Question

One number is 1 more than 7 times the other number. If their sum is 73 more than

twice the smaller number. Find the numbers.​

Answer 1

`\boxed{\boxed{\pink{\bf The \ two \ numbers \ are \ 12 \& \ 85 . }}}`

Explanation:

Given that , One number is 1 more than 7 times the other number. If their sum is 73 more than

twice the smaller number. And we need to find the numbers. So let us take the

• First Number be x .
• Second number be 7x+1 .

`\underline{\red{\boldsymbol{ According \ to \ the \ Question :- }}}`

`\bf\implies 2( Small \ no.) + 73 = Bigger \ no. + Smaller \ no.\\\\\bf\implies 2x + 73 = x + 7x + 1\\\\\bf\implies 2x + 73 = 8x + 1 \\\\\bf\implies 8x - 2x = 73-1 \\\\\bf\implies 6x = 72 \\\\\bf\implies x =(72)/(6) \\\\\bf\implies \boxed{\red{\bf x = 12 }}`

Hence the smaller number is 12 and the bigger number will be :-

`\bf \implies Number_(bigger) = 7x + 1 \\\\\bf\implies Number_(bigger) =7(12)+1\\\\\bf\implies Number_(bigger) = 84 + 1 \\\\\bf\implies \boxed{\red{\bf Number_(bigger)=85 }}`

Hence the bigger number is 85 and the smaller number is 12.

Answer 2

Explanation:

let us say the number: x

7x+1=y

x+y=73+2x

y-73=2x-x

y-73=x or

x=y-73

then, 7x+1=y

x=y-73

then after,we'll substitute

7x+1=y

7(y-73)+1=y

7y-511+1=y

7y-510=y

7y-y=510

6y=510

y=510/6

y=85

x=y-73

x=85-73

x=12

Therefore y=85 & x=12