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Explanation:
You have the right idea by saying that 2MO = KL which leads to the equation 2(6x-1) = 13x-10. That equation correctly solves to x = 8. In short, everything on your paper so far looks good.
Now use that x value to find the lengths of MO and KL
The jump from MO = 47 to KL = 94 is exactly "times 2". This helps confirm you have the correct x solution.
Because the midsegment MO doubles to KL, the same can be said about the other midsegments. Midsegment MN = 49 doubles to LJ = 98. Also, midsegment NO = 42 doubles to JK = 84. Each midsegment is parallel to the side it doubles to.
Once we know that,
we then add up the three sides to get the perimeter of triangle JKL
perimeter = JK + KL + LJ
perimeter = 84 + 94 + 98
perimeter = 276
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Here's another approach:
Use the value of x to find side MO
MO = 6x-1 = 6*8-1 = 47
Then we can say,
perimeter of MNO = MN+NO+OM = 49 + 42 + 47 = 138
We double this smaller perimeter value to get 2*138 = 276 which is the perimeter of triangle JKL.
This works because we're doubling each midsegment to get the corresponding longer parallel side
perimeter of JKL = JK + KL + LJ
perimeter of JKL = 2*NO + 2*MO + 2*MN
perimeter of JKL = 2*(NO + MO + MN)
perimeter of JKL = 2*(perimeter of MNO)