Solving simultaneous equations involving a quadratic

Question

I have the question

Solve the simultaneous equation pair

$$x^2 + y^2 = 25\tag1$$ $$2x - y = 5\tag2$$

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I have found the value of $y$ from the second equation which is $2x-5$ and substituted this into the first equations $y$ value.

I get $x^2 + (2x -5)^2 = 25$

When I expand the brackets I get the equation $$x^2+4x^2-20 =0\tag3$$

However, when I checked the solutions the equation should simplify to $x^2 - 4x = 0$ and I do not understand how this is achieved.

Answer 1

$$x^2 + y^2 = 25 $$

$$2x - y = 5 \Leftrightarrow y=2x-5$$

$$ x^2 + (2x-5)^2 = 25 $$

$$ x^2 +4x^2 -20x +25 = 25 $$

$$ 5x^2 -20x=0 $$

Answer 2

$$x^2 + (2x-5)^2 = x^2 + (2x)^2 - 2\cdot (2x)\cdot 5 + 5^2$$

What do you get when you simplify this further?

Answer 3

$$ \begin{cases} \text{x}^2+\text{y}^2=25\\ 2\text{x}-\text{y}=5 \end{cases}\space\Longleftrightarrow\space \begin{cases} \text{x}^2+\left(2\text{x}-5\right)^2=25\\ \text{y}=2\text{x}-5 \end{cases} $$

Solving:

$$\text{x}^2+\left(2\text{x}-5\right)^2=25$$

Gives us $x=0$ or $x=4$