the box is $154. If there are x series of 1-dollar coins in a box, form an equation in x. Hence find how most 50-cent coins there are in a box.
Please uncover me minute stairs on how to do it? THANKS!(:
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This post was written by dollarcoinworld on December 19, 2009
This is simpler to follow if we operate dual variables. Let them be H as well as D, with viewable significance. We are since which H = 2D, as well as 0. 5H + D = 154. Substituting a initial equation in to a second, we get 0. 5(2D) + D = 154, so D = 77 as well as H = 154.
For each a single dollar coin (x) there are 2 fifty-cent coins.
A dollar is next to to 1. 00 as well as fifty cents is . 50.
So there are 1(x)+. 5(2x)=154
x+x=154
2x=154
x=77
There are 77 a single dollar coins. Since there are twice as most fifty-cent coins there contingency be 154 fifty-cent coins in a box.
Lets stand in check a work. Does 1(77)+. 5(154)=154?
77+77=154
154=154
There are 154 fifty-cent coins in a box.
call F a series of fifty cent coins
call x a series of Dollars
the worth of all a fifty cent coins is 1/2F
the worth of all a dollars is x
we know which F=2x
and which 1/2F+x=154
substitute F=2x in to a equation 1/2F+X=154 as well as get:
1/2(2x)+x=154
2x=154
x=77
so F=154
there are 77 a single dollalr coins as well as 154 half dollar coins
simple,
let X be series of dol coins
and Y be series of 50cent coins
2(no of 50c coins)=number of dol coins
so
2y=X
x-2y=0. . . . . . . . . . . . . . . . . . . (1)
total volume in box is 154
i,e. ,addtion of all values of coins in a box
so
X+2Y=154. . . . . . . . . . . . . . . . . (2)
solving (1)&(2)
X-2Y=0
X+2Y=154
. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2x=154
x=77
now we know no of dol in box
given which y is twice x
so
y=154 coins
x be series 1 dollar coin
2x be series of 50cents coin
100x+100x=15400
200x=15400
x=77
77 a single dollar coin as well as 154 fifty cents coins