Solve: x^2 + y^2 = 34 2x - y = 1
Solve the second one for Y first ;-)
Then plug back into the first equation, and solve for X
You'll then have x = something Got back to your y= equation and substitute the now known value for X back in
Make sense?
Well sine it's y^2 wouldn't i have to do y= (1 - 2x)(1 - 2x) ?
Subtract 1 from both side and add +y to both sides 2x-1 = y
Yep, got to FOIL the binomial (2x-1)^2
\[(a -b)^2 = a^2 + b^2 - 2ab\]
@LinuxCharmd make sure you have 2x-1 though, not -2x+1 (which is the same as saying 1-2x)
Order matters in subtraction and division
So, then I would now plug 4x^2 - 4x - 1 = 0 into the y^2 to then solve for y?
yep! Combine like terms and solve \[x^2 + (2x-1)^2 = 34\] \[x^2 + 4x^2 - 4x +1 = 34\]
Indeed you'll have two answers for a second-order equation (aka. a quadratic), even if it's one double root or two imaginary roots (hope not).
So I then get -3x62 - 4x - 35 = 0 which I will take an pop in the quadratic form?
Sorry -3x^2 - 4x - 35. Lawl.
And why 2 x answers and 2 y answers? Because that first equation you have is a circle centered at (0,0) with a radius of \(\sqrt{34}\), having a line ( y=2x-1 ) cutting across it. The line crosses the circle's perimeter at two (x,y) coordinates. :-) |dw:1340989056683:dw|
Got it now @LinuxCharmd ? :-)
I think so. That would make the answer: -2/3 +- sqrt101/3 i ?
Read my above post, 2 x answers, 2 y answers. Look at the awesome sketch too :D
See if your answers make sense
Hmm... I think the one I got makes sense, but I have to do something else to it, which I'm not getting.
Watch you order of operations on the quadratic equation: \[x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\] for \[ax^2+bx+c=0\] If it's -x\(^2\)-x-2=0 for example... a = -1 b = -1 c = -2
watch your*
Hmm, can't find the answer. Guess I'll hunt my book up.
Order of operations mnemonic = PELDMAS ;-)
Or PELMDAS or whatever
It's Pemdas. Please excuse my dear aunt Sally. Parenthesis, exponents, multiplication, division, addition, subtraction. :)
solve it by quadratic equation
x^2 + y^2 = 34 2x - y = 1 y = 2x - 1 x^2 + (2x-1)^2 = 34 Now you can solve it from there
5x^2 - 4x - 33 = 0 Use Quadratic Formula
@JBrightman3 The way I remember it (operands that undo each other) 1. Parenthesis 2. Exponents & Logarithms 3. Multiplication # Division 4. Addition & Subtraction Everybody's got their own thing, but after you've done math for long enough it become second nature. You just have to watch out for moments your neurons in your brain decide to take a nap, or a misplaced typo when typing it in isn't messing you up. That's specifically what I meant. @Hero Same as above, aye? @shkak She knows that already I believe. ^_^
x1=3 x2=-11/5
@agentx5 Well it's almost the same as mine anyways. Of course, as long as it gets us the answer, it isn't wrong. Different strokes for different folks. :)
Indeed!
I guess nobody liked my sketch :-(
She's gone anyways.
Sorry, I went to look it up in my book. I figured it out. Thanks guys!
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