divide 36 into four parts so that if 2 is added to the first part, 2 is subtraced from the second part, the third part is multiplied by 2,and the fourth part is divided by 2, then the resulting number is the same in each case
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let a,b,c,d be 4 numbers after dividing in to 4 parts.
now a+b+c+d=36
again a+2=b-2=c×2=d÷2
=>b-a=4,b=a+4
c=(a+2)/2
d=2(a+2)
now putting all these values in main eqn
a+a+4+(a+2)/2+2a+4=36
=>4a+8+(a+2)/2=36
=>9a+18=72
=>9a=54
*a=6
now b=a+4=6+4=10
again c=(a+2)/2=(6+2)/2=4
now d=2(a+2)=2(6+2)=2×8=16
______hope it will help you
now a+b+c+d=36
again a+2=b-2=c×2=d÷2
=>b-a=4,b=a+4
c=(a+2)/2
d=2(a+2)
now putting all these values in main eqn
a+a+4+(a+2)/2+2a+4=36
=>4a+8+(a+2)/2=36
=>9a+18=72
=>9a=54
*a=6
now b=a+4=6+4=10
again c=(a+2)/2=(6+2)/2=4
now d=2(a+2)=2(6+2)=2×8=16
______hope it will help you
SHREYANSHISIKARIA:
Thank you for your answer but we need to find it in 2 variables one because the chapter is simultaneous linear equations in two variables
but thanks for your effort and this method can also be considered valid
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