Question 1104672: Find the values of x and y in parallelogram PQRS. PT=y,TR=5x+1,QT=2y,TS=6x+10
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
The point T isn't mentioned. However, I'm assuming that T is the intersection of the two diagonals PR and QS.
If that assumption is correct, then the diagonals bisect each other. This means they cut each other in half.
The diagonal PR is cut into PT and TR, both of which are congruent or equal. So PT = TR.
Similarly, the other diagonal QS is split in half as well. The two equal pieces are QT and TS. So QT = TS.
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We're given
PT=y,TR=5x+1,QT=2y,TS=6x+10
Plug those values into the equations set up in the previous section to get
PT = TR
y = 5x+1
and
QT = TS
2y = 6x+10
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Start with 2y = 6x+10
Replace every y with 5x+1. We can do this because y = 5x+1 was found earlier (based on PT = TR)
So,
2y = 6x+10
2( y ) = 6x+10
2( 5x+1 ) = 6x+10
2*5x+2*1 = 6x+10
10x+2 = 6x+10
10x+2-6x = 6x+10-6x
4x+2 = 10
4x+2-2 = 10-2
4x = 8
4x/4 = 8/4
x = 2
If x = 2, then y is...
y = 5x+1
y = 5*x+1
y = 5*2+1
y = 10+1
y = 11
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In summary:
x = 2 and y = 11
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