Question

divide 543 into three part such that the second part will be 32 less than twice the largest one be 18 more than the first.

Answer 1

the question is not clear at all
bt I concept it as the following solution
let the three parts be a,b,c
where b=2a-32 and c=a+18...taking the first no 'a' instead of largest ..
now by question a+b+c=543
implies a+2a-32+a+18=543
4a-14=543 implies a=139.25,
b=2a-32=246.50
c=a+18=157.25

Answer 2

$Please \: Mark \: Me \: As \: BRAINLIEST. \\ .................................................. \\ Also \: Don't \: Forgot \: to \: Follow \: me. \\ .............. \\ As \: we \: know \: about \: Equations. \\ So, \\ Hello \: Guys, \\ ................ \\ Answer \: is$
$Let \: first \: no. \: be \: x. \\ \: \: \: \: Second \: no. = 2 \times x - 32 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2x - 32 \\ \: \: \: \: \: \: \: Third \: no. = x + 18 \\ ATQ, \\ x + 2x - 32 + x + 18 = 534 \\ 4x - 14 = 534 \\ 4x = 534 + 14 \\ 4x = 548 \\ x = \frac{548}{4} \\ x = 137 \\ \\ First \: no. = 137 \\ Second \: no. = 137 \times 2 - 32 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 274 - 32 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 242 \\ Third \: no. = 137 + 18 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 155$