To find the solution for the given system of linear equations:
[tex]\[
\begin{cases}
5p - 3r = 1 \\
8p + 6r = 4
\end{cases}
\][/tex]
we need to determine the values of [tex]\( p \)[/tex] and [tex]\( r \)[/tex] that satisfy both equations simultaneously.
### Step 1: Write down the equations
1. [tex]\( 5p - 3r = 1 \)[/tex]
2. [tex]\( 8p + 6r = 4 \)[/tex]
### Step 2: Normalize one of the equations, if possible
From the second equation, we can divide every term by 2 to simplify it:
[tex]\[
4p + 3r = 2
\][/tex]
### Step 3: Rewrite the system
This changes our system to:
[tex]\[
\begin{cases}
5p - 3r = 1 \\
4p + 3r = 2
\end{cases}
\][/tex]
### Step 4: Add the equations
To eliminate [tex]\( r \)[/tex], add the two equations together:
[tex]\[
(5p - 3r) + (4p + 3r) = 1 + 2
\][/tex]
The [tex]\( -3r \)[/tex] and [tex]\( +3r \)[/tex] cancel each other out:
[tex]\[
5p + 4p = 3
\][/tex]
[tex]\[
9p = 3
\][/tex]
### Step 5: Solve for [tex]\( p \)[/tex]
Divide both sides of the equation by 9:
[tex]\[
p = \frac{3}{9} = \frac{1}{3}
\][/tex]
### Step 6: Substitute [tex]\( p \)[/tex] back into one of the original equations
Use the first equation to solve for [tex]\( r \)[/tex]:
[tex]\[
5p - 3r = 1
\][/tex]
Substitute [tex]\( p = \frac{1}{3} \)[/tex]:
[tex]\[
5\left(\frac{1}{3}\right) - 3r = 1
\][/tex]
[tex]\[
\frac{5}{3} - 3r = 1
\][/tex]
### Step 7: Solve for [tex]\( r \)[/tex]
Isolate [tex]\( r \)[/tex] by moving the fraction to the other side:
[tex]\[
\frac{5}{3} - 1 = 3r
\][/tex]
[tex]\[
\frac{5}{3} - \frac{3}{3} = 3r
\][/tex]
[tex]\[
\frac{2}{3} = 3r
\][/tex]
Divide both sides by 3:
[tex]\[
r = \frac{2}{3 \cdot 3} = \frac{2}{9}
\][/tex]
### Step 8: Conclusion
The solution to the system of equations is:
[tex]\[
\left( p, r \right) = \left( \frac{1}{3}, \frac{2}{9} \right)
\][/tex]
So, the correct ordered pair is [tex]\( \left( \frac{1}{3}, \frac{2}{9} \right) \)[/tex]. This matches the option: [tex]\( \left( \frac{1}{3}, \frac{2}{9} \right) \)[/tex].
