Question

2^x+2^y=10 and 4^x+4^y=68, solve x and y

Answer 1

let z=2^x and w=2^y

Answer 2

the first equation becomes z+w=10 with that substitution

Answer 3

now, replace 4 with 2^2, so the second equation becomes

2^2x+2^2y=68

now, 2^2x is that same as 2^x2, which is also the same as (2^x)^2...using a similar argument, for y 4^y is really (2^y)^2

but we called 2^x z and 2^y we called w, so we substitute in and get

z^2+w^2=68

Answer 4

now using the first equation z+w=10, solve for z and then substitute into the second equation.

so z=10-w

and the second equation is then (10-w)^2 +w^2=68

expand the (10-w)^2 with foil to get 100-20w+w^2+w^2=68 and then group like terms and move everything over

Answer 5

2w^2-20w+32=0

dividing out by 2, w^2-10w+16=0

(w-8)(w-2)=0

w=2 and w=8

now, plug back in for w=2^y

2^y=2 and 2^y=8...so y=1 and y=3

plugging back in for y=1 gives x=3 and y=3 gives x=1

so the 2 solutions are (1,3) and (3,1)

Answer 6

hoped that helped!

Answer 7

thank you