Answer :
Certainly! Let's solve the system of equations both algebraically and graphically.
### System of equations:
[tex]\[ \begin{cases} 1. \quad x + y = 4 \\ 2. \quad x^2 - 2xy - 3 = 0 \end{cases} \][/tex]
### Step-by-Step Algebraic Solution:
1. Express [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex] from Equation 1:
From the first equation:
[tex]\[ y = 4 - x \][/tex]
2. Substitute [tex]\( y \)[/tex] in Equation 2:
Substitute [tex]\( y = 4 - x \)[/tex] into the second equation:
[tex]\[ x^2 - 2x(4 - x) - 3 = 0 \][/tex]
Simplify this equation:
[tex]\[ x^2 - 2x(4 - x) - 3 = x^2 - 8x + 2x^2 - 3 = 0 \][/tex]
[tex]\[ 3x^2 - 8x - 3 = 0 \][/tex]
3. Solve the quadratic equation:
The quadratic equation is:
[tex]\[ 3x^2 - 8x - 3 = 0 \][/tex]
Use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ a = 3, \quad b = -8, \quad c = -3 \][/tex]
[tex]\[ x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 3 \cdot (-3)}}{2 \cdot 3} \][/tex]
[tex]\[ x = \frac{8 \pm \sqrt{64 + 36}}{6} \][/tex]
[tex]\[ x = \frac{8 \pm \sqrt{100}}{6} \][/tex]
[tex]\[ x = \frac{8 \pm 10}{6} \][/tex]
So, we get two solutions:
[tex]\[ x = \frac{8 + 10}{6} = 3 \quad \text{or} \quad x = \frac{8 - 10}{6} = -\frac{1}{3} \][/tex]
4. Find the corresponding [tex]\( y \)[/tex] values:
For [tex]\( x = 3 \)[/tex]:
[tex]\[ y = 4 - 3 = 1 \][/tex]
For [tex]\( x = -\frac{1}{3} \)[/tex]:
[tex]\[ y = 4 - \left(-\frac{1}{3}\right) = 4 + \frac{1}{3} = \frac{13}{3} \][/tex]
Thus, the solutions are:
[tex]\((x, y) = (3, 1)\)[/tex] and [tex]\((x, y) = \left(-\frac{1}{3}, \frac{13}{3}\right)\)[/tex]
### Verification Algebraically:
- For [tex]\( (3, 1) \)[/tex]:
[tex]\[ 3 + 1 = 4 \quad \text{which satisfies the first equation} \][/tex]
[tex]\[ 3^2 - 2 \cdot 3 \cdot 1 - 3 = 9 - 6 - 3 = 0 \quad \text{which satisfies the second equation} \][/tex]
- For [tex]\( \left(-\frac{1}{3}, \frac{13}{3}\right) \)[/tex]:
[tex]\[ -\frac{1}{3} + \frac{13}{3} = \frac{12}{3} = 4 \quad \text{which satisfies the first equation} \][/tex]
[tex]\[ \left(-\frac{1}{3}\right)^2 - 2 \cdot \left(-\frac{1}{3}\right) \cdot \frac{13}{3} - 3 = \frac{1}{9} + \frac{26}{9} - 3 = \frac{1}{9} + \frac{26}{9} - \frac{27}{9} = 0 \quad \text{which satisfies the second equation} \][/tex]
### Graphical Solution:
1. Equation [tex]\( x + y = 4 \)[/tex]:
Rearrange to find [tex]\( y \)[/tex]:
[tex]\[ y = 4 - x \][/tex]
This is a linear equation, and its graph is a straight line.
2. Equation [tex]\( x^2 - 2xy - 3 = 0 \)[/tex]:
Rearrange to express [tex]\( y \)[/tex]:
[tex]\[ y = \frac{x^2 - 3}{2x} \][/tex]
This is a non-linear equation and its graph is a hyperbola.
3. Plot these equations:
- The straight line [tex]\( y = 4 - x \)[/tex]
- The hyperbola [tex]\( y = \frac{x^2 - 3}{2x} \)[/tex]
4. Locate the points of intersection:
Graphically, the points where the line [tex]\( y = 4 - x \)[/tex] intersects the hyperbola [tex]\( y = \frac{x^2 - 3}{2x} \)[/tex] are the solutions to the system of equations. Specifically, the intersections are at points [tex]\((3, 1)\)[/tex] and [tex]\(\left(-\frac{1}{3}, \frac{13}{3}\right)\)[/tex].
### Final Answer:
The solutions to the system of equations are:
[tex]\[ \boxed{(3, 1) \text{ and } \left(-\frac{1}{3}, \frac{13}{3}\right)} \][/tex]
### System of equations:
[tex]\[ \begin{cases} 1. \quad x + y = 4 \\ 2. \quad x^2 - 2xy - 3 = 0 \end{cases} \][/tex]
### Step-by-Step Algebraic Solution:
1. Express [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex] from Equation 1:
From the first equation:
[tex]\[ y = 4 - x \][/tex]
2. Substitute [tex]\( y \)[/tex] in Equation 2:
Substitute [tex]\( y = 4 - x \)[/tex] into the second equation:
[tex]\[ x^2 - 2x(4 - x) - 3 = 0 \][/tex]
Simplify this equation:
[tex]\[ x^2 - 2x(4 - x) - 3 = x^2 - 8x + 2x^2 - 3 = 0 \][/tex]
[tex]\[ 3x^2 - 8x - 3 = 0 \][/tex]
3. Solve the quadratic equation:
The quadratic equation is:
[tex]\[ 3x^2 - 8x - 3 = 0 \][/tex]
Use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ a = 3, \quad b = -8, \quad c = -3 \][/tex]
[tex]\[ x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 3 \cdot (-3)}}{2 \cdot 3} \][/tex]
[tex]\[ x = \frac{8 \pm \sqrt{64 + 36}}{6} \][/tex]
[tex]\[ x = \frac{8 \pm \sqrt{100}}{6} \][/tex]
[tex]\[ x = \frac{8 \pm 10}{6} \][/tex]
So, we get two solutions:
[tex]\[ x = \frac{8 + 10}{6} = 3 \quad \text{or} \quad x = \frac{8 - 10}{6} = -\frac{1}{3} \][/tex]
4. Find the corresponding [tex]\( y \)[/tex] values:
For [tex]\( x = 3 \)[/tex]:
[tex]\[ y = 4 - 3 = 1 \][/tex]
For [tex]\( x = -\frac{1}{3} \)[/tex]:
[tex]\[ y = 4 - \left(-\frac{1}{3}\right) = 4 + \frac{1}{3} = \frac{13}{3} \][/tex]
Thus, the solutions are:
[tex]\((x, y) = (3, 1)\)[/tex] and [tex]\((x, y) = \left(-\frac{1}{3}, \frac{13}{3}\right)\)[/tex]
### Verification Algebraically:
- For [tex]\( (3, 1) \)[/tex]:
[tex]\[ 3 + 1 = 4 \quad \text{which satisfies the first equation} \][/tex]
[tex]\[ 3^2 - 2 \cdot 3 \cdot 1 - 3 = 9 - 6 - 3 = 0 \quad \text{which satisfies the second equation} \][/tex]
- For [tex]\( \left(-\frac{1}{3}, \frac{13}{3}\right) \)[/tex]:
[tex]\[ -\frac{1}{3} + \frac{13}{3} = \frac{12}{3} = 4 \quad \text{which satisfies the first equation} \][/tex]
[tex]\[ \left(-\frac{1}{3}\right)^2 - 2 \cdot \left(-\frac{1}{3}\right) \cdot \frac{13}{3} - 3 = \frac{1}{9} + \frac{26}{9} - 3 = \frac{1}{9} + \frac{26}{9} - \frac{27}{9} = 0 \quad \text{which satisfies the second equation} \][/tex]
### Graphical Solution:
1. Equation [tex]\( x + y = 4 \)[/tex]:
Rearrange to find [tex]\( y \)[/tex]:
[tex]\[ y = 4 - x \][/tex]
This is a linear equation, and its graph is a straight line.
2. Equation [tex]\( x^2 - 2xy - 3 = 0 \)[/tex]:
Rearrange to express [tex]\( y \)[/tex]:
[tex]\[ y = \frac{x^2 - 3}{2x} \][/tex]
This is a non-linear equation and its graph is a hyperbola.
3. Plot these equations:
- The straight line [tex]\( y = 4 - x \)[/tex]
- The hyperbola [tex]\( y = \frac{x^2 - 3}{2x} \)[/tex]
4. Locate the points of intersection:
Graphically, the points where the line [tex]\( y = 4 - x \)[/tex] intersects the hyperbola [tex]\( y = \frac{x^2 - 3}{2x} \)[/tex] are the solutions to the system of equations. Specifically, the intersections are at points [tex]\((3, 1)\)[/tex] and [tex]\(\left(-\frac{1}{3}, \frac{13}{3}\right)\)[/tex].
### Final Answer:
The solutions to the system of equations are:
[tex]\[ \boxed{(3, 1) \text{ and } \left(-\frac{1}{3}, \frac{13}{3}\right)} \][/tex]
