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Question

If x+3z+42y76a10b3210=063y2632c+22b+4210
Find the values of a,b,c,x,y and z.

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Solution

Given:x+3z+42y76a10b3210=063y2632c+22b+4210
Comparing corresponding elements,
x+3=0x=03
x=3
2y7=3y22y3y=2+7
y=5
z+4=6z=64
z=2
a1=3a=3+1
a=2
b3=2b+4b2b=4+3
b=7
2c+2=0
c=1
a=2,b=7,c=1,x=3,y=5 and z=2.

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