Question

(2x² + 2y = 5, -x + 3y = 6;​

Answer 1

To solve the system of equations $2x^2 + 2y = 5$ and $-x + 3y = 6$, we can use either the substitution method or the elimination method. I'll demonstrate the elimination method here.

First, let's multiply the second equation by 2 to make it easier to eliminate one of the variables:

Original equations:

$2x^2 + 2y = 5$ ...(1)

$-x + 3y = 6$ ...(2)

Multiply equation (2) by 2:

$-2x + 6y = 12$ ...(3)

Now, let's add equation (1) and equation (3) to eliminate $x$:

$2x^2 + 2y - 2x + 6y = 5 + 12$

Simplify the equation:

$2x^2 + 8y - 2x = 17$ ...(4)

Next, we can rearrange equation (4) to isolate $x$:

$2x^2 - 2x + 8y = 17$

$2x^2 - 2x = 17 - 8y$

$x^2 - x = \frac{17 - 8y}{2}$

$x^2 - x = 8.5 - 4y$ ...(5)

Now, let's focus on equation (2) again:

$-x + 3y = 6$ ...(6)

We can rearrange equation (6) to isolate $x$:

$-x = -3y + 6$

$x = 3y - 6$ ...(7)

Now that we have expressions for $x$ in terms of $y$ from equation (7) and equation (5), we can set them equal to each other:

$3y - 6 = \sqrt{8.5 - 4y} \quad \text{or} \quad 3y - 6 = -\sqrt{8.5 - 4y}$

Solving for $y$ in each case will give us possible values for $y$. After that, we can use these values to find the corresponding $x$ values. Please note that this equation may result in real or complex solutions depending on the specific values involved.