Select the correct answer.

What is the solution of this linear system?
[tex]$
\left\{
\begin{array}{c}
p + q + r = 32 \\
p - r = 4 \\
2p + q = 36
\end{array}
\right.
$[/tex]

A. [tex][tex]$(16, 8, 8)$[/tex][/tex]
B. [tex][tex]$(14, 10, 8)$[/tex][/tex]
C. [tex][tex]$(14, 8, 10)$[/tex][/tex]
D. [tex][tex]$(10, 12, 10)$[/tex][/tex]
E. [tex][tex]$(10, 15, 10)$[/tex][/tex]



Answer :

To solve this system of linear equations:
[tex]\[ \left\{ \begin{array}{c} p + q + r = 32 \\ p - r = 4 \\ 2p + q = 36 \\ \end{array} \right. \][/tex]
we will solve it step-by-step.

### Step 1: Use the second equation to express [tex]\( p \)[/tex] in terms of [tex]\( r \)[/tex]
From the second equation [tex]\( p - r = 4 \)[/tex], we can solve for [tex]\( p \)[/tex]:
[tex]\[ p = r + 4 \][/tex]

### Step 2: Substitute [tex]\( p = r + 4 \)[/tex] into the third equation
The third equation is [tex]\( 2p + q = 36 \)[/tex]. Substitute [tex]\( p = r + 4 \)[/tex]:
[tex]\[ 2(r + 4) + q = 36 \][/tex]
Simplify this equation:
[tex]\[ 2r + 8 + q = 36 \][/tex]
[tex]\[ 2r + q = 28 \][/tex]
[tex]\[ q = 28 - 2r \][/tex]

### Step 3: Substitute [tex]\( p = r + 4 \)[/tex] and [tex]\( q = 28 - 2r \)[/tex] into the first equation
The first equation is [tex]\( p + q + r = 32 \)[/tex]. Substitute [tex]\( p = r + 4 \)[/tex] and [tex]\( q = 28 - 2r \)[/tex]:
[tex]\[ (r + 4) + (28 - 2r) + r = 32 \][/tex]
Combine like terms:
[tex]\[ r + 4 + 28 - 2r + r = 32 \][/tex]
Simplify:
[tex]\[ 32 = 32 \][/tex]

Since 32 equals 32, consistency is verified, and we can find the values of [tex]\( r \)[/tex].

### Step 4: Solve for [tex]\( r \)[/tex]
At this point, the values satisfy the equations, and we can determine each variable. Recall our equations for [tex]\( p \)[/tex] and [tex]\( q \)[/tex]:
[tex]\[ p = r + 4 \][/tex]
[tex]\[ q = 28 - 2r \][/tex]

### Step 5: Substitute the possible value of [tex]\( r \)[/tex] back to find specific solutions
To find the values of [tex]\( p \)[/tex], [tex]\( q \)[/tex], and [tex]\( r \)[/tex], let's assume [tex]\( r = 10 \)[/tex] from possible values series given:

For [tex]\( r = 10 \)[/tex]:
[tex]\[ p = 10 + 4 = 14 \][/tex]
[tex]\[ q = 28 - 2 \cdot 10 = 28 - 20 = 8 \][/tex]

### Conclusion
The possible values satisfying all the equations are [tex]\( p = 14 \)[/tex], [tex]\( q = 8 \)[/tex], [tex]\( r = 10 \)[/tex].

Therefore, the correct answer is:
[tex]\[ \boxed{(14, 8, 10)} \][/tex]

So, the solution is given in option:
C. [tex]\((14, 8, 10)\)[/tex]