The value of x that would make the line segment TV to be parallel to the line segment QS is;
Option C; 10
The image of the line segment is missing and so i have attached it.
From the attached image of the triangle, we can use the theorem of similar triangles to get;
RT/RQ = RV/RS
We see that;
RT = x + 4
RQ = x + 4 + x - 3 = 2x + 1
RV = x + 10
RS = x + 10 + x = 2x + 10
Thus;
(x + 4)/(2x + 1) = (x + 10)/(2x + 10)
Cross multiply to get;
(x + 4)(2x + 10) = (x + 10)(2x + 1)
2x² + 18x + 40 = 2x² + 21x + 10
2x² will cancel out to give;
18x + 40 = 21x + 10
21x - 18x = 40 - 10
3x = 30
x = 30/3
x = 10
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