system of equations

Question

bob has 6 more dimes than quarters. he has $1.65 in all. how many coins of each type does he have?

Answer 1

You start with the equations that you know. The trick with this question is you have to know how much each coin is worth. Surely, you know this. We'll also label our dimes, D, and quarters, Q.

D = Q + 6

0.1D + 0.25Q = 1.65

Next we will plug the dimes equation into the total worth equation and distribute to solve for Q by combining like terms.

0.1(Q + 6) + 0.25Q = 1.65

0.1Q + 0.6 + 0.25Q = 1.65

0.35Q + 0.6 = 1.65

0.35Q + 0.6 - 0.6 = 1.65 - 0.6

0.35Q = 1.05

0.35 0.35

Q = 3

Now we can plug Q = 3 into our first equation for the dimes.

D = 3 + 6

D = 9

Answer 2

We have 2 unknowns, so we need to find 2 equations to solve this.

Let x = # of dimes

y = # of quarters

Eq. 1: Six more dimes than quarters : x = y + 6

Eq. 2: Total amount adds up to $1.65 : .10x + .25y = 1.65

Using the value for "x" from Eq. 1, put it into Eq. 2 and solve for "y":

.10x + .25y = 1.65 Eq. 2

.10(y + 6) + .25y = 1.65 substitute for "x"

.10y + .60 + .25y = 1.65 use distribution

.35y + .60 = 1.65 simplify

.35y = 1.65 - .60 subtract .60 from both sides

.35y = 1.05 simplify

.35y/.35 = 1.05/.35 divide both sides by .35

y = 3 result

Take this value of "y" and put it into Eq. 1 to find the value of "x":

x = y + 6 Eq. 1

x = 3 + 6 substitute for "y"

x = 9 result

Therefore, the number of dimes, x, is 9 and the number of quarters, y, is 3.

Hope this helps!

Answer 3

Let x = number of quarters

so x + 6 would be the number of dimes

Our equation would be .25x + .10(x + 6) = $1.65

Distribute .25x + .10x + .6 = 1.65

.35x = 1.65 - .6

.35x = 1.05

divide each side by .35

x = 3 or number of quarters

x + 6 = 9 or number of dimes

Check 3(.25) + 9(.10) = .75 + .90 = 1.65