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Solve the simultaneous equation?
y= 3x+2
x²+y²=20
Thanks to anyone who can help :)
3 Answers
- ?Lv 78 years agoFavorite Answer
y= 3x+2 square THIS equation and subtract the OTHER equation
x²+y²=20
y^2 = 9x^2 +12x +4
y^2 = -x^2 +20
0 = 10x^2 + 12x -16 divide by 2
5x^2 + 6x -8 = 0
(5x -4)(x + 2) = 0
2 solutions
5x-4= 0
5x=4
x= 4/5 y = 22/5
OR
x+2= 0
x = -2 y = 4
- Marley KLv 78 years ago
substituting the first (linear) equation into the second equation:
x² + (3x + 2)² = 20
x² + 9x² + 12x + 4 = 20
10x² + 12x - 16 = 0
5x² + 6x - 8 = 0
(5x - 2)(x + 2) = 0
x = 2/5 or x = -2
when x = 2/5,
y = 3(2/5) + 2 = 16/5
when x = -2,
y = 3(-2) + 2 = -4
two solutions:
(2/5, 16/5) and (-2,-4)
that's it! ;)
- 8 years ago
x squared * (3x+2)(3x+2)=20
xsquared * 9xsquared+12x+4=20
10x(s)+12x-16=0
use a calculator use quadratic formula
then subsitute x value in to y=3x+2
