Solve simullaneously for $$ x $$ and $$ y $$ $$ 3 y=1+x+x^{2} $$ and $$ \sum_{k=3}^{4} x(k-2)^{2}=5 y $$

Question

UpStudy AI · Accepted Answer

We are given the following system of equations:

1. $$ 3y = 1 + x + x^2 $$
2. $$ \sum_{k=3}^{4} x(k-2)^2 = 5y $$

Let's solve them step by step.

### Step 1: Simplify the Summation

First, evaluate the summation in the second equation:

$$
\sum_{k=3}^{4} x(k-2)^2 = x(3-2)^2 + x(4-2)^2 = x(1)^2 + x(2)^2 = x + 4x = 5x
$$

So, the second equation becomes:

$$
5x = 5y \implies y = x
$$

### Step 2: Substitute $$ y = x $$ into the First Equation

Substitute $$ y $$ with $$ x $$ in the first equation:

$$
3x = 1 + x + x^2
$$

Rearrange the equation:

$$
x^2 + x + 1 - 3x = 0 \implies x^2 - 2x + 1 = 0
$$

### Step 3: Solve the Quadratic Equation

The quadratic equation is:

$$
x^2 - 2x + 1 = 0
$$

Calculate the discriminant ($$ \Delta $$):

$$
\Delta = (-2)^2 - 4 \cdot 1 \cdot 1 = 4 - 4 = 0
$$

Since the discriminant is zero, there is exactly one real solution:

$$
x = \frac{2}{2} = 1
$$

### Step 4: Find $$ y $$

Since $$ y = x $$:

$$
y = 1
$$

### Conclusion

The solution to the system of equations is:

$$
x = 1 \quad \text{and} \quad y = 1
$$

Answer: $$x = 1$$ and $$y = 1$$
$$ x = 1 $$ and $$ y = 1 $$.

Solve simullaneously for \( x \) and \( y \) \( 3 y=1+x+x^{2} \) and \( \sum_{k=3}^{4} x(k-2)^{2}=5 y \)