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To solve the system of equations given by [tex]\(x + y - 4 = 0\)[/tex] and [tex]\(x^2 - 2xy - 3 = 0\)[/tex] algebraically and graphically, let's proceed step by step.
### Step 1: Algebraic Solution
1. Solve the first equation for [tex]\(y\)[/tex]:
[tex]\[ x + y = 4 \][/tex]
[tex]\[ y = 4 - x \][/tex]
2. Substitute [tex]\(y = 4 - x\)[/tex] into the second equation:
[tex]\[ x^2 - 2x(4 - x) - 3 = 0 \][/tex]
Simplify the expression inside the equation:
[tex]\[ x^2 - 2x \cdot 4 + 2x^2 - 3 = 0 \][/tex]
[tex]\[ x^2 - 8x + 2x^2 - 3 = 0 \][/tex]
[tex]\[ 3x^2 - 8x - 3 = 0 \][/tex]
3. Solve the quadratic equation [tex]\(3x^2 - 8x - 3 = 0\)[/tex] using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
Here, [tex]\(a = 3\)[/tex], [tex]\(b = -8\)[/tex], and [tex]\(c = -3\)[/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]
- For the positive square root:
[tex]\[ x = \frac{8 + 10}{6} = \frac{18}{6} = 3 \][/tex]
- For the negative square root:
[tex]\[ x = \frac{8 - 10}{6} = \frac{-2}{6} = -\frac{1}{3} \][/tex]
4. Substitute these values of [tex]\(x\)[/tex] back into [tex]\(y = 4 - x\)[/tex]:
- When [tex]\(x = 3\)[/tex]:
[tex]\[ y = 4 - 3 = 1 \][/tex]
- When [tex]\(x = -\frac{1}{3}\)[/tex]:
[tex]\[ y = 4 - \left(-\frac{1}{3}\right) = 4 + \frac{1}{3} = \frac{12}{3} + \frac{1}{3} = \frac{13}{3} \][/tex]
Therefore, the algebraic solutions are:
[tex]\[ (x, y) = (3, 1) \quad \text{and} \quad \left(-\frac{1}{3}, \frac{13}{3}\right) \][/tex]
### Step 2: Graphical Solution
To graphically solve the equations, we need to plot both equations and find their intersections.
1. Equation 1:
[tex]\[ x + y = 4 \quad \text{or} \quad y = 4 - x \][/tex]
This is a straight line with slope [tex]\(-1\)[/tex] and y-intercept 4.
2. Equation 2:
[tex]\[ x^2 - 2xy - 3 = 0 \][/tex]
We can rewrite this as:
[tex]\[ y = \frac{x^2 - 3}{2x} \][/tex]
This represents a hyperbola, as there are [tex]\((x, y)\)[/tex] values which are obtained through the same algebraic manipulation as shown above.
3. Plot the equations:
- The straight line [tex]\(y = 4 - x\)[/tex]
- The hyperbola derived from [tex]\(x^2 - 2xy - 3 = 0\)[/tex]
By plotting both, you'll find the intersection points, confirming the algebraic solutions [tex]\((3, 1)\)[/tex] and [tex]\(\left(-\frac{1}{3}, \frac{13}{3}\right)\)[/tex].
The following graphical representation shows the system:
[tex]\[ \begin{array}{c} \text{Graph for } y = 4 - x \text{ and } y = \frac{x^2 - 3}{2x}: \\ \text{(Graph not actually shown here) } \end{array} \][/tex]
This completes the algebraic and graphical solutions for the system of equations. The intersection points observed graphically would match the solutions found algebraically: [tex]\((3, 1)\)[/tex] and [tex]\(\left(-\frac{1}{3}, \frac{13}{3}\right)\)[/tex].
### Step 1: Algebraic Solution
1. Solve the first equation for [tex]\(y\)[/tex]:
[tex]\[ x + y = 4 \][/tex]
[tex]\[ y = 4 - x \][/tex]
2. Substitute [tex]\(y = 4 - x\)[/tex] into the second equation:
[tex]\[ x^2 - 2x(4 - x) - 3 = 0 \][/tex]
Simplify the expression inside the equation:
[tex]\[ x^2 - 2x \cdot 4 + 2x^2 - 3 = 0 \][/tex]
[tex]\[ x^2 - 8x + 2x^2 - 3 = 0 \][/tex]
[tex]\[ 3x^2 - 8x - 3 = 0 \][/tex]
3. Solve the quadratic equation [tex]\(3x^2 - 8x - 3 = 0\)[/tex] using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
Here, [tex]\(a = 3\)[/tex], [tex]\(b = -8\)[/tex], and [tex]\(c = -3\)[/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]
- For the positive square root:
[tex]\[ x = \frac{8 + 10}{6} = \frac{18}{6} = 3 \][/tex]
- For the negative square root:
[tex]\[ x = \frac{8 - 10}{6} = \frac{-2}{6} = -\frac{1}{3} \][/tex]
4. Substitute these values of [tex]\(x\)[/tex] back into [tex]\(y = 4 - x\)[/tex]:
- When [tex]\(x = 3\)[/tex]:
[tex]\[ y = 4 - 3 = 1 \][/tex]
- When [tex]\(x = -\frac{1}{3}\)[/tex]:
[tex]\[ y = 4 - \left(-\frac{1}{3}\right) = 4 + \frac{1}{3} = \frac{12}{3} + \frac{1}{3} = \frac{13}{3} \][/tex]
Therefore, the algebraic solutions are:
[tex]\[ (x, y) = (3, 1) \quad \text{and} \quad \left(-\frac{1}{3}, \frac{13}{3}\right) \][/tex]
### Step 2: Graphical Solution
To graphically solve the equations, we need to plot both equations and find their intersections.
1. Equation 1:
[tex]\[ x + y = 4 \quad \text{or} \quad y = 4 - x \][/tex]
This is a straight line with slope [tex]\(-1\)[/tex] and y-intercept 4.
2. Equation 2:
[tex]\[ x^2 - 2xy - 3 = 0 \][/tex]
We can rewrite this as:
[tex]\[ y = \frac{x^2 - 3}{2x} \][/tex]
This represents a hyperbola, as there are [tex]\((x, y)\)[/tex] values which are obtained through the same algebraic manipulation as shown above.
3. Plot the equations:
- The straight line [tex]\(y = 4 - x\)[/tex]
- The hyperbola derived from [tex]\(x^2 - 2xy - 3 = 0\)[/tex]
By plotting both, you'll find the intersection points, confirming the algebraic solutions [tex]\((3, 1)\)[/tex] and [tex]\(\left(-\frac{1}{3}, \frac{13}{3}\right)\)[/tex].
The following graphical representation shows the system:
[tex]\[ \begin{array}{c} \text{Graph for } y = 4 - x \text{ and } y = \frac{x^2 - 3}{2x}: \\ \text{(Graph not actually shown here) } \end{array} \][/tex]
This completes the algebraic and graphical solutions for the system of equations. The intersection points observed graphically would match the solutions found algebraically: [tex]\((3, 1)\)[/tex] and [tex]\(\left(-\frac{1}{3}, \frac{13}{3}\right)\)[/tex].
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