We know that: X*Y =56

What could they be? trial and error…

2*28=56

4*14=56

7*8=56

Doubble one of them, because, “twice another”

so using the second numbers we have

28+28= 56

14+14 = 28

8+8=16

Now the first number has to be 9 less than one of these.

56-9=47 which is OBVIOUSLY not 2 so these cant be the two integers you are looking for.

28-9= 19 Which is not 4. so these cant be your pair.

16-9= 7 which is obviously 7. so our two integers are 7 and 8.

Make sense?

Lets check. using the quesition…

One integer (we are saying its 7) is 9 less than twice another (we are saying 8). If their product is 56, find the two integers.

so algebra form is:

7=(2*8)-9

does this hold true?

yep.

=)

let us say one integer is A and other is B

1. one integer is 9 less than twice other,

so A= 2B-9

2. product is 56

so A*B=56

substitute value of A from (1) in (2)

so (2B-9) * B = 56

so 2B*B – 9B – 56 = 0

this becomes a quadratic equation

you can convert it to

2B*B – 16B + 7B – 56=0

so

2B(B-8) + 7(B-8) = 0

(2B+7)(B-8)=0

so either 2B+7=0 OR B-8=0

if 2B+7=0 it means that B=-7/2. This cannot be true since B is an integer therefore it cannot be a fraction.

if B-8=0 it means B=8. This is true.

now B=8 and A*B=56 it means A*8=56 so A=56/8 =7

+9 +9

_______

2a = 65

Divide each side by 2 to isolate the variable “a”

a = 32.5