Q24. - Algebra 2

Question

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.

Step-by-Step Solution

Answer

The three numbers are $3,12,$ and .5

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 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:

.$\begin{array}{c}x+y+z=20\\ 4x-y=0\\ x+z=8\end{array}$

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

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

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

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

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

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

Step 3 – Find the values of x , y and z .

Solve $4x=12$ for $x$:

$\begin{array}{c}4x=12\\ \frac{4x}{4}=\frac{12}{4}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Divide both sides by }4\\ x=3\end{array}$

Substitute $x=3 in 4x-y=0$ and find the value of $y$

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

Substitute$x=3 in x+z=8$ and find the value of$z$

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

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 $3,12,and 5$.

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