Solve the following system of equations:
3x² + 4xy - 2y² = 2
3x + 2y = -10
Solve the second equation for either of the unknowns.
I'll pick x:
3x + 2y = -10
add -2y to both sides
3x = -10 - 2y
Divide both sides by 3
x = (-10 - 2y)/3
factor out -2
x = -2(5 + y)/3
In the first original equation,
3x² + 4xy - 2y² = 2
substitute -2(5 + y)/3 for x
3[-2(5 + y)/3]² + 4[-2(5 + y)/3]y - 2y² = 2
3[4(5 + y)²/9] + 4y[-2(5 + y)/3] - 2y² = 2
4(5 + y)²/3 - 8y(5 + y)/3 - 2y² = 2
4(25 + 10y + y²)/3 - (40y + 8y²)/3 - 2y² = 2
Multiply through by 3 to clear of fractions:
4(25 + 10y + y²) - (40y + 8y²) - 6y² = 6
100 + 40y + 4y² - 40y - 8y² - 6y² = 6
Collect terms
100 - 10y² = 6
Get 0 on the right
94 - 10y² = 0
Divide thru by 2 47 - 5y² = 0
-5y² = -47
y² = 47/5
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y = ±Ö47/5
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Rationalize denominator y = ±Ö235/5
Substitute into 3x + 2y = -10
Using the + value
___
3x + 2(Ö235/5) = -10
Multiply through by 5
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15x + 2Ö235 = -50
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15x = -50 - 2Ö235
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x = (-50 - 2Ö235)/15
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x = -2(25 + Ö235)/15
So one solution is
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(x, y) = ( -2(25 + Ö235)/15, Ö235/5 )
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Using the - value
___
3x + 2(-Ö235/5) = -10
Multiply through by 5
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15x - 2Ö235 = -50
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15x = -50 + 2Ö235
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x = (-50 + 2Ö235)/15
___
x = -2(25 - Ö235)/15
So the other solution is
___ ___
(x, y) = ( -2(25 - Ö235)/15, -Ö235/5 )
Tell your teacher I said he or she is a meany for giving you such a
messy problem. It is messy enough as it is. However, if you solve for
y first instead of x, the problem is even messier, in fact a lot
messier. That was the first way I did it!! Also there is no way to
tell in advance which way is going to be messier. Your teacher needs
a spanking!! :-)
Edwin
