Question 176170
lets call the cost of 1m and 1.5m radiators, r and s , respectively
:
4r+3s=159.....eq 1
5r+2s=134.....eq 2
:
multiply eq 1 by 2 and eq 2 by -3 
:
::8r+6s=318....eq 1 revised
-15r-6s=-402....eq 2 revised
:
 add the two equations together. as you can observe the s terms are eliminated because 6s-6s=0. We are left with 8r-15r=318-402.
:
-7r=-84
:
{{{highlight(r=12)}}}per 1m radiator
:
now plug the value of r just found into any of the equations. I choose eq 1
:
4(12)+3s=159--->3s=111
:
{{{highlight(s=37)}}}per 1.5m radiator