Question

5. 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
sont 3y = 92;
49-7= 2
by 2.​

Answer 1

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Answer 2

$\sf\large\underline\purple{Let:-}$

$\tt{\implies The\:first\:_{(number)}=x}$

$\tt{\implies The\:second\:_{(number)}=y}$

$\sf\large\underline\purple{To\: Find:-}$

$\sf{\implies The\:number=?}$

$\sf\large\underline\purple{Solution:-}$

• To calculate the number ,at first we have to set up equation as per the given clue in the question:

$\sf\small\underline\green{Given\:in\:case\:(i):-}$

• Two numbers such that the sum of twice the first and thrice the second is 92:-]

$\tt{\implies First*2+second*3=92}$

$\tt{\implies 2x+3y=92------(i)}$

$\sf\small\underline\green{Given\:in\:case\:(ii):-}$

• 4 times the first exceeds 7 times of second by 2:-]

$\tt{\implies 4x=7y+2}$

$\tt{\implies 4x-7y=2-----(ii)}$

• In eq (i) multiply by 4 and (ii) by 2 then subtract

$\tt{\implies 8x+12y=368}$

$\tt{\implies 8x-14y=4}$

• By solving we get, here:-]

$\tt{\implies 26y=364}$

$\tt{\implies y=14}$

• Putting the value of y=14 in (I):-]

$\tt{\implies 2x+3y=92}$

$\tt{\implies 2x+3*14=92}$

$\tt{\implies 2x+42=92}$

$\tt{\implies 2x=92-42}$

$\tt{\implies 2x=50}$

$\tt{\implies x=25}$

$\sf\large{Hence,}$

$\tt\blue{\implies The\:first\:_{(number)}=25}$

$\tt\red{\implies The\:second\:_{(number)}=14}$