The sum of 3 numbers is 12. The first number is twice the sum of the second and third. The third number is 5 less than the first. Find the numbers.

Let the three numbers be$x,y,$and $z$.

The sum of three numbers is 12, so, mathematically it can be written as $x+y+z=12$.

The first number is twice the second and third.

This can be mathematically written as $\$x=2\backslash left(y+z\backslash right)$$\backslash Rightarrowx-2y-2z=0$.

The third number is 5 less than the first.

This can be mathematically written as $z=x-5\Rightarrow -x+z=-5$.

So, the system of equations will be:

$\begin{array}{c}x+y+z=12\\ x-2y-2z=0\\ -x+z=-5\end{array}$

Add the $equationx+y+z=12$to the equation$-x+z=-5$.

$\begin{array}{l}\underset{\_}{\begin{array}{l}x+y+z=12\\ -x++z=-5\end{array}}\\ y+2z=7\end{array}$

So, the resultant equation is $y+2z=7$

Subtract the equation$x-2y-2z=0$from the equation $-x+z=-5$

$\begin{array}{l}\underset{\_}{\begin{array}{l}x-2y-2z=0\\ -x++z=-5\end{array}}\\ -2y-z=-5\end{array}$

So, the resultant equation is $-2y-z=-5$.

Multiply $y+2z=7$by 2 and add the new resultant equation to $-2y-z=-5.$

$\begin{array}{l}\underset{\_}{\begin{array}{l}y+2z=7\\ -2y-z=-5\end{array}}\begin{array}{c}\text{multiplyby2}\\ \end{array}\underset{\_}{\begin{array}{l}2y+4z=14\\ -2y-z=-5\end{array}}\\ 0+3z=9\end{array}$

Solve $3z=9forx$

$\begin{array}{c}3z=9\\ \frac{3z}{3}=\frac{9}{3}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}dividebothsidesby3}\\ z=3\end{array}$

Substitute $z=3$in $y+2z=7$and find the value of $y$.

$\begin{array}{c}y+2z=7\\ y+2\left(3\right)=7\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Substitute}3\text{for}y\\ y+6=7\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Simplify}\\ y=1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Subtract}6\text{frombothsides}\end{array}$

Substitute $z=3$in $-x+z=-5$and find the value of $x$.

$\begin{array}{c}-x+z=-5\\ -x+3=-5\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}substitute}3\text{for}z\\ -x=-8\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Subtract}3\text{\hspace{0.17em}frombothsides}\\ x=8\text{Dividebothsidesby}-1\end{array}$

Hence, the solution of the given system of equations is$\left(x,y,z\right)=\left(8,1,3\right)$.

So, the three numbers are $8,1,$and $3$.