Hi, Im really confused on one of the problems that I have for homework. Solve the system: x^2 + 2y^2 =33 x^2 + y^2 + 2x =19 Thanks!
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Multiply the bottom equation by 2. Subtract the bottom from the top and you should get a quadratic which you can solve.
okay, i did it out and got -1 + square route of six, and -1 - square route of six. Did i do it right?
$\displaystyle x^2 + 2y^2 =33$ $\displaystyle 2x^2+2y^2+4x=38$ $\displaystyle -x^2-4x+5=0$ $\displaystyle x^2+4x-5=0$ $\displaystyle (x+5)(x-1)=0$ take it from there.
ohhh ok thanks i get it now i did some stupid multiplication thing before
Be sure you finish the problem! The reason rtblue said "take it from there" is that a solution is a pair of numbers, x and y. This problem has four separate solutions.
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