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Solve the following simultaneous equations:
[tex]\[
\begin{array}{l}
\left\{
\begin{array}{l}
4p + q = 19 \\
-5p + 4q = -29
\end{array}
\right. \\
p = \square \quad q = \square
\end{array}
\][/tex]

Sagot :

GinnyAnswer
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]
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