Question

The sum of one-third of a number and three-fourths of the number exceeds that number by one. Which equation could be used to find the number? n = n + 1 n + n = n - 1 n + n = n + 1

Answer 1

$\frac{n}{3} + \frac{3n}{4} = n+1$
$4n + 9n = n+1$
$13n = n+1$
$13n-n = 1$
$12n = 1$

Answer 2

Answer:

$\frac{n}{3}$ + $\frac{3n}{4}$ = n +1.

Number = 12.

Step-by-step explanation:

Given : The sum of one-third of a number and three-fourths of the number exceeds that number by one.

To find : Which equation could be used to find the number.

Solution : We have given a statement

Let a number  = n

According to question

One third of a number = $\frac{n}{3}$.

Three - fourths of the number  =  $\frac{3n}{4}$.

Sum of their is  exceeds that number by one.

$\frac{n}{3}$ + $\frac{3n}{4}$ = n +1.

Taking common denominator

$\frac{4n+9n}{12}$ = n+1 .

$\frac{13n}{12}$ = n+1.

On subtracting n both sides.

$\frac{13n}{12}$ -n = 1

Taking common denominator

$\frac{13n -12n}{12}$ = 1

$\frac{n}{12}$ = 1

On multiplying by 12 both sides

n= 12.

Therefore, $\frac{n}{3}$ + $\frac{3n}{4}$ = n +1.

Number = 12.