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2 Answers

Hi, Mario!
 
Is the system supposed to be the following?
 
x2 + 2y2 = 2
x - y = 2
 
If so, graphing the system shows an ellipse with a line outside of it; they do not intersect, therefore no solution. 
 
To solve it algebraically, solve the linear equation for either variable: x = y + 2
 
Substitute y+2 into the first equation:
 
(y + 2)2 + 2y2 = 2
y2 + 4y + 4 + 2y2 = 2
3y2 + 4y + 2 = 0
 
Using the quadratic formula results in a negative number under the square root sign; therefore, there is no real number solution.
 
Hope this helps!
Kathye P.
No solution as written; the first is an ellipse, the second a line, and they don't intersect.
 
Using substitution you can generate 3y2 + 4y + 2 = 0, which has no real roots. There are imaginary (complex, actually) roots where
 
y = (1/3)(-2 ± i√2)