Question

परोपकारिणी परोपकार आद​

Answer 1

Solution :-

Let the two numbers be x and y.

According to the first condition :-

: \implies \sf \dfrac{x}{y} = \dfrac{4}{3}:⟹

y

x

=

3

4

: \implies \sf x = \dfrac{4}{3}y \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ................(1):⟹x=

3

4

y ................(1)

According to the second condition :-

: \implies \sf \dfrac{x+2}{y-6} = \dfrac{7}{4}:⟹

y−6

x+2

=

4

7

: \implies \sf 4(x+2)=7(y-6):⟹4(x+2)=7(y−6)

: \implies \sf 4x+8 = 7y-42:⟹4x+8=7y−42

: \implies \sf 4x-7y = -50 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ................(2):⟹4x−7y=−50 ................(2)

Substitute the value of x in eq(2) :-

: \implies \sf 4 \times \dfrac{4}{3}y-7y = -50:⟹4×

3

4

y−7y=−50

: \implies \sf \dfrac{16y}{3}-7y = -50:⟹

3

16y

−7y=−50

: \implies \sf \dfrac{16y-21y}{3}=-50:⟹

3

16y−21y

=−50

: \implies \sf 16y-21y = -50 \times 3:⟹16y−21y=−50×3

: \implies \sf -5y = -150:⟹−5y=−150

: \implies \bf y = 30:⟹y=30

Substitute y = 30 in eq(1) :-

: \implies \sf x = \dfrac{4}{3} \times 30:⟹x=

3

4

×30

: \implies \sf x = 4 \times 10:⟹x=4×10

: \implies \bf x = 40:⟹x=40

The two numbers are 40 and 30.