Barbara has nickels, dimes, and quarters in her purse worth a total of $2.35. The number of dimes is three less than the sum of the number of nickels and quarters. There are 19 coins altogether. How many of each type of coin does Barbara have?
Can you please just write the equations? Thank you
Can you please just write the equations? Thank you
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First write the equation. I am using cents, hence 235 cents, instead of 2 dollars and 35 cents.
"a.)" means equation 1
"b.)" means equation 2, and so on
a.) 10x + 5y + 25z = 235 This is the first equation, which just says how many of each kind of coin you have. Leave this one alone for a bit, we will use it later.
b.) x = y + z - 3 We take into account the number of coins. We are told that the number of dimes is 3 less than the sum of nickles and quarters. x = # dimes, y = # nickles, z = # quarters
c.) 19 = x + y + z This says the total number of coins equals 19
Sub b.) into c.)
19 = [y + z - 3] + y + z Here I replaced the x with its equivalent from equation b.)
d.) z = 11 - y now solve for z to get
Now we have an equation for x and z which are in terms of y.
10x + 5y + 25z = 235
10(y + z - 3) + 5y + 25(11 - y) = 235 Here I replaced x with b and z with d.
10(y + (11 - y) - 3) + 5y + 275 - 25y = 235 Here replace the z with d
80 + 275 - 20y = 235 Keep working the equation
20y = 120
y = 6 This is what you should get for y, after some # crunching
Now just plug the y back into the other equations to solve for z and x
z = 11 - y
z = 5
x = y + Z - 3 plug in y and z
x = 8
Check with your initial equation
10x + 5y + 25z = 235
10(8) + 5(6) + 25(5) = 235
80 + 30 + 125 = 235
235 = 235
So in the end Barbara has 8 dimes, 6 nickels, and 5 quarters.
Hope this helps, sorry but its tough explaining this on here.
"a.)" means equation 1
"b.)" means equation 2, and so on
a.) 10x + 5y + 25z = 235 This is the first equation, which just says how many of each kind of coin you have. Leave this one alone for a bit, we will use it later.
b.) x = y + z - 3 We take into account the number of coins. We are told that the number of dimes is 3 less than the sum of nickles and quarters. x = # dimes, y = # nickles, z = # quarters
c.) 19 = x + y + z This says the total number of coins equals 19
Sub b.) into c.)
19 = [y + z - 3] + y + z Here I replaced the x with its equivalent from equation b.)
d.) z = 11 - y now solve for z to get
Now we have an equation for x and z which are in terms of y.
10x + 5y + 25z = 235
10(y + z - 3) + 5y + 25(11 - y) = 235 Here I replaced x with b and z with d.
10(y + (11 - y) - 3) + 5y + 275 - 25y = 235 Here replace the z with d
80 + 275 - 20y = 235 Keep working the equation
20y = 120
y = 6 This is what you should get for y, after some # crunching
Now just plug the y back into the other equations to solve for z and x
z = 11 - y
z = 5
x = y + Z - 3 plug in y and z
x = 8
Check with your initial equation
10x + 5y + 25z = 235
10(8) + 5(6) + 25(5) = 235
80 + 30 + 125 = 235
235 = 235
So in the end Barbara has 8 dimes, 6 nickels, and 5 quarters.
Hope this helps, sorry but its tough explaining this on here.
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let no. of nickel coins=x
let no. of dimes=y
let no. of quarters=z
x+y+z=19...............eq1
y=x+z-3.................eq2
x*value of nickel + y*value of dime + z*value of quarter = $2.35.............eq3
from eq1 & eq2
since x+z = y+3
substitute dis in eq1 to get
y+y+3 = 19
2y = 16
y = 8
substitute the value of y in the equations to obtain the remaining no. of coins
let no. of dimes=y
let no. of quarters=z
x+y+z=19...............eq1
y=x+z-3.................eq2
x*value of nickel + y*value of dime + z*value of quarter = $2.35.............eq3
from eq1 & eq2
since x+z = y+3
substitute dis in eq1 to get
y+y+3 = 19
2y = 16
y = 8
substitute the value of y in the equations to obtain the remaining no. of coins