Question 50052
Let x be the libaray books, y be the textbooks and z be the telephone books
EQUATION 1:
x+y+z=36
EQUATION 2:
x+z=2y ---> {{{y=(x+z)/2}}}
EQUATION 3:
y+z=3x


SUb EQUATION 2 into 1 and 3
{{{x+(x+z)/2+z=36}}}
3x+3z=72 (EQUATION 4)


{{{(x+z)/2+z=3x}}}
{{{x+3z=6x}}}
3z-5x=0 (EQUATION 5)


Subtract equation4 from 5 and simplfy:
3x-(-5x)=72
3x+5x=72
8x=72
x=9


3z=5x
3z=5(9)
z=15


y=(x+z)/2
y=12
so hence, there are 9 libaray books, 12 textbooks and 15 telephone books in the locker.
Paul.