Let
x------> amount of <span> big marbles in the box
y------> </span>amount of small marbles in the box
R----> amount of small marbles that are red in the box
B----> amount of small marbles that are blue in the box
we know that
x=45+(3/4)*x----> x-(3/4)*x=45----> x/4=45-----> x=180
R+B=y-----> equation 1
R=(2/5)*y---> equation 2
B=(3/5)*y---> equation 3
B=R+24----> equation 4
substitute equation 2 in equation 1
[(2/5)*y]+B=y----> equation 5
substitute equation 2 in equation 4
B=[(2/5)*y]+24----> equation 6
resolve the system
[(2/5)*y]+B=y---> multiply by 5----> 2y+5B=5y----> 3y=5B
B=[(2/5)*y]+24----> multiply by 5----> 5B=2y+120
using a graph tool
see the attached figure
the solution is
B=72
y=120
R=y-B----> R=48
total marbies=x+y-----> 180+120-----> 300
tota; big marbies=180
<span>percent of the marbles are big marbles
</span>if 100%-----------> 300
X----------------> 180
x=180*100/300-----> x=60%
the answer is
60%
x------> amount of <span> big marbles in the box
y------> </span>amount of small marbles in the box
R----> amount of small marbles that are red in the box
B----> amount of small marbles that are blue in the box
we know that
x=45+(3/4)*x----> x-(3/4)*x=45----> x/4=45-----> x=180
R+B=y-----> equation 1
R=(2/5)*y---> equation 2
B=(3/5)*y---> equation 3
B=R+24----> equation 4
substitute equation 2 in equation 1
[(2/5)*y]+B=y----> equation 5
substitute equation 2 in equation 4
B=[(2/5)*y]+24----> equation 6
resolve the system
[(2/5)*y]+B=y---> multiply by 5----> 2y+5B=5y----> 3y=5B
B=[(2/5)*y]+24----> multiply by 5----> 5B=2y+120
using a graph tool
see the attached figure
the solution is
B=72
y=120
R=y-B----> R=48
total marbies=x+y-----> 180+120-----> 300
tota; big marbies=180
<span>percent of the marbles are big marbles
</span>if 100%-----------> 300
X----------------> 180
x=180*100/300-----> x=60%
the answer is
60%