Simultaneous equations and negative exponents

Question

I am an elderly maths ignoramus! But trying to help my granddaughter with a problem. The simultaneous equations are: $x^2 + y^2 = 25$ and $y - 3x = 13$. I have tried to get a value for y from equation b) in order to substitute it into a). But then I get to $10x^2 = -144$ and I do not think x^2 can be a negative number! Is this correct? Can someone solve this simple (!) problem? Many thanks!

Answer 1

it is $$x^2+(3x+13)^2=25$$ $$x^2+9x^2+169+78x=25$$ Can you proceed? write for $y$: $$y=3x+13$$ simplifying and factorizing we obtain $$2\, \left( 5\,x+24 \right) \left( x+3 \right) =0$$ from here you will get two cases: $$5x+24=0$$ or $$x+3=0$$

Answer 2

It can happen that a system of equations has no solutions but somewhere along the way, probably something went wrong in the calculations. Substitution of $y=3x+13$ into $x^2+y^2=25$ gives: $$x^2+(3x+13)^2=25 \iff 10 x^2 + 78 x + 144 = 0$$ which is a (standard) quadratic equation.

Although it's possible that this equation has no (real) solutions, that's not the case here.

Answer 3

$x^2$ can be negative if imaginary numbers are allowed, e.g., $x^2 = -1$ means $x = \pm i$. But we don't need to worry about that here.

$y - 3x = 13$ means $y = 3x + 13$. Substitute this into the other equation. \begin{align*} x^2 + y^2 &= 25\\ x^2 + (3x + 13)^2 &= 25\\ x^2 + 9x^2 + 78x + 169 &= 25\\ 10x^2 + 78x + 144 &= 0 \end{align*} Divide both sides by $2$ to get $5x^2 + 39x + 72 = 0$. This can be factored but since the numbers are bordering on the unpleasant, the quadratic formula may be a good way to go instead.

Answer 4

I think you may have made a small arithmetical error. You used the second equation to get $y = 3x+13$, and then insertet this into the first equation to get $x^2 + (3x+13)^2 = 25$. Then from here, I suspect that you went directly for $x^2 + 9x^2 + 169 = 25$. This is a mistake. $(3x+13)^2$ and $9x^2 + 169$ are not the same thing.

I think this is best seen if you write $(3x+13)^2$ as $(3x+13)\cdot (3x+13)$ (this is what $^2$ means, after all), and then multiply the parentheses the long way. Because squaring is so common, many books teach you to just look at $(3x + 13)^2$ and immediately see that it's equal to $9x^2 + 26x + 169$ (usually through a formula like $(a+b)^2 = a^2 + 2ab + b^2$). I suggest you only start to do that after having squares a few different brackets and seen how everything work.

So the actual equation you should solve is $10x^2 + 26x = -144$, or $10x^2 + 26x + 144 = 0$.

Side note: It is entirely possible for equations to have no solutions, particularily when the highest power the unknown is raised to is even. $10x^2 = -144$ is one example of this, as you rightly point out. There are number systems invented (or maybe discovered?) to deal with this. By far the most popular are the so-called complex numbers, which didn't get their name because they're difficult, but because they're put together from two parts. But for now it's safe to just say that there is no solution.