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leonardthegreat · May 27, 2009, 01:03 PM

how do you find intersections in two equations such as x^2/4+y^2=1 and x+y=1

Perito · May 27, 2009, 02:45 PM

The intersection is where the two lines/curves meet. If you graph them, there will be an intersection, and you can read it from the graph. Otherwise, you do something like this (I'll solve this using "substitution". There are other methods):

$x+y=1$

therefore

$y=1-x$

$\frac {x^2}{4} + y^2 = 1$

So, substituting 1-x for y, you get

$\frac {x^2}{4} + (1-x)^2 = 1$

Simplifying and collecting terms

$x^2 + 4(1-x)^2 = 4$

$x^2 +4x^2 -8x +4=4$

$5x^2 -8x =0$

factor

$x(5x-8)=0$

x=0 and x=8/5 are the roots (solutions) of the equation. You can find the corresponding y values by substituting into the equations. If x=0 then y=1; if x=8/5 then y=-3/5. The two points of intersection are, therefore (0,1) and (8/5, -3/5)