Question #bf493

2 Answers
Jun 14, 2017

#x=500#
#y=700#

Explanation:

One number is less than 3 times the other.

Let us say that the "One number" is #x#, and the other is #y#.
We get the following equations

#x=3y-800#
#x+y=1200#

Subtract #-3y# from the first equation.

#x-3y=-800#

Multiply the first equation by #-1#.

#-x+3y=800#

Now, add the two equations together.

#-x+3y=800#
#x+y=1200#

#4y=2000#

Divide both sides by #4#.

#y=500#

Substitute #y=500# into either of the 2 equations. This time, I'll use the second one simply because it's easier.

#x+500=1200#

#x=700#

Oh but wait, it says the smaller number is #x#!
No problem, just change the variables.

#x=500#
#y=700#

Check :

#500*3-800# does indeed equal #y#, which is #700#.
#800+700# does indeed equal #1200#.

Jun 17, 2017

The numbers are #500 and 700#

Explanation:

We can define both the numbers using one variable.

Let the one number be #color(blue)(x)#
.
The other is #800# less than #3# times #x#

The other number is therefore: #color(blue)(3x-800)#

Their sum is #1200#

#x + 3x -800 = 1200#

#4x = 1200+800#
#4x = 2000#

#x = 500 and 3x -800 = 700#

The numbers are #500 and 700#