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# When four consecutive integers are added, the sum is $46$. Find the integers.

Last updated date: 16th Jul 2024
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Hint: In order to find the four consecutive integers whose sum would be $46$, we must be considering the first integer as any variable, let’s say $x$. Since the numbers are consecutive which means side by side integers or the difference between one another would be one, we will be adding one more to the previous integer and equate it to $46$ . And then solving this equation will give us required integers.

Complete step-by-step solution:
Now let us have a brief regarding linear equations. Now let us learn about linear equations. A linear equation can be expressed in the form any number of variables as required. As the number of the variables increase, the name of the equation simply denotes it. The general equation of a linear equation in a single variable is $ax+b=0$. We can find the linear equation in three major ways. They are: point-slope form, standard form and slope-intercept form.
Now let us find the four consecutive integers whose sum would be $46$.
Let us consider one of them as $x$.
Since the integers are consecutive, the remaining three integers would be $x+1,x+2,x+3$.
Now let us add all the four integers obtained and equate them to $46$.
The equation formed would be,
$\Rightarrow x+x+1+x+2+x+3=46$
Upon solving we get,
\begin{align} & \Rightarrow x+x+1+x+2+x+3=46 \\ & \Rightarrow 4x+6=46 \\ & \Rightarrow 4x=40 \\ & \Rightarrow x=10 \\ \end{align}
We obtain one of the integers as $10$.
The other three integers would be,
\begin{align} & x=10 \\ & \left( x+1 \right)=10+1=11 \\ & \left( x+2 \right)=10+2=12 \\ & \left( x+3 \right)=10+3=13 \\ \end{align}
$\therefore$ The four consecutive integers whose sum is $46$ is $10,11,12,13$

Note: We must have a note while assigning the variable values to the integers because we are supposed to find the consecutive integers. So we must take that point into consideration for sure and then assign. If not considered the answer obtained will not be the required one as there could exist a number of sets of four integers that give us the sum as $46$.