Question 8 (4 points) Find the values of x and y if NO = x + 2y, MO = 3x, PQ = 6x + 4, and OQ = 2y + 4.
1 answer:
Answer:
<em>x = 12 and y = 16</em>
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
<em>Hence x = 12 and y = 16</em>
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