how we should do it

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how we should do it

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According to the question x+y = 80

It is given that 5x = 3y

x=3y÷5

x+y=3y÷5+y=80

8y÷5=80

8y=80×5=400

y=400÷8=50

x+y=80

x+50=80

x=80-50=30

The 2 numbers are 30 and 50.

x+y=80 (eqn 1)

and 5x=3y (eqn 2)

Therefore, x=3y/5 (from eqn 2)

substituting value for x from above to eqn 1

3y/5 + y = 80

3y + 5y = 80 × 5

8y = 80 × 5

y = 50

Therefore,

x=80-y

x=30

The numbers are 30 and 50.