Astore is having a sale on jelly beans and almonds. for 6 pounds of jelly beans and 2
pounds of almonds, the total cost is $16 . for 3 pounds of jelly beans and 5 pounds of almonds, the total cost is $13. find the cost for each pound of jelly beans and each pound of almonds.

The cost of each pound of jelly beans is $ 2.25 and cost of each pound of almond is $ 1.25

Solution:

Let the cost of each pound of jelly beans is "a"

Let the cost of each pound of almond is "b"

For 6 pounds of jelly beans and 2  pounds of almonds, the total cost is $16

Therefore, we frame a equation as:

6 pounds of jelly beans x cost of each pound of jelly beans + 2  pounds of almonds x cost of each pound of almond = 16

6a + 2b = 16 --------- eqn 1

For 3 pounds of jelly beans and 5 pounds of almonds, the total cost is $13

Therefore, we frame a equation as:

3 pounds of jelly beans x cost of each pound of jelly beans + 5 pounds of almonds x cost of each pound of almond = 13

3a + 5b = 13 ----------- eqn 2

Let us solve eqn 1 and eqn 2

Multiply eqn 2 with 2

6a + 10b = 26 --------- eqn 3

Subtract eqn 1 from eqn 3

6a + 10b = 26

6a + 2b = 16

( - ) -------------------------

8b = 10

Substitute b = 1.25 in eqn 1

6a + 2(1.25) = 16

6a + 2.5 = 16

6a = 13.5

Thus cost of each pound of jelly beans is $ 2.25 and cost of each pound of almond is $ 1.25

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)