confused on this ALGEBRA math problem!! PLEASE HELP!!?

All you need to do is set up equations as directed by the question.

A=Alan, B=Betty, and C=Chris.

A+B+C = 31...(i)

B = A-3...(ii)

C = A+B+1...(iii)

Rewrite, equation (iii) in terms of A, using (ii),

C = A+A-3+1

C = 2A - 2

Plug in the value of B and the value of V into equation (i)

A+B+C = 31

A+A-3+2A-2 = 31

4A-5 = 31

4A = 36

A = 9...so, we know that Alan has $9.00

Plug the value of A in equation (ii)

B = A-3

B = 9-3

B = 6...so, Betty has $6.00

Now, for Chris, plug in the value of A and B into equation (iii)

C = A+B+1

C = 9+6+1

C = 16...so, Chris has $16.00

CHECK:

Together they have:

9+6+16 = $31...agrees with the question;

Betty has $3 less than Alan ($9 vs. $6)...agrees with the question; and

Alan and Betty combined have: 9+6 = $15, which is $1 less than Chris...this agrees with the question as well. So you're good to go.

Together Alan, Betty, and Chris have $31.

A+B+C = 31: Eqn 1

Find the amount each has if Betty has $3 less than Alan,

B = A-3 : Eqn 2

and Chris has $1 more than Alan and Betty combined.

C = A + B + 1 : Eqn3

put eqn 2 into eqn 1 and into eqn 3

A + A -3 + C = 31

2 A + C = 34 : eqn 4

C = A + A-3+ 1

2 A - C = 2 : Eqn 5

add eqn 4 & eqn 5

4 A = 36

A = 9

B = A -3 = 9 -3 = 6

A+B+C = 31

9+6 +C = 31

15+ C = 31

C = 16

Okay, let's assign some values. Let's let the amount of money that Alan has be A, Betty be B, and Chris be C.

They have $31 altogether:

A+B+C=31

Betty has $3 less than Alan:

B+3=A ☺

And Chris has $1 more than Alan and Betty combined:

C=A+B+1 ☺

Let's substitute for Alan with the two equations up there with smiley faces:

C=(B+3)+B+1

C=2B+4

Now, we use our new values:

A=B+3

C=2B+4

And plug it into A+B+C=31:

(B+3)+B+(2B+4)=31

4B+7=31

4B=24

B=6

So Betty has $6. Alan has $3 more, which is $9. And Chris has the leftovers, 31-6-9 = $16.

Call the amounts a, b and c. Write down the equations you read. Then solve them:

(1) a+b+c=31

(2) b=a-3

(3) c=a+b +1

Substitute (3) in (1) and then (2) in (1);

a+b+a+b+1 = 31 or 2a+ 2b = 30 or a + b = 15

Using (2) gives: a + a -3 = 15, 2a=18 or a = 9

Substitute this in (2): b = 9-3= 6

Finally c = 15+1 =16.

Let x be the amount of money that Alan has

Let y be the amount of money that Betty has

Let z be the amount that Cris has

y + 3 = x

x + y +1 = z (rearrange to x + y = z-1)

z+x+y = 31

Substitute equation 2 in to equation 3

z + z-1 = 31

2z = 32

z = 16

Rearrange first equation to y = x-3

Substitute equation 1 and z = 16 into equation 2

x + x-3 + 1 = 16

2x = 18

x = 9

Substitute x = 9 into first equation

y + 3 = 9

y = 6

Chris has 16 dollars, Betty has 6 dollars and Alan has 9 dollars.

First, reduce the words to equations

1. A + B + C = 31

2. A - B = 3

3. C - A - B = 1

Three variables, three equations. Easy. Add 1 and 3 above to get

2C = 32 and C = 16. Therefore, from 1

4. A + B = 15, Now add 2 to this to get 2A = 18, so A = 9 and therefore B = 6

Melanie's cookies: M Felipe: 2M Hilary: 2M - 3 Nicholas: 6M - 9 Parents: 2 Had Nicholas eaten 5 cookies there would have been 6 cookies left, but there's only 2, so Nico ate 5 + 4 = 9 cookies So 6M - 9 = 9 <=> 6M = 18 <=> M = 3 Total amount: 11M -12 + 2 = 11M-10 = 33 - 10 = 23

betty has $6

Alan has $9

and chris has $16

6+9+16=31

I got Alan=$12, Betty=$9, and Chris=$22

The only problem is it doesn't check.....

Good luck!

2(b-3=a)+1=c

31-1=30

30/4=7.5

a=7.5

b=7.5-3=4.5

c=15+1=16

hope its helped