How do I solve for the two values of x's by substitution. (formula attached)

- anonymous

How do I solve for the two values of x's by substitution. (formula attached)

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- anonymous

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- anonymous

The second equation is already solved for y so substitute (x+1) for y in the first equation.

- anonymous

i did that and I got 5x^2+8X=0

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## More answers

- anonymous

Yes, that looks right. Now solve for x by factoring.

- anonymous

x(5x+8)=0

- anonymous

And set each factor equal to 0. Solve each equation for x.

- anonymous

hmm this really doesn't look right.

- anonymous

Why?

- anonymous

it just looks weird

- anonymous

You will have a fraction for an answer. There is nothing wrong with that.

- anonymous

so how do I solve from x(5x+8)=0 guess and check?

- phi

gypsy told you

- phi

when you have A*B=0 that means either A is zero or B is zero

- anonymous

there is only one factor thats already = to zero

- phi

yes x=0 is one of the solutions

- phi

but (5x+8) could be zero (with the right x value)

- anonymous

x = 0 and 5x+8=0
Set each factor equal to zero separately.

- anonymous

x=-8/5?

- anonymous

-8/5

- phi

yes, you have two solutions for x

- anonymous

There are two answers, that is only one of them.

- anonymous

x=0 and x=-8/5

- 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

are the solutions suppose to work for both problems?

- anonymous

Yes.

- 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

with x=0 you get 0 + 4y^2 = 4
y^2 = 1
y=1

- phi

But I would use the second equation to solve for y, because the square root in the first will introduce ambiguity

- anonymous

if y=1 than it would make the equation =16 not 4

- phi

? how ?

- phi

Here is a graph of what is going on

##### 1 Attachment

- anonymous

(0)^2+4(1)^2
0+4^2=16 not 4

- phi

the first line is ok
but it means
0*0 + 4*1*1 is 4

- anonymous

PEMDAS first multiply 4(1) b/c paranthesis and then use the exponents

- 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

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

ooooooooo sorry bout that, but wat about (-8/5) subbed into x^2+4y^2=4 actually equals 64/5

- anonymous

ooo wait sorry I keep on subbing x into the y

- 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

\(x^2+4y^2=4\) 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

wow so much more clear, thank you guys!

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