Work has to be shown my teacher said that I made some mistakes please find them and fix
Problem
Jamil always throws loose change into a pencil holder on his desk and
takes it out every two weeks. This time it is all nickels and dimes. There
are 2 times as many dimes as nickels, and the value of the dimes is $1.65
more than the value of the nickels. How many nickels and dimes does
Jamil have?
My work
Let the number of dimes Jamil throws into the pencil holder be Y
Let the number of nickel Jamil throws into the pencil holder be X
Since there are 2 times as many dimes as nickel’s then we would have;
Y=2X
A dimes value is $0.05
A nickel’s value is $ 0.10
Since we have twice as many dimes as nickles then this would imply;
The sum of availables dimes and nickle is equal to $1.65
Converting Y=2X to a simple equation we would have;
Y- 2X=0 ……………………………..i
0.10X-0.05Y=1.65 …………….ii multiplying equation ii by 100 to get whole numbers then it will be;
Y- 2X=0 ………………………………i
10X+5Y = 165 ………………………iii
The least common multiple of 2 and 5 is 10. Multiply the first equation by 5 and the second equation by 2, by doing so that will have the y variable eliminated when the second equation is subtracted from the first equation.
(5)Y- (5)2X=(5)0 which solves to 5Y-10X=0
(2)10X+(2)5Y = (2)165 which solves to 20X+10Y=230
Subtract the second equation from the first and solve for x.
5Y-10X=0
20Y-10X=330
-15x= –330divide by sides by 12 to solve for x
-15 -15
Y = 22
From the equation we can be able to conclude that jamil has 22 dimes. We can go ahead and replace the number of dimes into 1 of the equations so that we can solve for the number of nickles. Taking the second equation.
20X+10Y=230 since we have computed for y=22
20(22)-10X=330then subtract 29.92 from both sides of the equation
440 – 10X=330
10X= 440-330
10X=110 dividing both sides by 10 we get;
10 10
X= 11 therefore jamil had 11 nickles
This would imply that jamil throwings in the pencil holder was 11 nickles and 22 dimes