What was the value of x and y for the following equations? Equation 1: 2/3 (x-6) = 6 Equation 2: 2/3y - 6 = 6 x = 15, y = 15 x = 18, y = 15 x = 18, y = 18 x = 15, y = 18

The value of x is 15 and value of y is 18

Solution:

Given equations are:

From first equation,

From second equation,

Thus value of x is 15 and value of y is 18

m a t h w a y. c o m

step-by-step explanation:

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)