Solving a system of linear equations

[elysium]

I forgot how to do this. I don't understand how to solve for x and y individually, or how they got:

The problem can now be written as a system of linear equations:

x+y = 75
20x+25y = 1675

where x and y are the values to be found.

Solving the equations, we find x= 40 and y= 35.

 

[Denis]

I forgot how to do this. I don't understand how to solve for x and y individually, or how they got:
The problem can now be written as a system of linear equations:
x+y = 75
20x+25y = 1675
where x and y are the values to be found.
Solving the equations, we find x= 40 and y= 35.

x + y = 75 [1]
20x + 25y = 1675 [2]

from [1]: y = 75 - x ; substitute that in [2]:
20x + 25(75 - x) = 1675

solve that for x; then get y.

 

[brendan3eb]

You could also multiply that one equation by -20 and solve by elimination.

 

[tkhunny]

I detect a hint of insanity, here.

20x + 25y = 1675

Common factor of 5

4x + 5y = 335

Kind of a sneaky trick.

x + y = 75
4(x+y) = 300

4x + 5y = 335
4x + 4y + y = 335
4(x + y) + y = 335
300 + y = 335

It's really not any different from simply subtracting.

4x + 5y = 335
-4x - 4y = -300

I'm just trying to get you to think about what is going on. There is not necessarily just one way to proceed. Do what seems most effective. I'm particularly fond of one certain goal while working such problems, "The worst numbers you should see are the ones you are given." One cannot always accomplish that goal, but it serves well to keep it in mind.