Let the first number be represented by "x".
Let the second number be represented by "y".
"Four times a number..." = 4x
"... minus 5..." = 4x - 5
"... times another number..." = y(4x - 5)
"... is equal to 21." = y(4x - 5) = 21
y(4x - 5) = 21
"Three times the sum of the two numbers (x & y) is 36." = 3(x +y) = 36
3(x + y) = 36
y(4x - 5) = 21
1) solve for "y"
1) y(4x - 5) = 21
2) 4xy - 5y = 21
3) 4xy/4x - 5y = 21/4x
(what you do to one side of the equals sign, you must do to the other so 4x cancels out on the left side, and on the right side you have to divide 21 by 4x)
4) y - 5y/-5 = 5.25x/-5
5) y + y = -1.05x
6) 2y = -1.05x
7) 2y/2 = -1.05x/2
8) y = -.525x
2) substitute "y" in for the second equation
1) 3(x + (-.525x)) = 36
(notice how you now only have x's in this equation)
3) solve for x
1) 3x - 1.575x = 36
2) 1.425x = 36
3) 1.425x/1.425 = 36/1.425
4) x = 25.3
4) finish finding "y" in either one of the equations using the answer you got for "x"
1) y(4(25.3) - 5) = 21
2) y(101.2 - 5) = 21
3) y(96.2) = 21
4) 96.2y = 21
5) 96.2y/96.2 = 21/96.2
6) y = .22
5) check your work by substituting the two answers back into one of the equations
1) .22(4(25.3) - 5) = 21
2) .22(101.2 - 5) = 21
3) .22(96.2) = 21
4) 21.164 = 21
(round down to get the answer)
