Prove that for integers m, n, p if m n and n p are both odd then m p must be an even integer.(Direct)

Answer and Step-by-step explanation:

Solution:

Given:

m, n, and p are integers

m + n and n + p are odd integers.

Then prove that m + p is even integer.

Proof:

Properties of odd and even integer:

If x is an odd integer, then exist an integer y such that x = 2y + 1.

If x is an even integer, then there exist an integer y such that x = 2y.

Let m + n and n + p be odd integers, then there exist integer y and z such that :

       m + n = 2y + 1

and   n + p = 2z + 1

let add the two odd integers:

m + n+ n + p = 2y + 1 + 2z + 1

m + 2n + p = 2y + 2z + 2

we are interested in m + p , so subtract 2n from both sides:

m + 2n – 2n + p = 2y + 2z+ 2 – 2n

m + p = 2y + 2z – 2n -2

by taking common 2 from right side:

m + p = 2 ( y + z – n + 1)

since y, z and n are integers, y + z – n is also an integer and thus m + p is an even integer.

Hence proved if m, n, and p are integers and m + n and n + p are odd integers. Then, m + p is even integer..

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18