You have a total of 52 dimes and quarters. You have 10 more quarters than dimes. Which system of equations can you use to find the number z of quarters and number y of dimes you have? Use the system to
determine how much money you have in quarters and dimes.

9514 1404 393

  z + y = 52; z - y = 10; z = 31; y = 21; $9.85 total

Step-by-step explanation:

The variables are defined in the problem statement. The equations could be ...

  z + y = 52 . . . . . . total number of coins

  z - y = 10 . . . . . . . 10 more quarters than dimes

These equations can be solved by adding them together.

  (z +y) +(z -y) = (52) + (10)

  2z = 62 . . . . . . . simplify

  z = 31 . . . . . . . . . divide by 2

  31 -10 = y = 21 . . . . find the number of dimes

There are 31 quarters and 21 dimes.

The value of the coins is ...

  31($0.25) +21($0.10) = $7.75 +2.10 = $9.85

hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)

no because mercury has a density of 13.6 grams per centimeter squared while gold has 19.32 grams centimeter squared

step-by-step explanation: