Certainly! Let's solve the given system of linear equations using the method of cross multiplication.
The system of equations is:
1. [tex]\( x - y = 1 \)[/tex]
2. [tex]\( 2x + y = 8 \)[/tex]
First, write the equations in the standard form [tex]\( ax + by + c = 0 \)[/tex]:
1. [tex]\( x - y - 1 = 0 \)[/tex]
2. [tex]\( 2x + y - 8 = 0 \)[/tex]
The standard form for the equations is:
[tex]\[
\begin{aligned}
&1. \quad x - y - 1 = 0 \\
&2. \quad 2x + y - 8 = 0
\end{aligned}
\][/tex]
With these equations, we now identify the coefficients:
[tex]\[
\begin{aligned}
&1. \quad a_1 = 1, \quad b_1 = -1, \quad c_1 = -1 \\
&2. \quad a_2 = 2, \quad b_2 = 1, \quad c_2 = -8
\end{aligned}
\][/tex]
Using the method of cross multiplication, we set up the following determinant formula:
[tex]\[
\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}
\][/tex]
Substitute the coefficients into the formula:
[tex]\[
\frac{x}{(-1)(-8) - (1)(-1)} = \frac{y}{(-1)(2) - (-8)(1)} = \frac{1}{(1)(1) - (2)(-1)}
\][/tex]
Simplify each part of the ratios:
[tex]\[
\frac{x}{8 + 1} = \frac{y}{-2 + 8} = \frac{1}{1 + 2}
\][/tex]
This simplifies further to:
[tex]\[
\frac{x}{9} = \frac{y}{6} = \frac{1}{3}
\][/tex]
Now, set each ratio equal to [tex]\(\frac{1}{3}\)[/tex]:
[tex]\[
\begin{aligned}
&\frac{x}{9} = \frac{1}{3} \\
&\frac{y}{6} = \frac{1}{3}
\end{aligned}
\][/tex]
Solve these equations for [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[
\begin{aligned}
&x = \frac{9}{3} = 3 \\
&y = \frac{6}{3} = 2
\end{aligned}
\][/tex]
Therefore, the solutions to the given system of equations are:
[tex]\[
x = 3 \quad \text{and} \quad y = 2
\][/tex]
So, [tex]\(x = 3\)[/tex] and [tex]\(y = 2\)[/tex] is the solution.
