Q25. - Algebra 2

Question

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.

Step-by-Step Solution

Answer

The three numbers are$8,1,$ and .3

Step 1 – First construct a system of equations corresponding to the given situation.

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 Rightarrow x-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:

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

Step 2 – Use the elimination method to get the system of equations in two variables.

Add the $equation x+y+z=12$ 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 $-2y-z=-5$.

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Step 3 – Solve the system of equations in two variables

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

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

Solve $3z=9for x$

$\begin{array}{c}3z=9\\ \frac{3z}{3}=\frac{9}{3}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} divide both sides by 3}\\ z=3\end{array}$

Step 4 – Find the values of y and z .

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

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

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

$\begin{array}{c}-x+z=-5\\ -x+3=-5\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}substitute }3\text{ for }z\\ -x=-8\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Subtract }3\text{\hspace{0.17em}from both sides}\\ x=8\text{ Divide both sides by }-1\end{array}$

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 $3$.