Find the smallest positive integer that satisfies both of the following equations: x≡3(mod4) and x≡5(mod6).

Step-by-step explanation:

x≡3(mod4) and x≡5(mod6)

=> x≡-1(mod4) and x≡-1(mod6)

Since LCM of 4 and 6 is 12,

=> x≡-1(mod12)

=> x≡11(mod12)

The smallest positive integer for x is 11.

11

Step-by-step explanation:

Because 4 and 6 aren't co prime we can't start the chinese remainder theorem

so first we check

because 4 = 2*2

and the other one

so now we have

but the mod 2 we don't need it

so now we have

for the first one we can say that

so we plug in that in the second one

we can say that

so for x

so if k=0

a solution is 11

we check if it works

so it works so the smallest solution is 11

The answer is circle

Step-by-step explanation: The explanation is that you can see by looking at the shape you can easily make a rectangle, etc but you can not make a circle :).