Question

One positive integer is 6 less than twice another. The sum of their squares is 377. Find the integer.

Answer 1

start with
\[x\] and
\[2x-6\]

Answer 2

then
\[x^2+(2x-6)^2=377\] solve for x

Answer 3

\[x^2+(2x-6)^2=377\]
\[x^2+4x^2-24x+36=377\]
\[5x^2-24x-341=0\]
\[(x-11)(5x+31)=0\] etc

Answer 4

thank you thank you thank you

Answer 5

yw