Answer
The numbers are $\left\{ -1,2,3 \right\}$.
Work Step by Step
Let us consider the first number as x, the second number as y, and the third number as z.
So, we get the three statements:
Thrice the first number, the second number, and two times the third number adds to 5.
Simplify it to get the equation:
$3x+y+2z=5$
Mark it as (I)
Thrice the second number when subtracted from the addition of the first number plus thrice the third number gives 2. Simplify to get:
$\begin{align}
& \left( x+3z \right)-3y=2 \\
& x-3y+3z=2
\end{align}$
Mark it as (II)
Third number when subtracted from twice the first number plus thrice the second number gives 1. It implies
$\begin{align}
& \left( 2x+3y \right)-z=1 \\
& 2x+3y-z=1
\end{align}$
Mark it as (III)
So, the resultant system of equations in variables a, b and c is as shown below:
$\begin{align}
& 3x+y+2z=5 \\
& x-3y+3z=2 \\
& 2x+3y-z=1
\end{align}$
By multiply equation (I) by 3 and eliminating y from equations (I) and (II) we get:
$\begin{align}
& \text{9}x+3y+6z=15 \\
& x-3y+3z=2
\end{align}$
$10x+9z=17$ (IV)
By eliminating y from equations (II) and (III), we get:
$\begin{align}
& x-3y+3z=2 \\
& \text{2}x+3y-z=1
\end{align}$
$3x+2z=3$ (V)
Now, multiply equation (IV) by 3, equation (V) by −10, and then add both resulting equations to get the value of z:
$\begin{align}
& \text{30}x+27z=51 \\
& -30x-20z=-30
\end{align}$
$7z=21$ (VI)
By dividing equation (VI) by 7, we get:
$\begin{align}
& \frac{7z}{7}=\frac{21}{7} \\
& \text{ }z=3 \\
\end{align}$
Putting the value of z in equation (V), we get
$\begin{align}
& 3x+2\left( 3 \right)=3 \\
& 3x+6=3
\end{align}$
Now, subtract 6 from both sides to get the value of x:
$\begin{align}
& 3x+6-6=3-6 \\
& 3x=-3
\end{align}$
By dividing both sides by 3, we get
$\begin{align}
& \frac{3x}{3}=\frac{-3}{3} \\
& \text{ }x=-1 \\
\end{align}$
Substitute the values of x and z in equation (III) to get the value of y:
$\begin{align}
& 2\left( -1 \right)+3y-3=1 \\
& -2+3y-3=1 \\
& 3y-5=1
\end{align}$
Now, add $5$ to both sides of the equation to get the value of y:
$\begin{align}
& 3y-5+5=1+5 \\
& 3y=6
\end{align}$
By dividing both sides by 3, we get,
$\begin{align}
& \frac{3y}{3}=\frac{6}{3} \\
& \text{ }y=2 \\
\end{align}$
Thus, the numbers are $\left\{ -1,2,3 \right\}$.
