Solve for x and y simultaneously: 3x - y = 4 and x^2 + 2xy - y = 0; 2x + y = 5 and 4x^2 - y^2 = 0.

Steps to Solve

1. Solve the first linear equation for y

Starting with the first linear equation from 2.1:
$$ 3x - y = 4 $$
Rearranging this gives us:
$$ y = 3x - 4 $$

2. Substitute y in the quadratic equation

Now, substitute (y) in the second equation from 2.1:
$$ x^2 + 2xy - y = 0 $$
Substituting (y) gives:
$$ x^2 + 2x(3x - 4) - (3x - 4) = 0 $$
This simplifies to:
$$ x^2 + 6x^2 - 8x - 3x + 4 = 0 $$
So we have:
$$ 7x^2 - 11x + 4 = 0 $$

3. Solve the quadratic equation for x

Now we can use the quadratic formula:
$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$
Here, (a = 7), (b = -11), and (c = 4):
$$ x = \frac{11 \pm \sqrt{(-11)^2 - 4(7)(4)}}{2(7)} $$
$$ x = \frac{11 \pm \sqrt{121 - 112}}{14} $$
$$ x = \frac{11 \pm \sqrt{9}}{14} $$
Which results in:
$$ x = \frac{11 \pm 3}{14} $$

4. Calculate the possible values for x

Solving this gives two values:
$$ x_1 = \frac{14}{14} = 1 $$
$$ x_2 = \frac{8}{14} = \frac{4}{7} $$

5. Find corresponding y values

For each (x), find (y) using (y = 3x - 4):

• For (x = 1):
  $$ y = 3(1) - 4 = -1 $$
• For (x = \frac{4}{7}):
  $$ y = 3\left(\frac{4}{7}\right) - 4 = \frac{12}{7} - \frac{28}{7} = -\frac{16}{7} $$

6. Solve the second set of equations

Now we take the second set from 2.2:

1. $$ 2x + y = 5 $$

2. $$ 4x^2 - y^2 = 0 $$

3. Solve the first linear equation for y

From the first equation:
$$ y = 5 - 2x $$

8. Substitute y in the second equation

Now substitute in the second equation:
$$ 4x^2 - (5 - 2x)^2 = 0 $$
Expanding gives:
$$ 4x^2 - (25 - 20x + 4x^2) = 0 $$
This simplifies to:
$$ 20x - 25 = 0 $$

9. Solve for x

Solving this gives:
$$ 20x = 25 \implies x = \frac{25}{20} = \frac{5}{4} $$

10. Find corresponding y value

Now substitute (x) back to find (y):
$$ y = 5 - 2\left(\frac{5}{4}\right) $$
$$ y = 5 - \frac{10}{4} = 5 - \frac{5}{2} = \frac{10}{2} - \frac{5}{2} = \frac{5}{2} $$