Question

find 2 nos. such that the sum of twice the first and thrice the second is 92 and four times the first exceed seven times the second by 2.​

Answer 1

Answer:

y = 14

x = 25

Step-by-step explanation:

• twice the first number + thrice the second should be = 92

• Four times first number should exceed seven times the second.

☞ Now , let's assume numbers to be x and y.

☞ From the given data , we can infer that :

$\implies \sf \: 2x + 3y = 92 \: \longrightarrow \: first \: equation.$

$\implies \sf \: 4x - 7y = 2 \: \longrightarrow \: second \: equation$

Now , multiplying first equation by 2.

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

$\implies \sf \: 4x + 8y = 184$

☞ Now, subtracting first equation from second.

$\: \: \: \: \: 4x - 7y = 2 \\ \\ {\red{- }} \: 4x + 6y = 184 \\ \\ {\boxed{ \: \: 13y = 182 \: \: }}$

$\implies \sf \: 13y = 182$

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

$\boxed{\implies \sf \: y = 14}$

Finding x :

☞ Multiply the first equation by 7.

$\implies \sf \: \green7(2x + 3y) = \green7(92)$

$\implies \sf \: 14x + 21y = 644$

☞ Multiply the second equation by 3.

$\implies \sf \blue3(4x - 7y) = \blue3(2)$

$\implies \sf \: 12x - 21y = 6$

☞ Adding both of the newly formed equation :-

$\: \: \: \: 14x + 21y = 644 \\ \\ \green+ 12x - 21y = 6 \: \\ \\ {\boxed{26x = 650}}$

$\implies \sf \: 26x = 650$

$\implies \sf \: x = \dfrac{650}{26}$

${\sf {\boxed {\sf{ \implies \: x = 25}}}}$