document.write( "Question 615654: How do you find the system of equations:\r
\n" ); document.write( "\n" ); document.write( "x^2+y^2=9\r
\n" ); document.write( "\n" ); document.write( "4x^2-4y^2=16 \n" ); document.write( "

Algebra.Com's Answer #387286 by jsmallt9(3758) $\"\"$ $\"About$

You can put this solution on YOUR website!
$\"x%5E2%2By%5E2=9\"$
\n" ); document.write( " $\"4x%5E2-4y%5E2=16\"$
\n" ); document.write( "As usual, there is more than one way to solve this. One way, which might be the easiest, is to use the Substitution Method. We can quickly solve the first equation for $\"x%5E2\"$ by subtracting $\"y%5E2\"$ from each side:
\n" ); document.write( " $\"x%5E2=9-y%5E2\"$
\n" ); document.write( "Note: Since the other equation has only an $\"x%5E2\"$ term, we do not have to solve for x, just $\"x%5E2\"$ .

\n" ); document.write( "Next we substitute the expression we now have for $\"x%5E2\"$ into the other equation:
\n" ); document.write( " $\"4%289-y%5E2%29-4y%5E2=16\"$
\n" ); document.write( "Distributing the 4 we get:
\n" ); document.write( " $\"36+-+4y%5E2-4y%5E2=16\"$
\n" ); document.write( "Combining like terms we get:
\n" ); document.write( " $\"36+-+8y%5E2=16\"$
\n" ); document.write( "Subtracting 36:
\n" ); document.write( " $\"-8y%5E2=-20\"$
\n" ); document.write( "Dividing by -8:
\n" ); document.write( " $\"y%5E2=%28-20%29%2F%28-8%29\"$
\n" ); document.write( "Since I'm about to find square roots, I'm only going to reduce the fraction to:
\n" ); document.write( " $\"y%5E2=10%2F4\"$
\n" ); document.write( "so that I have a perfect square denominator. Now we find the square root of each side. (Remember the negative square roots!)
\n" ); document.write( " $\"y+=+sqrt%2810%2F4%29\"$ or $\"y+=+-sqrt%2810%2F4%29\"$
\n" ); document.write( "which simplify as follows:
\n" ); document.write( " $\"y+=+sqrt%2810%29%2Fsqrt%284%29\"$ or $\"y+=+-sqrt%2810%29%2Fsqrt%284%29\"$
\n" ); document.write( " $\"y+=+sqrt%2810%29%2F2\"$ or $\"y+=+-sqrt%2810%29%2F2\"$

\n" ); document.write( "We have two y values. So we will have at least two solutions! We can use the \"solved for\" equation, $\"x%5E2=9-y%5E2\"$ , to find the x value(s) for these y values:
\n" ); document.write( "For $\"y+=+sqrt%2810%29%2F2\"$ :
\n" ); document.write( " $\"x%5E2+=+9-%28sqrt%2810%29%2F2%29%5E2\"$
\n" ); document.write( " $\"x%5E2+=+9-%2810%2F4%29\"$
\n" ); document.write( " $\"x%5E2+=+36%2F4-%2810%2F4%29\"$ Again, keeping the denominator of 4 since a square root is coming)
\n" ); document.write( " $\"x%5E2+=+26%2F4\"$
\n" ); document.write( "Square root (w/ the negative square roots!):
\n" ); document.write( " $\"x+=+sqrt%2826%29%2F2\"$ or $\"x+=+-sqrt%2826%29%2F2\"$
\n" ); document.write( "For this one y value, $\"sqrt%2810%29%2F2\"$ we get two x values! This means that both ( $\"sqrt%2826%29%2F2\"$ , $\"sqrt%2810%29%2F2\"$ ) and ( $\"-sqrt%2826%29%2F2\"$ , $\"sqrt%2810%29%2F2\"$ ) are solutions.

\n" ); document.write( "For $\"y+=+-sqrt%2810%29%2F2\"$ :
\n" ); document.write( " $\"x%5E2+=+9-%28-sqrt%2810%29%2F2%29%5E2\"$
\n" ); document.write( "I hope you can see that squaring $\"-sqrt%2810%29%2F2\"$ will result in the same thing as squaring $\"sqrt%2810%29%2F2\"$ did above. And since this is true the rest of what follows will be the same as what we did above. So we will end up with the exact same x values, $\"x+=+sqrt%2826%29%2F2\"$ or $\"x+=+-sqrt%2826%29%2F2\"$ , for $\"y+=+-sqrt%2810%29%2F2\"$ as we did for $\"y+=+sqrt%2810%29%2F2\"$ . So we have two more solutions:
\n" ); document.write( "( $\"sqrt%2826%29%2F2\"$ , $\"-sqrt%2810%29%2F2\"$ ) and ( $\"-sqrt%2826%29%2F2\"$ , $\"-sqrt%2810%29%2F2\"$ )

\n" ); document.write( "All together we have four solutions:
\n" ); document.write( "( $\"sqrt%2826%29%2F2\"$ , $\"sqrt%2810%29%2F2\"$ ) and ( $\"-sqrt%2826%29%2F2\"$ , $\"sqrt%2810%29%2F2\"$ ) and ( $\"sqrt%2826%29%2F2\"$ , $\"-sqrt%2810%29%2F2\"$ ) and ( $\"-sqrt%2826%29%2F2\"$ , $\"-sqrt%2810%29%2F2\"$ )

\n" ); document.write( "Getting 4 solutions should not be a surprise. The first equation is the equation of a circle centered on the origin. And the second equation is the equation of a hyperbola centered on the origin. If you picture this it should not be hard to imagine that there could be 4 symmetric points of intersection.
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