Question 167397
1.{{{x^2+2y^2=18}}}
2.{{{x=2y}}}
Substitute eq. 2 into eq. 1, directly substitute for x and solve for y.
Then go back and find x.
1.{{{x^2+2y^2=18}}}
{{{(2y)^2+2y^2=18}}}
{{{4y^2+2y^2=18}}}
{{{6y^2=18}}}
{{{y^2=3}}}
{{{y=0 +- sqrt(3)}}}
Then from eq. 2,
2.{{{x=2y}}}
{{{x=2(0 +- sqrt(3))}}}
{{{x=0 +- 2*sqrt(3)}}}
.
.
.
When we graph the equations, you can see the solutions more clearly.
Equation 1 is the ellipse.
Equation 2 is the straight line.
.
.
.
(1.73,3.46) and (-1.73,-3.46)
.
.
.
{{{ graph( 300, 300, -5, 5, -5, 5, x/2, sqrt((18-x^2)/2), -sqrt((18-x^2)/2)) }}}