To solve the given system of equations:
[tex]\[
\begin{cases}
x + y = 4 \\
x^2 - 2xy - 3 = 0
\end{cases}
\][/tex]
we will follow a systematic approach to find the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy both equations.
### Step-by-Step Solution:
1. Solve the First Equation for [tex]\( y \)[/tex]:
The first equation is:
[tex]\[
x + y = 4
\][/tex]
We can express [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]:
[tex]\[
y = 4 - x
\][/tex]
2. Substitute [tex]\( y = 4 - x \)[/tex] into the Second Equation:
The second equation is:
[tex]\[
x^2 - 2xy - 3 = 0
\][/tex]
Substitute [tex]\( y = 4 - x \)[/tex] into this equation:
[tex]\[
x^2 - 2x(4 - x) - 3 = 0
\][/tex]
Expand and simplify:
[tex]\[
x^2 - 2x(4 - x) - 3 = x^2 - 8x + 2x^2 - 3 = 3x^2 - 8x - 3 = 0
\][/tex]
3. Solve the Quadratic Equation:
The quadratic equation is:
[tex]\[
3x^2 - 8x - 3 = 0
\][/tex]
To solve this, we can use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 3 \)[/tex], [tex]\( b = -8 \)[/tex], and [tex]\( c = -3 \)[/tex].
Calculate the discriminant ([tex]\( \Delta \)[/tex]):
[tex]\[
\Delta = b^2 - 4ac = (-8)^2 - 4(3)(-3) = 64 + 36 = 100
\][/tex]
Since [tex]\( \Delta \)[/tex] is positive, we have two real solutions:
[tex]\[
x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{8 \pm \sqrt{100}}{6} = \frac{8 \pm 10}{6}
\][/tex]
So, the solutions for [tex]\( x \)[/tex] are:
[tex]\[
x = \frac{18}{6} = 3 \quad \text{and} \quad x = \frac{-2}{6} = -\frac{1}{3}
\][/tex]
4. Find Corresponding [tex]\( y \)[/tex] Values:
For each [tex]\( x \)[/tex] value, find the corresponding [tex]\( y \)[/tex] value using [tex]\( y = 4 - x \)[/tex]:
- If [tex]\( x = 3 \)[/tex]:
[tex]\[
y = 4 - 3 = 1
\][/tex]
- If [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]
### Solutions:
The solutions to the system of equations are:
[tex]\[
(x, y) = (3, 1) \quad \text{and} \quad \left(-\frac{1}{3}, \frac{13}{3}\right)
\][/tex]
### Verification:
Algebraically:
1. For [tex]\( (x, y) = (3, 1) \)[/tex]:
[tex]\[
x + y = 3 + 1 = 4 \quad \text{(True)}
\][/tex]
[tex]\[
x^2 - 2xy - 3 = 3^2 - 2(3)(1) - 3 = 9 - 6 - 3 = 0 \quad \text{(True)}
\][/tex]
2. For [tex]\( \left(-\frac{1}{3}, \frac{13}{3}\right) \)[/tex]:
[tex]\[
x + y = -\frac{1}{3} + \frac{13}{3} = \frac{12}{3} = 4 \quad \text{(True)}
\][/tex]
[tex]\[
x^2 - 2xy - 3 = \left(-\frac{1}{3}\right)^2 - 2\left(-\frac{1}{3}\right)\left(\frac{13}{3}\right) - 3 \\
= \frac{1}{9} + \frac{26}{9} - 3 = \frac{27}{9} - 3 = 3 - 3 = 0 \quad \text{(True)}
\][/tex]
Graphically:
You can graph the two equations:
- [tex]\( x + y = 4 \)[/tex]: This is a straight line.
- [tex]\( x^2 - 2xy - 3 = 0 \)[/tex]: This is a hyperbola or another complex curve.
The points [tex]\( (3, 1) \)[/tex] and [tex]\( \left(-\frac{1}{3}, \frac{13}{3}\right) \)[/tex] are the intersections of these two graphs, confirming that our solutions are correct.
Thus, both algebraically and graphically, we have verified the solutions [tex]\( (3, 1) \)[/tex] and [tex]\( \left(-\frac{1}{3}, \frac{13}{3}\right) \)[/tex].
