Absolutely, let’s solve the system using the substitution method step-by-step:
We have the system of equations:
[tex]\[
\left\{
\begin{array}{l}
y = x - 1 \\
xy = 6
\end{array}
\right.
\][/tex]
### Step 1: Substitute [tex]\( y \)[/tex] from the first equation into the second equation.
From the first equation,
[tex]\[ y = x - 1 \][/tex]
Substitute [tex]\( y \)[/tex] in the second equation:
[tex]\[ x(x - 1) = 6 \][/tex]
### Step 2: Simplify the equation.
The equation becomes:
[tex]\[ x^2 - x = 6 \][/tex]
Rearrange it:
[tex]\[ x^2 - x - 6 = 0 \][/tex]
### Step 3: Solve the quadratic equation.
To solve the quadratic equation [tex]\( x^2 - x - 6 = 0 \)[/tex], we will use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For the equation [tex]\( x^2 - x - 6 = 0 \)[/tex]:
[tex]\[ a = 1, \; b = -1, \; c = -6 \][/tex]
Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = (-1)^2 - 4 \cdot 1 \cdot (-6) = 1 + 24 = 25 \][/tex]
Now, solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{-(-1) \pm \sqrt{25}}{2 \cdot 1} = \frac{1 \pm 5}{2} \][/tex]
This gives us two solutions for [tex]\( x \)[/tex]:
[tex]\[ x_1 = \frac{1 + 5}{2} = 3 \][/tex]
[tex]\[ x_2 = \frac{1 - 5}{2} = -2 \][/tex]
### Step 4: Find the corresponding values of [tex]\( y \)[/tex].
For [tex]\( x_1 = 3 \)[/tex]:
[tex]\[ y_1 = x_1 - 1 = 3 - 1 = 2 \][/tex]
For [tex]\( x_2 = -2 \)[/tex]:
[tex]\[ y_2 = x_2 - 1 = -2 - 1 = -3 \][/tex]
### Step 5: Compile the solutions.
The solutions to the system, with [tex]\( x_1 < x_2 \)[/tex], are:
[tex]\[ x_1 = -2, \; y_1 = -3 \][/tex]
[tex]\[ x_2 = 3, \; y_2 = 2 \][/tex]
So, the solutions of the system are:
[tex]\[
\left( x_1 = -2, y_1 = -3 \right) \text{ and } \left( x_2 = 3, y_2 = 2 \right)
\][/tex]
