. Pencils are sold in packages of 10, and erasers are sold in packages of 6. What is the least number of pencils and erasers you can buy so that there is one pencil for each eraser with none left over?​

3 packs of pencils

5 packs of erasers

Step-by-step explanation:

To solve this problem, find the least common multiple. The least common multiple is the smallest number that both terms are a factor of. In this case, that number is 30. Both 10 and 6 are a factor of 30. So, the answer is the other factor needed to get 30 through multiplication. 10's other factor is 3 and 6's other factor is 5. Therefore, the answers are 3 and 5 respectively.

step-by-step explanation:

try remembering the steps

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)