System of Nonlinear Eqn (Word Prob): diff. of cubes of 2 numbers is 37; sum of square

[thebeanstalk]

Hello,

I am stuck at this word problem

"The difference of the cubes of two numbers is 37. The sum of their squares is 25. Find the numbers"

So Let X = be the higher number
Let Y = be the lower number

x^3 - y^3=37
x^2 + y^2=25

I am stuck at here

x^2+y^2=25
x^2 = 25-y^2

Please help me.

 

[stapel]

Hello,

I am stuck at this word problem

"The difference of the cubes of two numbers is 37. The sum of their squares is 25. Find the numbers"

So Let X = be the higher number
Let Y = be the lower number

x^3 - y^3=37
x^2 + y^2=25

I am stuck at here

It's messy, but it's quite solvable. I would suggest solving the second equation for "y=", cubing the result, and setting this equal to the y^3 in the first equation. To get rid of the radical, you'll need to square both sides. You should end up with a sixth-degree polynomial.

Do a quick graph on your graphing calculator to see which x-value you should use in the synthetic division.

 

[lookagain]

"The difference of the cubes of two numbers is 37. The sum of their squares is 25. Find the numbers"

So Let x = be the higher number
Let y = be the lower number

x^3 - y^3 = 37
x^2 + y^2 = 25

Suppose x and y are integers.

\(\displaystyle x^3 - y^3 = 37 \ \ \implies\)

\(\displaystyle (x - y)(x^2 + xy + y^2) = 1\cdot37 \ \implies\)

\(\displaystyle (x - y) \ = \ 1 \ \implies \)

\(\displaystyle x \ = \ y + 1\)

Now, substitute that expression for x into either

\(\displaystyle x^2 + xy + y^2 = 37 \ \ \ \) or

\(\displaystyle x^2 + y^2 = 25. \ \ \ \)


I recommend the latter option, because there are fewer operations to carry out.
Then you still will have to find the x-value to go with your y-value.


There will be two sets (containing x and y) of valid answers to this problem.
The quadratic equation will provide the roots (both y-values), one positive and
the other negative.

Use x = y + 1 (or related equation) to find the corresponding x-values.