1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ If I had to guess, you're not expanding $(2x-5)^2$ correctly. Could you include your expansion? $\endgroup$ – Michael Burr Nov 10 '16 at 11:51
1
$\begingroup$

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

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

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

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

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

$\endgroup$
1
$\begingroup$

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

What do you get when you simplify this further?

$\endgroup$
1
$\begingroup$

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

Solving:

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

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

$\endgroup$
  • $\begingroup$ How is X^2 - 4X = 0 achieved from this ? $\endgroup$ – Dan Nov 10 '16 at 11:49
  • 1
    $\begingroup$ @Dan notice that $\left(2x-5\right)^2=4x^2-20x+25$ $\endgroup$ – Jan Nov 10 '16 at 11:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.