calculating the Least Common Multiplier :

3.1 Find the Least Common Multiple (L.C.M)

The left denominator is : x

The right denominator is : 154m

|———————————————|—————————————————————————————————————|
| | Number of times each prime factor |
| | appears the factorization of: |
| | |
| Prime | Left Right L.C.M = Max |
| Factors |Denominator Denominator {Left,Right}|
| —————————— |——————————— ——————————— ————————————|
| 2 | 0 1 1 |
| 7 | 0 1 1 |
| 11 | 0 1 1 |
|———————————————|—————————————————————————————————————|
|Product of all | |
| Prime Factors | 1 154 154 |
|———————————————|—————————————————————————————————————|
| | # of times each Algebraic Factor |
| | appears the factorization of: |
| | |
| Algebraic | Left Right L.C.M=Max |
| Factors |Denominator Denominator {Left,Right}|
| —————————— |——————————— ——————————— ————————————|
| x | 1 0 1 |
| m | 0 1 1 |
|———————————————|—————————————————————————————————————|

Least Common Multiple: 154xm

Calculating Multipliers :

3.2 Calculate multipliers for the two fractions

L.C.M
L. Multiplier = —————————————— = 154m
L. Denominator


L.C.M
R. Multiplier = —————————————— = x
R. Denominator

Making Equivalent Fractions :

3.3 Rewrite the two fractions into equivalent fractions

L. Mult. • L. Num. 154m
—————————————————— = —————
L.C.M 154xm

R. Mult. • R. Num. (25m+154) • x
—————————————————— = —————————————
L.C.M 154xm

Adding fractions that have a common denominator :

3.4 Adding up the two equivalent fractions

154m - ((25m+154) • x) -25xm - 154x + 154m
—————————————————————— = ———————————————————
154xm 154xm

Pulling out like terms :

3.5 Pull out like factors :

-25xm - 154x + 154m = -1 • (25xm + 154x - 154m)
Equation at the end of step 3 :

-1 • (25xm + 154x - 154m)
————————————————————————— = 0
154xm

Step 4 :

-1•(25xm+154x-154m)
Solve ——————————————————— = 0
154xm

When a fraction equals zero :

4.1 When a fraction equals zero ...

Where a fraction equals zero, its nominator, 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:

-1•(25xm+154x-154m)
——————————————————— • 154xm = 0 • 154xm
154xm

Now, on the left hand side, the 154xm cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
-1 • (25xm+154x-154m) = 0
Equations which are never true :

4.2 Solve : -1 = 0

This equation has no solution.
A a non-zero constant never equals zero.
Solving a Single Variable Equation :

4.3 Solve 25xm+154x-154m = 0

In this type of equations, having more than one variable (unknown), you have to specify for which variable you want the equation solved.

We shall not handle this type of equations at this time.

Answer 7

ok thanks