Question 282766
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Let x = the number of highlighters
Let y = the number of three-ring binders
Let z = the number of hole punchers

We have this system

{{{system(x+y+z=100,"$0.50"x+"$3"y+"$10"z="$100")}}}

Multiply the second equation by 10 to clear the decimal,
and drop the dollar signs:

{{{system(x+y+z=100,5x+30y+100z=1000)}}}

Let's eliminate x:

Multiply the first equation by -5

{{{system(-5x-5y-5z=-500,5x+30y+100z=1000)}}}

Add the first equation to the second:

{{{25y+95z=500}}}

Divide the second equation through by 5

{{{5y+19z=100}}}

Solve for y:

{{{5y=100-19z}}}

{{{y=20-19z/5}}}

z must be a multiple of 5, since the 
fraction must equal to an integer:

Also since {{{y>=1}}}

{{{20-19z/5>=1}}}

{{{100-19z>=5}}}

{{{-19z>=-95}}}

{{{z<=5}}}

Since z is a multiple of 5 and also {{{x<=5}}}

the only possibility is that {{{z=5}}}, choice (E).

Substituting back we find that the complete solution is

94 highlighters, 1 three-ring binder, and 5 hole punchers

Edwin</pre>