Solve the following simultaneous linear congruences.
a) x ? 1 (mod 3), x ? 2 (mod 4), x ? 3 (mod 5).
b) x ? 4 (mod 10), x ? 8 (mod 12), x ? 6 (mod 18).

a. The moduli are coprime, so you can apply the Chinese remainder theorem directly. Let

By the CRT, we have

i.e. any number (where is an integer) satisifes the system.

b. The moduli are not coprime, so we need to check for possible contradictions. If and , then we need to have . This basically amounts to checking that if , then we should also have .

The last congruence conflicts with the previous one modulo 3, so there is no solution to this system.

x= 149. 24+7=31. 180-31=149

your welcome

the answer is the option b

the range is all real numbers

step-by-step explanation:

we know that

the absolute value parent function is equal to

see the attached figure to better understand the problem

so

a. it is v-shaped > is true

b. the range is all real numbers > is not true

because the range is the > [0,∞)

all real numbers greater than or equal to zero

c. it goes through the origin > is true

d. the domain is all real numbers > is true

step-by-step explanation: