The sum of 3 numbers is 20. The second number is 4 times the first and the sum of the first and third is 8. Find the numbers.

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

The sum of three numbers is 20, so, mathematically it can be written as $x+y+z=20$

The second number is 4 times the first number.

This can be mathematically written as $y=4x\Rightarrow 4x-y=0$.

The sum of first and third number is 8.

This can be mathematically written as $x+z=8$.

So, the system of equations will be:

Add the$equationx+y+z=20$to the equation$4x-y=0$.

localid="1647998459082" $\begin{array}{l}\underset{}{\begin{array}{l}x+y+z=20\\ 4x-y+=0\end{array}}\\ 5x+0+z=20\end{array}$

So, the resultant equation is $5x+z=20$

Subtract the equation$5x+z=20$from the equation $x+z=8$

$\begin{array}{l}\underset{\_}{\begin{array}{l}5x+z=20\\ (-)x+z=8\end{array}}\\ 4x+0=12\end{array}$

Solve $4x=12$for $x$:

$\begin{array}{c}4x=12\\ \frac{4x}{4}=\frac{12}{4}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Dividebothsidesby}4\\ x=3\end{array}$

Substitute $x=3in4x-y=0$and find the value of $y$

$\begin{array}{c}4x-y=0\\ 4\left(3\right)-y=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Substitute}3\text{for}x\\ 12-y=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Simplify}\\ -y=-12\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Subtract}12\text{frombothsides}\\ y=12\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Dividebothsidesby}-1\end{array}$

Substitutelocalid="1647998527720" $x=3inx+z=8$and find the value of$z$

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

Hence, the solution of the given system of equations is$\left(x,y,z\right)=\left(3,12,5\right)$.

So, the three numbers are $3,12,and5$.

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