My Math Forum Solution of y = x and y = -x

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 November 20th, 2010, 08:23 PM #1 Newbie   Joined: Sep 2008 Posts: 4 Thanks: 0 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
 November 20th, 2010, 09:46 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Solution of y = x and y = -x If you have y = x and y = -x, you did correctly to set x = -x Add x to both sides: 2x = 0 Divide through by 2 x = 0 Think of it as +0 = -0
November 20th, 2010, 10:09 PM   #3
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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]$\text{Because }\,+0 \,=\,-0$

 November 27th, 2010, 08:06 PM #4 Newbie   Joined: Sep 2008 Posts: 4 Thanks: 0 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
 November 27th, 2010, 08:21 PM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Solution of y = x and y = -x Using your examples: (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.

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