The sum of three numbers is 57. If twice the first number is 6 less than the second and the second number in 10 more than the third,find numbers.
Question from Ch-7(Linear Equation) Class VIII.Please help.
Answers
Answer:-
Given Information:-
Sum of the numbers = 57
Condition given: Twice the first number is 6 less than the second. The second one is 10 more than the third.
To Find:-
The numbers
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Let the first number be = (x)
As per the condition, the second one becomes:-
= (2x + 6)
[Since the second number is six more than twice the number]
THEREFORE, third one is
= (2x + 6) - 10
[Since the third number is 10 less than the second number]
= (2x - 4)
As per the Question:-
(First number) + (Second number) + (Third number) = 57
That is,
(x) + (2x + 6) + (2x - 4) = 57
(Equation formation as per the Question)
→ x + 2x + 6 + 2x - 4 = 57
(Removed the brackets)
→ 5x + 2 = 57
(Added the like terms)
→ 5x = 57 - 2 = 55
(Constant terms are moved to RHS)
→ x = 55/5
(Moved 5 to the RHS to fetch the value of x)
→ x = 11
(Value of x)
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Now, let us put the value:-
Required Answer:-
★ First Number = x = 11
★ Second Number = (2x + 6) = 28
★ Third Number = (2x - 4) = 18
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Verification of the Conditions:-
Sum of the numbers = 11 + 28 + 18 = 57 (✓)
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TheAssassin's Reply:-
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Answer:
Dude your answer is here:-
Step-by-step explanation:
Let three number x, y, z
then we have given
x+y+z=57
and
twice of first number is 6 less then second no.
so,
2x+6=y
and second is 10 more then third
so,
y-10=z
from these equations
we come that..
put y and z in eq 1
x+2x+6+2x-4=57
5x+2=57
5x=55
x=11
and then 2x+6=y
22+6=y
y=28
z=y-10
z=28-10
z=18
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