Question

there are two number. four times of the first one exceeds by 2 from 7 times of the second one . Sum of two times of the first and three times of the second is 92 . Find the number​

Answer 1

Step-by-step explanation:

Correct Question:-

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.

$\bf\underline{\underline{\red{Given:-}}}$

• Two numbers are such that sum of twice the first and thrice the second is 92.

• Four times the first exceeds seven times the second by 2.

$\bf\underline{\underline{\blue{To\:Find:-}}}$

• The two numbers.

$\bf\underline{\underline{\green{Solution:-}}}$

$\sf Let \: the \: first \: number \: be \: x$

$\sf And, \: second \: number \: be \: y$

$\bf {\underline{\pink{Case\: 1:-}}}$

Sum of twice the first and thrice the second is 92.

$\sf 2( First \: number) + 3(Second\: number) = 92$

$\sf \implies 2x + 3y = 92.....(i)$

$\bf{\underline{\orange{Case\: 2:-}}}$

Four times the first exceeds seven times the second by 2.

$\sf 4(First\: number) - 7(Second\: number) = 2$

$\sf \implies 4x - 7y = 2.....(ii)$

$\sf Multiply \: the \: equation\: (i) \: with\: 2$

$\sf \implies 2(2x + 3y = 92)$

$\sf \implies 4x + 6y = 184.....(iii)$

$\sf\underline{ Subtracting \: equation\: (ii) \: from \: (iii)}$

$\sf \implies 4x + 6y - (4x - 7y) = 184 - 2$

$\sf \implies 4x + 6y - 4x + 7y = 182$

$\sf \implies 13y = 182$

$\sf \implies y = \dfrac{182}{13} = 14$

$\sf Substituting\: y = 14\: in \: equation \: (i)$

$\sf\implies 2x + 3y = 92$

$\sf \implies 2x + 3(14) = 92$

$\sf \implies 2x + 42 = 92$

$\sf \implies 2x = 92 - 42$

$\sf\implies 2x = 50$

$\sf \implies x = 25$

$\sf \underline{We\: have:-}$

$\sf \dashrightarrow First \: number = x = 25$

$\sf \dashrightarrow Second \: number = y = 14$

$\underline{\boxed{\purple{\rm \therefore The\: two\: numbers \: are \: 25 \: and \: 14}}}$