the numbers.
Find two numbers such that the sum of twice the first and thrice the
second is 92, and four times the first exceeds seven times the second
by 2.
Answers
Step-by-step explanation:
Given :-
The sum of twice the first and thrice the second is 92, and four times the first exceeds seven times the second number .
To find :-
Find the two numbers ?
Solution :-
Let the two numbers be X and Y
Let the first number = X
Let the second number = Y
Twice the first number = 2X
Thrice the second number = 3Y
Given that
The sum of twice the first number and thrice the
second number = 92
=> 2X+3Y = 92 ------------(1)
On multiplying with 2 both sides
=> 4X+6Y = 184 ------------(2)
Four times the first number = 4X
Seven times the second number = 7Y
Given that
Four times the first exceeds seven times the second by 2.
=> 4X = 7Y +2
=> 4X - 7Y = 2 ---------------(3)
On Subtracting (3) from (2)
4X+6Y = 184
4X - 7Y = 2
(-)
__________
0 + 13Y = 182
___________
=> 13Y = 182
=> Y = 182/13
=> Y = 14
On Substituting the value of Y in (1) then
=>2X+3(14) = 92
=> 2X +42 = 92
=> 2X = 92-42
=> 2X = 50
=>X = 50/2
=> X = 25
Therefore, X = 25 and Y = 14
The first number = 25
The second number = 14
Answer:-
The required two numbers for the given problem are 25 and 14
Check :-
The two numbers = 25 and 14
The sum of twice the first number and thrice the
second number
=> 2(25)+3(14)
=> 50+42
=>92
and
=> 4(25)-7(14)
=> 100 -98
=> 2
Four times the first exceeds seven times the second by 2.
Verified the given relations in the given problem.