Thread: Algebra Problem Solving

Here's the part of the problem I cannot solve:

You have two numbers. The difference between the two numbers is 12. The product of the two numbers is 17.

I've been trying and trying but all the numbers have added up to just over or under 17.

I've narrowed it down to somewhere around:

1.2801 x 13.2801 = 16.99985601

This damn homework is due tomorrow and I got nothing.
Maybe I've misread the question or something..Any help please?

set your first number to be x and your second number to be y;

therefore from the question;
x-y=12 ----(1)
xy=17 ----(2)

rearrange eq (2)
y=17/x ----(3)

and sub in to (1)
x-(17/x)=12

multiply by x
x^2 -12x - 17 =0

use the quadratic formula you get
x= 6 +/- SQRT(29)

then substitute both of those values of x back in to eq (3) and you get two values of y. Now check the corresponding values of x and y in the original two equations to see if both sets are solutions or only one set is a solution.

Originally Posted by RyderX

Here's the part of the problem I cannot solve:

You have two numbers. The difference between the two numbers is 12. The product of the two numbers is 17.

I've been trying and trying but all the numbers have added up to just over or under 17.

I've narrowed it down to somewhere around:

1.2801 x 13.2801 = 16.99985601

This damn homework is due tomorrow and I got nothing.
Maybe I've misread the question or something..Any help please?

You can set up system of equations:
x-y = 12
x*y = 17


Solve first equation for x:
x=12+y

Substitute x into second equation:
(12+y)*y=17
12y+y^2=17
y^2+12y-17=0

Solutions of this quadratic equation are approximately:
y_1 = 1.28
y_2 = -13.28

Substitute both values in first equation and you get:
x_1 = 13.28
x_2 = -1.28


So, numbers x and y can be (13.28, 1.28) and (-1.28, -13.28).

There isn't integer solution of that problem.

Ok Thanks Alot I knew up to the ~1.28/13.28 part already but I just figured out that the rest of the problem really made the extra 1000ths and 10000ths, etc irrelevent :S