Question 8 (4 points) Find the values of x and y if NO = x + 2y, MO = 3x, PQ = 6x + 4, and OQ = 2y + 4.

x = 12 and y = 16

Step-by-step explanation:

Given the following lengths;

NO = x + 2y,

MO = 3x,

PQ = 6x + 4, and

OQ = 2y + 4.

The following postulates are true;

MN = PQ

Since MO + ON = MN

MN = 3x + x+2y

MN = 4x + 2y

Recall that PQ = 6x+4

Equating both expressions;

4x + 2y=  6x+4

Collect like terms;

4x-6x + 2y = 4

-2x + 2y = 4

x - y = -4 1

Also MO = OQ

3x =  2y + 4

3x-2y= 4 2

from 1; x = -4 + y

Substitute into 2:

3x-2y= 4

3(-4+y)-2y = 4

-12+3y-2y = 4

-12+y = 4

y = 4 + 12

y = 16

Since x = -4+y

x = -4 + 16

x = 12

Hence x = 12 and y = 16

x+14

step-by-step explanation:

the correct answer is x/3+10≥32

hope this !