Answer
The solutions to the system of equations are:
1. \( x = \dfrac{1 + \sqrt{73}}{4} \), \( y = \dfrac{5 - \sqrt{73}}{2} \)
2. \( x = \dfrac{1 - \sqrt{73}}{4} \), \( y = \dfrac{5 + \sqrt{73}}{2} \)
Solution
Certainly! Let's solve the following system of equations:
1. \( 2x + y = 3 \)
2. \( y = \dfrac{-6}{x + 1} \)
### Step 1: Substitute \( y \) from Equation (2) into Equation (1)
From Equation (2):
\[ y = \dfrac{-6}{x + 1} \]
Substitute this expression for \( y \) into Equation (1):
\[ 2x + \left( \dfrac{-6}{x + 1} \right) = 3 \]
### Step 2: Eliminate the Fraction
Multiply both sides of the equation by \( (x + 1) \) to eliminate the denominator:
\[ 2x(x + 1) + (-6) = 3(x + 1) \]
Expand and simplify:
\[ 2x^2 + 2x - 6 = 3x + 3 \]
### Step 3: Rearrange the Equation into Standard Quadratic Form
Bring all terms to one side to form a quadratic equation:
\[ 2x^2 + 2x - 6 - 3x - 3 = 0 \]
\[ 2x^2 - x - 9 = 0 \]
### Step 4: Solve the Quadratic Equation
Use the quadratic formula:
\[ x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 2 \), \( b = -1 \), and \( c = -9 \).
Calculate the discriminant:
\[ \Delta = b^2 - 4ac = (-1)^2 - 4(2)(-9) = 1 + 72 = 73 \]
Thus:
\[ x = \dfrac{1 \pm \sqrt{73}}{4} \]
### Step 5: Find Corresponding \( y \) Values
Using \( y = \dfrac{-6}{x + 1} \), substitute each \( x \) value.
1. **First Solution:**
\[
x = \dfrac{1 + \sqrt{73}}{4}
\]
\[
y = \dfrac{-6}{\dfrac{1 + \sqrt{73}}{4} + 1} = \dfrac{-6}{\dfrac{1 + \sqrt{73} + 4}{4}} = \dfrac{-6 \times 4}{5 + \sqrt{73}} = \dfrac{-24}{5 + \sqrt{73}}
\]
To rationalize the denominator:
\[
y = \dfrac{-24}{5 + \sqrt{73}} \times \dfrac{5 - \sqrt{73}}{5 - \sqrt{73}} = \dfrac{-24(5 - \sqrt{73})}{25 - 73} = \dfrac{-24(5 - \sqrt{73})}{-48} = \dfrac{5 - \sqrt{73}}{2}
\]
2. **Second Solution:**
\[
x = \dfrac{1 - \sqrt{73}}{4}
\]
\[
y = \dfrac{-6}{\dfrac{1 - \sqrt{73}}{4} + 1} = \dfrac{-6}{\dfrac{1 - \sqrt{73} + 4}{4}} = \dfrac{-6 \times 4}{5 - \sqrt{73}} = \dfrac{-24}{5 - \sqrt{73}}
\]
Rationalizing the denominator:
\[
y = \dfrac{-24}{5 - \sqrt{73}} \times \dfrac{5 + \sqrt{73}}{5 + \sqrt{73}} = \dfrac{-24(5 + \sqrt{73})}{25 - 73} = \dfrac{-24(5 + \sqrt{73})}{-48} = \dfrac{5 + \sqrt{73}}{2}
\]
### Step 6: Summary of Solutions
The system has two solutions:
1. **First Solution:**
\[
x = \dfrac{1 + \sqrt{73}}{4}, \quad y = \dfrac{5 - \sqrt{73}}{2}
\]
2. **Second Solution:**
\[
x = \dfrac{1 - \sqrt{73}}{4}, \quad y = \dfrac{5 + \sqrt{73}}{2}
\]
### Verification
To ensure the solutions are valid, substitute them back into the original equations.
For example, let's verify the first solution:
1. **Check in Equation (1):**
\[
2x + y = 2\left(\dfrac{1 + \sqrt{73}}{4}\right) + \dfrac{5 - \sqrt{73}}{2} = \dfrac{2(1 + \sqrt{73})}{4} + \dfrac{5 - \sqrt{73}}{2} = \dfrac{1 + \sqrt{73}}{2} + \dfrac{5 - \sqrt{73}}{2} = \dfrac{6}{2} = 3
\]
It satisfies Equation (1).
2. **Check in Equation (2):**
\[
y = \dfrac{-6}{x + 1} = \dfrac{-6}{\dfrac{1 + \sqrt{73}}{4} + 1} = \dfrac{-6}{\dfrac{5 + \sqrt{73}}{4}} = \dfrac{-24}{5 + \sqrt{73}} = \dfrac{5 - \sqrt{73}}{2}
\]
It satisfies Equation (2).
Similarly, the second solution can be verified in both equations.
Reviewed and approved by the UpStudy tutoring team
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