Question about MathRescue Word Problems Of Algebra Lite

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Jamie = x

Luiz = x+4

Ramon = x+7

3x+11=29

3x=18

x=6

Jamie = 6 , Luiz = 6+4 = 10 , Ramon = 6+3 = 13

Posted on Oct 13, 2010

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Posted on Jan 02, 2017

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Let Adjoa's age be x and Kwame's y.

- Adjoa is 3 years older than Kwame: x=y+3
- Seven years ago sum of their ages was 13: (x-7)+(y-7)=13
- Now you substitute first equation into second equation: (y+3-7)+(y-7)=13
- 2y-4-7=13
- 2y=24
- y=24/2=12 years
- x=y+3=12+3=15 years
- sum of their present ages is x+y=12+15=27 years

Sep 27, 2011 | Office Equipment & Supplies

Let m be Marian's age, d Diana's age and c Carlos age.

I hope I helped :)

- sum of ages is 17: m+d+c=17
- two years from now Marian will be twice as old as Diana: m+2=2*(d+2) so m+2=2d+4 so 2d=m-2 or d=(m-2)/2=m/2-1
- Carlos will be half as old as Marian 4 years from now: c+4=1/2(m+4) or c+4=m/2+2 or c=m/2-2
- now: m+d+c=m+m/2-1+m/2-2=17
- 2m=17+3=20
- m=20/2=10 years
- d=m/2-1=5-1=4 years
- c=m/2-2=5-2=3 years

I hope I helped :)

Sep 18, 2011 | Office Equipment & Supplies

let c = cory's age now
let s = her sisters age now

cory's age now = sisters age + 6 years

=> c = s + 6

4 years ago, cory was 4 years younger ( i.e. c-4) and her sister was 4 years younger (s- 4)

4 years ago cory was 4 times older than her sister

=> c - 4 = 4*(s-4)

We can simplify this to c = 4s - 16 +4 c = 4s - 12

We can then use simultaneous equations to solve the two equations

(i) c = s + 6 (ii) c = 4s - 12

If we multiply both sides of equation (i) by 4 we get (iii) 4c = 4s + 24

We can then subtract equation (ii) from equation (iii) to eliminate the '4s' term => ( 4c = 4s + 24 ) - (c = 4s - 12) => 4c - c = 4s +24 - 4s +12 => 3c = 4s - 4s +36 => 3c = 36 => c = 36/3 => c = 12 => Cory's age now = 12

__For clarity:__
Her sister's age now is of course 6
4 years ago Cory would of been 8 and her sister would of been 2

I hope this helps and good luck! If you have more questions - ask away!

Please take the time to rate this answer

Many Thanks Don.

cory's age now = sisters age + 6 years

=> c = s + 6

4 years ago, cory was 4 years younger ( i.e. c-4) and her sister was 4 years younger (s- 4)

4 years ago cory was 4 times older than her sister

=> c - 4 = 4*(s-4)

We can simplify this to c = 4s - 16 +4 c = 4s - 12

We can then use simultaneous equations to solve the two equations

(i) c = s + 6 (ii) c = 4s - 12

If we multiply both sides of equation (i) by 4 we get (iii) 4c = 4s + 24

We can then subtract equation (ii) from equation (iii) to eliminate the '4s' term => ( 4c = 4s + 24 ) - (c = 4s - 12) => 4c - c = 4s +24 - 4s +12 => 3c = 4s - 4s +36 => 3c = 36 => c = 36/3 => c = 12 => Cory's age now = 12

I hope this helps and good luck! If you have more questions - ask away!

Please take the time to rate this answer

Many Thanks Don.

Sep 13, 2011 | Sunburst My Mathematical Life Single...

5 - 6 years difference is ok, once both parties are about 26 and 21. But a 20 year old dating a 14 year old is not to cool. Ether way, older boy or girl, there is a since that the older one can't communicate on their own level. If the younger of the two is far more mature for their age they will miss things that happen when dating a person of their own age. The mom and dad of the younger of the couple will loose some control; being the older one will feel as they can handle things on an adult level. It will be hard to set boundaries that the older one will stick with, but the younger should live by. Such as in by a set time, homework, projects for school. The older of the couple will push for their time "alone" which will take from the younger of the two tending to their duties. Sorry, put your foot down and save a lot of drama in your future, just say "NO, NO, NO".

Aug 30, 2011 | Tatung CLT-L50 15" Flat Panel LCD Monitor

Nov 09, 2010 | Health & Beauty

X= (1/2)M + 3
2 + X + 20 = M - 6

The father statement is not needed.

Two variables, with two equations. Solve for X.

The father statement is not needed.

Two variables, with two equations. Solve for X.

Sep 29, 2010 | Cameras

Three unknowns with three equasions:

cindy is 5 years older than alex. write it as c = a + 5

cindy is 7 years older than peter. write it as c = p + 7

The sum of their ages is 21. write it as c + a + p = 21

first solve for peters age in terms of alex's age.

c = p + 7 or

(a + 5) = p + 7 or

p = (a + 5) - 7 or

p = a - 2

substitute in and solve the equasion:

(a + 5) + a + (a - 2) = 21 or

3a + 3 = 21 or

3a = 24 or

a = 8

c = a + 5 or

c = 8 + 5 = 13

c = p + 7 or

p = c - 7 or

p = 13 - 7 = 6

cindy is 13, alex is 8 and peter is 6

This is called simultaneous equasions and will always work if you have the same number of unknowns as you have equasions (or relationships).

cindy is 5 years older than alex. write it as c = a + 5

cindy is 7 years older than peter. write it as c = p + 7

The sum of their ages is 21. write it as c + a + p = 21

first solve for peters age in terms of alex's age.

c = p + 7 or

(a + 5) = p + 7 or

p = (a + 5) - 7 or

p = a - 2

substitute in and solve the equasion:

(a + 5) + a + (a - 2) = 21 or

3a + 3 = 21 or

3a = 24 or

a = 8

c = a + 5 or

c = 8 + 5 = 13

c = p + 7 or

p = c - 7 or

p = 13 - 7 = 6

cindy is 13, alex is 8 and peter is 6

This is called simultaneous equasions and will always work if you have the same number of unknowns as you have equasions (or relationships).

Sep 20, 2009 | Puzzle Toys

X+X+28=4(X+X)

2X+28=4X+4X

28=8X-2X

28=6X

X=4

The twins are 4 yrs. old. The mother is 32.

Do I pass?

2X+28=4X+4X

28=8X-2X

28=6X

X=4

The twins are 4 yrs. old. The mother is 32.

Do I pass?

Apr 21, 2009 | Mathsoft StudyWorks! Mathematics Deluxe...

Standard deviation is the average root mean squared deviation from the average of the numbers.

Familiarize yourself with the wikipedia page on standard deviation: http://en.wikipedia.org/wiki/Standard_deviation

In this case, your question is easy. The standard deviation is 0, so this means the boys ages are all the same.

So they're all 24/3 = 8 years old.

Proof:

8 + 8 + 8 = 24 <-- satisfies the constraint that the boys ages must total to 24.

Get the standard deviation of these numbers;

8+8+8 / 3 = 8 <-- the average value is 8

The deviation of each boy's age from the average is:

boy 1

--------

8 years old, which deviates from 8 by 0

boy 2

--------

8 years old, which deviates from 8 by 0

boy 3

--------

8 years old, which deviates from 8 by 0

So the deviations are 0, 0 and 0.

To get the standard deviation, you sum the squares of the deviations of each boy, get the average, and square root. So:

0² + 0² + 0² = 0

Average is 0 / 3 = 0

Square root of 0 = 0

So the standard deviation is 0. Which shouldn't be much of a surprise to you. We just had to actually DO the work to show that it was in fact 0.

Familiarize yourself with the wikipedia page on standard deviation: http://en.wikipedia.org/wiki/Standard_deviation

In this case, your question is easy. The standard deviation is 0, so this means the boys ages are all the same.

So they're all 24/3 = 8 years old.

Proof:

8 + 8 + 8 = 24 <-- satisfies the constraint that the boys ages must total to 24.

Get the standard deviation of these numbers;

8+8+8 / 3 = 8 <-- the average value is 8

The deviation of each boy's age from the average is:

boy 1

--------

8 years old, which deviates from 8 by 0

boy 2

--------

8 years old, which deviates from 8 by 0

boy 3

--------

8 years old, which deviates from 8 by 0

So the deviations are 0, 0 and 0.

To get the standard deviation, you sum the squares of the deviations of each boy, get the average, and square root. So:

0² + 0² + 0² = 0

Average is 0 / 3 = 0

Square root of 0 = 0

So the standard deviation is 0. Which shouldn't be much of a surprise to you. We just had to actually DO the work to show that it was in fact 0.

Feb 08, 2008 | Audio Players & Recorders

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