Sure, let's solve the following system of linear equations step-by-step:
[tex]\[
\begin{cases}
4p + q = 19 \\
-5p + 4q = -29
\end{cases}
\][/tex]
Step 1: Write both equations in standard form:
[tex]\[
4p + q = 19 \quad \text{(Equation 1)}
\][/tex]
[tex]\[
-5p + 4q = -29 \quad \text{(Equation 2)}
\][/tex]
Step 2: Eliminate one variable by making the coefficients of that variable equal in magnitude. Let's eliminate \( q \). To do this, we'll multiply Equation 1 by 4:
[tex]\[
4(4p + q) = 4 \times 19
\][/tex]
[tex]\[
16p + 4q = 76 \quad \text{(Equation 3)}
\][/tex]
Step 3: Subtract Equation 2 from Equation 3:
[tex]\[
(16p + 4q) - (-5p + 4q) = 76 - (-29)
\][/tex]
[tex]\[
16p + 4q + 5p - 4q = 76 + 29
\][/tex]
[tex]\[
21p = 105
\][/tex]
Step 4: Solve for \( p \):
[tex]\[
p = \frac{105}{21}
\][/tex]
[tex]\[
p = 5
\][/tex]
Step 5: Substitute \( p = 5 \) back into Equation 1 to find \( q \):
[tex]\[
4(5) + q = 19
\][/tex]
[tex]\[
20 + q = 19
\][/tex]
[tex]\[
q = 19 - 20
\][/tex]
[tex]\[
q = -1
\][/tex]
Therefore, the solution to the system of equations is:
[tex]\[
p = 5, \quad q = -1
\][/tex]
