In the given figure, PR is 12 more than twice PQ, and QR is two more than four times PQ. If all three sides of the triangle have integer lengths, what is the largest possible value of x? PLS HELP ASAP

it’s -1.25

Step-by-step explanation:

To find out X you put 12x+12=4x+2 because there is no answer to equal it to. So next you would do 12x-4x which leaves you with 8x+12=2 so then you do 2-12 which is -10 which leaves you with 8=-10x and you divide and you get -1.25. I’m not for sure but that’s my guess!!

hello from mrbilldoesmath!

answer:

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

discussion:

the trick is to recognize that 343 = 7^3 and 216 = 6^3 and, in general,

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

in our case

343x^3+216y^3   =   (7x)^3 + (6y)^3

so take a = 7x and b = 6y in the above equation to get the answer.

you,

mrb