# Solving simultaneous equations involving a quadratic

I have the question

Solve the simultaneous equation pair

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

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.

• If I had to guess, you're not expanding $(2x-5)^2$ correctly. Could you include your expansion? – Michael Burr Nov 10 '16 at 11:51

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

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

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

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

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

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

What do you get when you simplify this further?

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

Solving:

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

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

• How is X^2 - 4X = 0 achieved from this ? – Dan Nov 10 '16 at 11:49
• @Dan notice that $\left(2x-5\right)^2=4x^2-20x+25$ – Jan Nov 10 '16 at 11:52