Rewrite the equations of the given boundary lines:
<em>y</em> = -<em>x</em> + 1 ==> <em>x</em> + <em>y</em> = 1
<em>y</em> = -<em>x</em> + 4 ==> <em>x</em> + <em>y</em> = 4
<em>y</em> = 2<em>x</em> + 2 ==> -2<em>x</em> + <em>y</em> = 2
<em>y</em> = 2<em>x</em> + 5 ==> -2<em>x</em> + <em>y</em> = 5
This tells us the parallelogram in the <em>x</em>-<em>y</em> plane corresponds to the rectangle in the <em>u</em>-<em>v</em> plane with 1 ≤ <em>u</em> ≤ 4 and 2 ≤ <em>v</em> ≤ 5.
Compute the Jacobian determinant for this change of coordinates:

Rewrite the integrand:

The integral is then

Answer:
13, 18, 72
Step-by-step explanation:
let x-5 represent the first number, since the first number is 5 less than the third number. let x represent the third number. let 4x represent the third number, since it's four times greater than the third number.
set up an equation, and make it equal to 103:
x-5+x+4x=103
solve:
6x-5=103
6x=108
x=18
so, the third number is 18.
the first number is equal to x-5, so:
18-5=13
the second number is equal to 4x, so:
4(18)=72
so, the three numbers are 13, 18, and 72.