My Math Forum (http://mymathforum.com/math-forums.php)
-   Algebra (http://mymathforum.com/algebra/)
-   -   Solution of y = x and y = -x (http://mymathforum.com/algebra/15888-solution-y-x-y-x.html)

 bogger57 November 20th, 2010 07:23 PM

Solution of y = x and y = -x

I was helping someone with math and embarrassed myself today. I asked that person to use basic algebra to compute the distance between the line y = -x and the point (4,4). I told the person to:
• Compute the slope of all lines perpendicular to y = -x (we determined this was 1)
• Find a line with slope 1 that runs through point (4,4) (we determined this was the line y = x)
• Find where y = x and y = -x intersect.
• Find the distance between (the point defined by the intersection of y = x and y = -x) and (4,4)

This process went pretty well until Step#3 above, where I embarrassed myself showing the person how to solve the linear system
y = x
y = -x
My first instinct was to solve the system by substituting the first equation y = x into the second equation y = -x giving:
x = -x
but this is complete nonsense right of the bat! How can x = -x? If you divide both sides by x, then you get 1 = -1 which is even more bizarre. I guess I could try this:
x = -x
x+x=-x+x
2x=0
2x/2=0/2
x=0
,but I still can't get over how x can ever be -x. Much to my embarrassment, this blows my mind. Can someone please explain to me how I can solve the system
y=x
y=-x
without using a plotting or matrix algebra technique?

Confused,

bogger57

 MarkFL November 20th, 2010 08:46 PM

Re: Solution of y = x and y = -x

If you have y = x and y = -x, you did correctly to set

x = -x

2x = 0

Divide through by 2

x = 0

Think of it as +0 = -0 :mrgreen:

 soroban November 20th, 2010 09:09 PM

Re: Solution of y = x and y = -x

Hello, bogger57!

Quote:
 I was helping someone with math and embarrassed myself today. I asked that person to use basic algebra to compute the distance between the line $y \,=\, -x$ and the point (4,4). I told the person to: [color=beige]. . [/color][1] Compute the slope of all lines perpendicular to $y \,=\, -x$[color=beige] . [/color](We determined this was 1) [color=beige]. . [/color][2] Find a line with slope 1 that runs through point (4,4)[color=beige] . [/color](we determined this was the line $y \,=\, x\$) [color=beige]. . [/color][3] Find where $y \,=\, x$ and $y \,=\, -x$ intersect. [color=beige]. . [/color][4] Find the distance between the point of intersection and (4,4). This process went pretty well until Step #3 above, where I embarrassed myself showing the person how to solve the linear system [color=beige]. . [/color]$\begin{array}{ccc}y=&x \\ y=&-x \end{array}=$ My first instinct was to solve the system by substituting the first equation $y \,=\, x$ into the second equation $y\,=\,-x$ [color=beige]. . [/color]giving:[color=beige] .[/color]$x \,=\, -x$ But this is complete nonsense right of the bat! [color=beige] . [/color] [color=blue]No, it isn't[/color] How can $x \,=\, -x$ ? [color=beige] . [/color] [color=blue]See my reply below.[/color] If you divide both sides by $x$, then you get $1 \,=\, -1$ which is even more bizarre. [color=beige]. . [/color][color=blue]You must NOT divide by x . . . ever![/color] I guess I could try this:[color=beige] .[/color]$x \:=\: -x \;\;\;\Rightarrow\;\;\;2x \:=\:0 \;\;\;\Rightarrow\;\;\;x\,=\,0$ [color=beige] . [/color] [color=blue]This is the correct method.[/color] But I still can't get over how $x$ can ever be $-x$.

$\text{The equation is: }\:x \:=\:-x$

$\text{It asks, "What number is equal to its negative?"}$

$\text{If you think about it, you'll see that the answer is }zero.$
[color=beige]. . [/color]$\text{Because }\,+0 \,=\,-0$

 bogger57 November 27th, 2010 07:06 PM

Re: Solution of y = x and y = -x

MarkFL and soroban:

I can see how +0 = -0 so both of your answers seem to make sense to me, but I am still a little confused. Soroban says “You must NOT divide by x . . . ever!” I'm not sure what this means, since I don't remember my math teachers telling me I can't divide by x. Can someone show me what rule says I can't divide both sides of the following equations by x?
yx = x
x*x = x
x+x+x+x=x+x+x

Thanks for the Help So Far!

Bogger 57

 MarkFL November 27th, 2010 07:21 PM

Re: Solution of y = x and y = -x

(1) $yx=x$

If we divide through by x:

$y=1$

As you can see y = 1 is a solution by the multiplicative identity, i.e $a\cdot 1=a$, however, we miss a solution, namely x = 0. A better way to solve is to subtract x from both sides:

$yx-x=0$

Factor.

$x(y-1)=0$

Now we get both solutions, x = 0, y = 1.

(2) $x^2=x$

Just like with (1) if we divide by x, we lose the solution x = 0.

(3) $4x=3x$

If we divide by x, we get no solution, however if we subtract 3x from each side, we get:

$x=0$

In general, if we divide by f(x), we lose the solution f(x) = 0. I was actually taught that if you do divide by f(x), note f(x) = 0 as a solution, then proceed. But strictly speaking, soroban is right, you should find another way to solve, finding all solutions at the end.

 All times are GMT -8. The time now is 01:22 PM.