SOLUTION: x + y = 31 (2/3)x = (5/8)y 16x = 15y x + y = 31 -------(1) 16x – 15y = 0------(2) Multiply equation (1) by 16 16x + 16y = (31)(16) 16x – 15y = 0 Subtract the two

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: x + y = 31 (2/3)x = (5/8)y 16x = 15y x + y = 31 -------(1) 16x – 15y = 0------(2) Multiply equation (1) by 16 16x + 16y = (31)(16) 16x – 15y = 0 Subtract the two       Log On

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Question 1097884: x + y = 31
(2/3)x = (5/8)y
16x = 15y
x + y = 31 -------(1)
16x – 15y = 0------(2)
Multiply equation (1) by 16
16x + 16y = (31)(16)
16x – 15y = 0
Subtract the two equation
31y = 31(16)
Y = 16
Input y = 16 in equation (1)
x + y = 31
x= 31 – y = 31 – 16 = 15
Thus x = 15 and y = 16
Therefore the two values are 15 and 16
Found 2 solutions by Sc_moore, ikleyn:
Answer by Sc_moore(1) About Me  (Show Source):
You can put this solution on YOUR website!
x + y = 31
(2/3)x = (5/8)y
16x = 15y
x + y = 31 -------(1)
16x – 15y = 0------(2)
Multiply equation (1) by 16
16x + 16y = (31)(16)
16x – 15y = 0
Subtract the two equation
31y = 31(16)
Y = 16
Input y = 16 in equation (1)
x + y = 31
x= 31 – y = 31 – 16 = 15
Thus x = 15 and y = 16
Therefore the two values are 15 and 16

Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
.
1.  Your solution is correct.


2.  You can EASILY check it on your own by substituting the found values into your original equations.


    All people do it, and checking is NECESSARY and it is REQUIRED when you solve any problem and 

    when you solve systems of linear equations, in particularly.