#### what1sth1s

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- Jul 22, 2006

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What I did was:

Let x+2 be Cheryl's age.

Let 7x be Brooklyn's age.

2) Lisa has twice as many dimes as nickels. If there are 42 coins in total, how much money does Lisa have?

- Thread starter what1sth1s
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- Joined
- Jul 22, 2006

- Messages
- 7

What I did was:

Let x+2 be Cheryl's age.

Let 7x be Brooklyn's age.

2) Lisa has twice as many dimes as nickels. If there are 42 coins in total, how much money does Lisa have?

We will let Cheryl's age be: \(\displaystyle \L \;x\)1) Cheryl is 26 years older than her daughter Brooklyn. Last year, Cheryl was 2 years older than 7 times Brooklyn's age. How old are Cheryl and Brooklyn now?

Hmm you just gotta think about these problems.

We will let Brooklyn's age be: \(\displaystyle \L \;b\)

Cheryl is 26 years older than her daughter Brooklyn: So \(\displaystyle \L \,x=b+26\)

Last year Cheryl was 2 years older than 7 times Brooklyn's age: \(\displaystyle \L \,x-1=7(b-1)+2\)

Since both are equal set them equal to each other: \(\displaystyle \L \;b+26=7b+2\)

Solve for \(\displaystyle \L \,b\) and then fill that into one of our original equations with \(\displaystyle \L \,x\) in it to get both Brooklyn's and Cheryl's age.

We will let the number of Nickels be: \(\displaystyle \L \;n\)2) Lisa has twice as many dimes as nickels. If there are 42 coins in total, how much money does Lisa have?

We will let the number of Dimes be: \(\displaystyle \L \;d\)

Lisa has twice as many dimes as nickels: \(\displaystyle \L \,n=2d\)

There are 42 coins in total: \(\displaystyle \L \, n+d=42\)

Subsitution: \(\displaystyle \L \,2d+d=42\)

Solve for \(\displaystyle \,d\) and then plug it back into \(\displaystyle \,n=2d\) and you will know how many dimes and nickles Lisa had.

let brooklyns current age be b

1) x=b+26

let cheryls age last year be x-1

let brooklyns age last year be b-1

2)x-1=7(b-1)+2

rearrange: x=7b-4

b+26=7b-4

26+4=7b-b

30=6b

b=30/6=5

substitute into 1 and 2 to check they give the same answer for x:

1)x=5+26=31

2)x-1=7(5-1)+2

x-1=7(4)+2

x-1=30

x=31

cherly is 31, brooklyn is 5.

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if she only have dimes nad nickels, and twice as many dimes as nickels, then 2/3 of her coins will be dimes and 1/3 will be nickels.

2/3*42=28 dimes *0.10 cents each = $2.80

1/3*42=14 nickels * 0.05 cents each = $0.70

$2.80+$0.70=$3.50

Next time, use a new thread for new problems...what1sth1s said:1) Cheryl is 26 years older than her daughter Brooklyn. Last year, Cheryl was 2 years older than 7 times Brooklyn's age. How old are Cheryl and Brooklyn now?

all i did was:

Let x+2 = Cheryl

Let 7x = Brooklyn

2) Lisa has twice as many dimes as nickels. If there are 42 coins in total, how much money does Lisa have?

1) Cheryl is 26 years older than her daughter Brooklyn. Last year, Cheryl was 2 years older than 7 times Brooklyn's age.

Go easy, step by step:

"Cheryl is 26 years older than her daughter Brooklyn."

B = Brooklyn, C = Cheryl ; then C = B + 26 : ok?

"Last year, Cheryl was 2 years older than 7 times Brooklyn's age."

Last year, Brooklyn = B - 1 and Cheryl = C - 1 : ok?

So C - 1 = 7(B - 1) + 2 : still with me?

C - 1 = 7B - 7 + 2

C - 1 = 7B - 5

C = 7B - 4 : understand?

And since C = B + 26, then:

7B - 4 = B + 26

Can you finish it?

2)Lisa has twice as many dimes as nickels. If there are 42 coins in total, how much money does Lisa have?

Step by step again:

"Lisa has twice as many dimes as nickels."

N = nickels; then dimes = 2N : easy nuff?

"If there are 42 coins in total, how much money does Lisa have?"

N + 2N = 42 : easy again?

Now solve for N, then figure out the total money: .05(N) + .10(2N)