sum of three numbers is 60 such that twice the first,three times the secondand four times the third together make 180,first number 3 more than the second number,find the numbers.
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Let no be x y z respectively
X+y +z = 60 - 1
2X + 3Y + 4Z = 180 - 2nd
X = y+3 - 3rd
Put value of x in 1&2 we get
2Y + Z = 57 - 4
5Y + 4Z = 174 - 5
Multiply eqn 4 by 4 both sides we get
8Y + 4z = 228
5Y + 4Z = 174
Now subtract both eqn
Y = 18 , X =21 , Z = 21
X+y +z = 60 - 1
2X + 3Y + 4Z = 180 - 2nd
X = y+3 - 3rd
Put value of x in 1&2 we get
2Y + Z = 57 - 4
5Y + 4Z = 174 - 5
Multiply eqn 4 by 4 both sides we get
8Y + 4z = 228
5Y + 4Z = 174
Now subtract both eqn
Y = 18 , X =21 , Z = 21
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