Question 170266
Solve:
{{{x^2+y^2 = 10}}}
{{{3y = x^2+8}}} Rewrite this as {{{x^2 = 3y-8}}} and substitute into the 1st equation.
{{{(3y-8)+y^2 = 10}}} Simplify.
{{{y^2+3y-18 = 0}}} Factor.
{{{(y-3)(y+6) = 0}}} Apply the zero product rule.
{{{y = 3}}} or {{{y = -6}}}
Use the positive solution and substitute into the second equation to solve for x.
{{{3y = x^2+8}}} Substitute y = 3.
{{{3(3) = x^2+8}}} Subtract 8 from both sides.
{{{9-8 = x^2}}} Take the square root of both sides.
{{{x = 1}}} or {{{x = -1}}}
The solution consists of two points of intersection:
(1, 3) and (-1, 3)
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Just for interest, since we had two solutions for y (y = 3 and y = -6), let's see what happens when we choose the second solution of y = -6 and substitute into the first equation to solve for x.
{{{3y = x^2+8}}} Substitute y = -6.
{{{3(-6) = x^2+8}}} Subtract 8 from both sides.
{{{x^2 = -26}}} Take the square root of both sides.
{{{x = sqrt(-26)}}} or {{{x = -sqrt(-26)}}} and, as you can see, the results are "imaginary" numbers, so there is no "real" solution for y = -6.
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Let's see what the graph of these two functions look like:
{{{graph(400,400,-5,5,-5,5,sqrt(-x^2+10),-sqrt(-x^2+10),(1/3)x^2+8/3)}}}