When the 3 blocks are rleased form test, they accelerate with a maginitude of 0.5 m/s^2.?

Question

Block 1 has mass M, block 2 has 2M, and block 3 has 2M. What is the coeffieent of kinetic friction between block 2 and the table?

Answer: Let the tensions of on the strings connecting m2 and m3 be T23, and that connectiong m2 and m1 be T12, respectively. Applying Newton's second law with FN= m2g in this case to the system we have

m3g - T = m3a
T23 - µk m2g - T12 = m2a
T12 - m1g = m1a

Adding up the three equations and using m1 = M, m2 = m3 = 2M, we obtain

2Mg- 2µk Mg - Mg = 5 Ma

With a = 0.5m/s^2 this yields µk = 0.372. Thus, the coefficient of kinetic friction is roughly µk = 0.37

If you can, please explain and show haw they solved for µk because I didn't understand that last part. Thank you

Answer 1

2Mg-2µMg-Mg=5Ma
Cancel all the M's

2g-2µg-g=5a

A=0.5
2g-2µg-g=2.5
g-2µg=2.5

group the g's
g(1-2µ)=2.5

Divide both sides by g
1-2µ=2.5/g

Bring 1 over to the other side
-2µ=(2.5/g)-1

Multiply both sides by -1
2µ= -(2.5/g)+1

Divide both sides by 2

µ=.3724489796