anonymous
  • anonymous
How do I solve for the two values of x's by substitution. (formula attached)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1378671397062:dw|
anonymous
  • anonymous
The second equation is already solved for y so substitute (x+1) for y in the first equation.
anonymous
  • anonymous
i did that and I got 5x^2+8X=0

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anonymous
  • anonymous
Yes, that looks right. Now solve for x by factoring.
anonymous
  • anonymous
x(5x+8)=0
anonymous
  • anonymous
And set each factor equal to 0. Solve each equation for x.
anonymous
  • anonymous
hmm this really doesn't look right.
anonymous
  • anonymous
Why?
anonymous
  • anonymous
it just looks weird
anonymous
  • anonymous
You will have a fraction for an answer. There is nothing wrong with that.
anonymous
  • anonymous
so how do I solve from x(5x+8)=0 guess and check?
phi
  • phi
gypsy told you
phi
  • phi
when you have A*B=0 that means either A is zero or B is zero
anonymous
  • anonymous
there is only one factor thats already = to zero
phi
  • phi
yes x=0 is one of the solutions
phi
  • phi
but (5x+8) could be zero (with the right x value)
anonymous
  • anonymous
x = 0 and 5x+8=0 Set each factor equal to zero separately.
anonymous
  • anonymous
x=-8/5?
anonymous
  • anonymous
-8/5
phi
  • phi
yes, you have two solutions for x
anonymous
  • anonymous
There are two answers, that is only one of them.
anonymous
  • anonymous
x=0 and x=-8/5
anonymous
  • anonymous
Yes, although you may need to answer in simplest terms so the improper fraction may need to be expressed as a mixed number.
anonymous
  • anonymous
are the solutions suppose to work for both problems?
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
plugging 0 in x^2+4y^2=4 i get 0 for "y" not 4 and (-8/5) in get 64/5 not 4
phi
  • phi
with x=0 you get 0 + 4y^2 = 4 y^2 = 1 y=1
phi
  • phi
But I would use the second equation to solve for y, because the square root in the first will introduce ambiguity
anonymous
  • anonymous
if y=1 than it would make the equation =16 not 4
phi
  • phi
? how ?
phi
  • phi
Here is a graph of what is going on
1 Attachment
anonymous
  • anonymous
(0)^2+4(1)^2 0+4^2=16 not 4
phi
  • phi
the first line is ok but it means 0*0 + 4*1*1 is 4
anonymous
  • anonymous
PEMDAS first multiply 4(1) b/c paranthesis and then use the exponents
phi
  • phi
Parens in pedmas means do the operations *inside* parens example 4(1+2) you do 1+2 first but here 4(1)^2 there are no operations. now do E exponents: 4* 1^2 = 4*1 now do multiply
phi
  • phi
remember the original equation is x^2 + 4y^2 = 4 which is the same as x*x + 4* y*y =4 now sub in x=0, y=1
anonymous
  • anonymous
ooooooooo sorry bout that, but wat about (-8/5) subbed into x^2+4y^2=4 actually equals 64/5
anonymous
  • anonymous
ooo wait sorry I keep on subbing x into the y
phi
  • phi
x^2+4y^2=4 solving for y is ambiguous (you should use the other equation ) but (-8/5)*(-8/5) + 4 y^2 = 4 64/25 + 4 y^2 = 4 4 y^2 = 4 - 64/25 4 y ^ 2= 100/25 - 64/25 4 y^2 = 36/25 y^2 = 9/25 y = -3/5 (we have to take the negative square root value) but if you use y =x +1 you get y = -8/5 + 5/5 y = -3/5
anonymous
  • anonymous
\(x^2+4y^2=4\) and \(y=x+1\) \(x^2+4(x+1)^2=4\) \(x^2+4(x^2+2x+1)=4\) \(x^2 + 4x^2+8x+4=4\) \(5x^2 +8x+4=4\) \(5x^2+8x=0\) \(x(5x+8)=0\) \(x=0\) and \(5x+8=0\implies x=-\dfrac{8}{5}\) Back to the second equation: \(y=x+1\) \(y=\left(-\dfrac{8}{5}\right)+1\) \(y=-\dfrac{3}{5}\) AND \(y=0+1\) \(y=1\) So your answers are: \(\left(-\dfrac{8}{5},-\dfrac{3}{5}\right)\) and \(\left(0,1\right)\) If you plug those values back into the first equation, you should get the same answers, but remember that when you take a square root, you get a negative and positive value. So, when using the first equation you will get 4 answers for y, only 2 of them will check, the other two must be eliminated.
anonymous
  • anonymous
wow so much more clear, thank you guys!

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