Let's solve the given system of linear equations step-by-step.
The equations provided are:
1. [tex]\( 37x + 29y - 95 = 0 \)[/tex]
2. [tex]\( 32x + 29y = 105 \)[/tex]
First, let’s rewrite the first equation in a standard form:
[tex]\[ 37x + 29y = 95 \][/tex]
Now, we have the system of equations:
[tex]\[ \begin{cases}
37x + 29y = 95 \\
32x + 29y = 105
\end{cases} \][/tex]
To solve this system, we can use the method of elimination to eliminate one of the variables.
Step 1: Subtract the second equation from the first equation:
[tex]\[ (37x + 29y) - (32x + 29y) = 95 - 105 \][/tex]
Simplify the equation:
[tex]\[ 5x = -10 \][/tex]
Step 2: Solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{-10}{5} \][/tex]
[tex]\[ x = -2 \][/tex]
Step 3: Substitute [tex]\( x = -2 \)[/tex] back into one of the original equations to solve for [tex]\( y \)[/tex]:
Using the second equation [tex]\( 32x + 29y = 105 \)[/tex]:
[tex]\[ 32(-2) + 29y = 105 \][/tex]
[tex]\[ -64 + 29y = 105 \][/tex]
[tex]\[ 29y = 105 + 64 \][/tex]
[tex]\[ 29y = 169 \][/tex]
Step 4: Solve for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{169}{29} \][/tex]
[tex]\[ y = 5 \][/tex]
Therefore, the solution to the system of equations is:
[tex]\[ x = \frac{1130}{343} \][/tex]
[tex]\[ y = \frac{-5}{343} \][/tex]
Thus, our solutions are:
[tex]\[ x = \frac{1130}{343}, \; y = \frac{-5}{343} \][/tex]
This is the exact solution to the given linear equations.
