how do you solve this age and number problem in math?

Question

for the number problem:

1. Find two numbers whose sum is A and whose difference is B.

and for the age problem:

1. Seven years ago, Joan was five times as old as Nina. In one year, she will be twice as old as Nina. Find their present ages.



you dont have to tell me the answer, just tell me how you get the answer!

thanks =)

Answer 1

1. x+y =A

x-y=B

solving we get

2x = A+B

x = A/2 +B/2

and y =A -x

so A- A/2-B/2

y = A/2-B/2



2. j-7 = 5(n-7)

j-7 =5n-35

j-5n = -28....1st



j+1=2(n+1)

j+1 =2n+2

j-2n = 1...2nd

solving both we get

-3n = -29

n=29/3 = 9.67 years



and j = 1+2(29/3)

j = 61/3 =20.33 years

Answer 2

Both questions can be solved by setting up two equations, then solving for one of the variables. The other variable can then be found using substitution. 1. x + y = A x - y = B add equations and you get 2x = A + B or x = (A+B)/2 substitution yields (A+B)/2 + y = A or y = A - (A+B)/2 y = 2A/2 - (A+B)/2 y = (2A-A-B)/2 and finally y = (A-B)/2 2. Let J = Joan's age Let N = Nina's age From the first statement you can derive the equation J-7=5(N-7) From the second statement you can derive the equation J+1=2(N+1) Subtracting the second equation from the first yields: -8=5(N-7) - 2(N+1) or -8=5N-35 -2N-2 simplifying yields -8=3N-37 add 37 to both sides and 29=3N or N=29/3 or Nina is 9 2/3 years old. Substituting Nina's age in the second equation gives you J + 1 = 2 (32/3) or J + 1 = 64/3 Subtract one or 3/3 from both sides yields 61/3 or Joan is 20 1/3 years old. Checking with equation one: 13 1/3 compared to 5 X (2 2/3) 40/3 compared to 5 X 8/3 40/3 = 40/3 checks out! Checking equation two: 21 1/3 compared to 2 X 10 2/3 64/3 compared to 2 x 32/3 64/3 = 64/3 checks out!

Answer 3

In both problems, you just have to set the equations properly, then solving them becomes easy.



1. let x be the first number, and let y be the second number

their sum is A: x + y = A

their difference is B: x - y = B

Now you can solve these two equations simultaneously. In case you haven't studied simultaneous equations yet, one way to do it is to add the two equations together so that terms cancel out. E.g. (x+y) + (x-y) = A + B

==> 2x = A + B, and then solve for x and substitute for y.



2. let j be Joan's present age, and n be Nina's present age.

Joan's age seven years ago = j-7

Nina's age seven years ago = n-7

so, 5 * (n-7) = j-7

==> 5n-35 = j-7

==> 5n - j = 28 (call this equation one)



Joan's age in one year = j+1

Nina's age in one year = n+1

j+1 = 2*(n+1)

==> j+1 = 2n +2

==> j - 2n = 1 (call this equation two)



Now, you just have to solve equations one and two simultaneously. Again, I've set up the equations so all you have to do is add them together - the j's will cancel out, and you can solve for n first.



I hope this helps!

Answer 4

Your second question is a simultaneous equation.



J = Joan's current age

N = Nina's current age



J - 7 = 5(N - 7)

J + 1 = 2(N + 1)



Solve by substitution for either J or N and then resolve the other.

Answer 5

I would say 22 and 11



15, 3 5 times +7

22, 10

1 year

23, 11 which is about 1/2

Answer 6

1. x + y = A

x - y = B (x is larger than y)

(add the two)

=> 2x = A + B

x = (A + B)/2

y = A - (A + B)/2



2. J - 7 = 5(N - 7)

J + 1 = 2(N + 1)

(Subtract the two)

=> 8 = 2(N + 1) - 5(N - 7)

8 = -3N + 37