Question 615654
{{{x^2+y^2=9}}}
{{{4x^2-4y^2=16}}}
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^2}}} by subtracting {{{y^2}}} from each side:
{{{x^2=9-y^2}}}
Note: Since the other equation has only an {{{x^2}}} term, we do not have to solve for x, just {{{x^2}}}.<br>
Next we substitute the expression we now have for {{{x^2}}} into the other equation:
{{{4(9-y^2)-4y^2=16}}}
Distributing the 4 we get:
{{{36 - 4y^2-4y^2=16}}}
Combining like terms we get:
{{{36 - 8y^2=16}}}
Subtracting 36:
{{{-8y^2=-20}}}
Dividing by -8:
{{{y^2=(-20)/(-8)}}}
Since I'm about to find square roots, I'm only going to reduce the fraction to:
{{{y^2=10/4}}}
so that I have a perfect square denominator. Now we find the square root of each side. (Remember the negative square roots!)
{{{y = sqrt(10/4)}}} or {{{y = -sqrt(10/4)}}}
which simplify as follows:
{{{y = sqrt(10)/sqrt(4)}}} or {{{y = -sqrt(10)/sqrt(4)}}}
{{{y = sqrt(10)/2}}} or {{{y = -sqrt(10)/2}}}<br>
We have two y values. So we will have at least two solutions! We can use the "solved for" equation, {{{x^2=9-y^2}}}, to find the x value(s) for these y values:
For {{{y = sqrt(10)/2}}}:
{{{x^2 = 9-(sqrt(10)/2)^2}}}
{{{x^2 = 9-(10/4)}}}
{{{x^2 = 36/4-(10/4)}}} Again, keeping the denominator of 4 since a square root is coming)
{{{x^2 = 26/4}}}
Square root (w/ the negative square roots!):
{{{x = sqrt(26)/2}}} or {{{x = -sqrt(26)/2}}}
For this one y value, {{{sqrt(10)/2}}} we get two x values! This means that both ({{{sqrt(26)/2}}}, {{{sqrt(10)/2}}}) and ({{{-sqrt(26)/2}}}, {{{sqrt(10)/2}}}) are solutions.<br>
For {{{y = -sqrt(10)/2}}}:
{{{x^2 = 9-(-sqrt(10)/2)^2}}}
I hope you can see that squaring {{{-sqrt(10)/2}}} will result in the same thing as squaring {{{sqrt(10)/2}}} 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(26)/2}}} or {{{x = -sqrt(26)/2}}}, for {{{y = -sqrt(10)/2}}} as we did for {{{y = sqrt(10)/2}}}. So we have two more solutions:
({{{sqrt(26)/2}}}, {{{-sqrt(10)/2}}}) and ({{{-sqrt(26)/2}}}, {{{-sqrt(10)/2}}})<br>
All together we have four solutions:
({{{sqrt(26)/2}}}, {{{sqrt(10)/2}}}) and ({{{-sqrt(26)/2}}}, {{{sqrt(10)/2}}}) and ({{{sqrt(26)/2}}}, {{{-sqrt(10)/2}}}) and ({{{-sqrt(26)/2}}}, {{{-sqrt(10)/2}}})<br>
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.