Question 109402
Let x=number of children tickets sold
And y=number of adult tickets sold
And z=number of senior tickets sold

Now the problem tells us the following:

5x+11y+6z=1750-------------------------------eq1
and
y=x+50------------------------------------------eq2
and
z=4x----------------------------------------------eq3

Three equations, three unknowns.  Lets solve by substitution.  We will substitute the value for y in eq2 and the value for z in eq3 into eq1 and we get:

5x+11(x+50)+6(4x)=1750 get rid of parens
5x+11x+550+24x=1750  subtract 550 from both sides

5x+11x+550-550+24x=1750-550  collect like terms

40x=1200   divide both sides by 40 
x=30------------------------------------number of children tickets sold
substitute x=30 into eq2:
y=30+50=80-----------------------------number of adult tickets sold

substitute x=30 into eq3 and we have:

z=4*30=120------------------------------number of senior tickets sold

CK

From eq1:

5*30+11*80+6*120=1750
150+880+720=1750
1750=1750


Hope this helps---ptaylor