Find three consecutive numbers such that the sum of one-fourth the first and one-fifth the second is five less than one-seventh the third

The numbers are -16,-15 and -14

Step-by-step explanation:

Let

x > the first number

x+1 > the second number

x+2 ---> the third number

we know that

(1/4)x+(1/5)(x+1)=(1/7)(x+2)-5

Multiply by (4*5*7) both sides

35x+28(x+1)=20(x+2)-140*5

35x+28x+28=20x+40-700

43x=-688

x=-16

so

x+1=16+1=-15

x+2=-16+2=-14

The numbers are -16,-15 and -14

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step-by-step explanation: