what is wrong here?

A

allegansveritatem

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I was trying today to find the intersection of two circles by solving the system of their equations. I had already found the point, (2,2), but I wanted to see if I could get the same result by the systems of equation technique. Here is what I did:
circles.PNG
I did this again and again and couldn't get it to work. What am I doing wrong here?
 
I assume from the scoring out in 3, you are subtracting the two equations to give one equation. However in 4 and 5 you are interpreting this as two equations.
The result of 3, is a single equation: -4y+4x=0, i.e. x=y, which you can then substitute into an original equation to get your answers (plural).
 
[MATH]x^2+(y-2)^2 = 4 \implies x^2 + y^2 - 4y = 0[/MATH]
[MATH](x-2)^2 + y^2 = 4 \implies x^2 - 4x + y^2 = 0[/MATH]
subtracting the second equation from the first ...

[MATH]4x - 4y = 0 \implies y = x \implies x^2 + x^2 - 4x = 0 \implies 2x^2 - 4x = 0 \implies 2x(x - 2) = 0 \implies x = 2 \text{ and } x = 0[/MATH]
points of intersection are (2,2) and (0,0)

2_circles.jpg
 
Hi allegansveritatem. Your step #3 is not a valid way of solving systems. If you subtract the left-hand side of an equation, you must also subtract the right-hand side.

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The two equations are symmetric so you can set y=x.
So [MATH]x^2+(y-2)^2 = 4\ becomes\ x^2+(x-2)^2 = 4\ and\ x^2 + x^2 -4x = 0 \implies x=0\ or\ x=2[/MATH]
 
Jomo said:
The two equations are symmetric so you can set y=x.
So [MATH]x^2+(y-2)^2 = 4\ becomes\ x^2+(x-2)^2 = 4\ and\ x^2 + x^2 -4x = 0 \implies x=0\ or\ x=2[/MATH]
Click to expand...
Nice solution, since [MATH]x=y[/MATH].

Note, equations being symmetric in y and x does not imply that [MATH]x=y[/MATH].
e.g. consider [MATH]xy=16, x+y=10[/MATH]or [MATH]x^2+2y^2=3, y^2+2x^2=3[/MATH]or simply [MATH]x^2+y^2=1[/MATH]...

[MATH]x=y[/MATH] is consistent with equations being symmetric in x and y, so there may be solutions where [MATH]x=y[/MATH].
 
lex said:
I assume from the scoring out in 3, you are subtracting the two equations to give one equation. However in 4 and 5 you are interpreting this as two equations.
The result of 3, is a single equation: -4y+4x=0, i.e. x=y, which you can then substitute into an original equation to get your answers (plural).
Click to expand...
Right. I had a premonition as I went through this process that I was mixing methods but I was caught in a rut and just kept repeating the same stumble again and again. It is a form of hysterical blindness that comes over me sometimes when working these exercises. I will copy your post and digest it later today and go back at it then. Thanks
 
skeeter said:
[MATH]x^2+(y-2)^2 = 4 \implies x^2 + y^2 - 4y = 0[/MATH]
[MATH](x-2)^2 + y^2 = 4 \implies x^2 - 4x + y^2 = 0[/MATH]
subtracting the second equation from the first ...

[MATH]4x - 4y = 0 \implies y = x \implies x^2 + x^2 - 4x = 0 \implies 2x^2 - 4x = 0 \implies 2x(x - 2) = 0 \implies x = 2 \text{ and } x = 0[/MATH]
points of intersection are (2,2) and (0,0)

View attachment 28177
Click to expand...
I see it. Thanks. A good way to go.
 
Otis said:
Hi allegansveritatem. Your step #3 is not a valid way of solving systems. If you subtract the left-hand side of an equation, you must also subtract the right-hand side.

?
Click to expand...
I knew something was not legal in what I was perpetrating but...yes, I see it now. Thanks
 
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