A 11 gram sample of a substance that's used to treat thyroid disorders has a k-value of 0.1247 find the substance's half-life, in days. round to the nearest tenths.

So the first thing we do is to multiply the top equation by 3. the reason i want to do this is i notice there will be a 6y on both equations and can possibly add/subtract them. so we get 6y = -3x + 27. adding that and the second equation we get 6x = 27-15 which is 12. divide by 6 and we get 2. ∴ x = 2. substituting 2 into the first equation we get 2y = -2 + 9 which is 7. divide by 2 and we get 7/2. ∴ y = 7/2, x = 2.

Ok here we go ..there are plenty of ways to solve such math problems but we will go for the smartest and easiest ..5x+2y=7  -2x+6y=9we rearrange them into the following ..by multiplying the 1st equation by 2 it becomes 10x + 4y = 14we rearrange it and so it becomes 10x = 14 - 4yas for the second equation we multiply it by 5 and rearrange it so it becomes as to the following ..-10x + 30y = 45rearranging so 10x = 30y -45now 10x equals both 14 - 4y and 30y -45so   14 - 4y = 30y -4531 = 26yy= 31/26 = 1.19 or around 1.2so by now we can calculate x easily as  -2x+6y=9so 2x = 6y -9 = 6 * 31/26 - 9 = -1.84  then x = - 0.92 or around -0.9best luck!