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Solve this Simultaneous equations?
x^2 + y^2 = 53
y = 3x - 1
im stuck this is what ive done so far
x^2 + (2x+1)^2=2
(2x+1)(2x+1) + x^2
(4x^2+2x+2x+1) + x^2
2 Answers
- Anonymous1 decade agoFavorite Answer
1. By substitution method, substitute the second equation for the first one and solve for x:
x² + (3x - 1)² = 53
x² + 9x² - 6x + 1 = 53
10x² - 6x + 1 = 53
10x² - 6x - 52 = 0
2(5x² - 3x - 26) = 0
2(5x - 13)(x + 2) = 0
5x - 13 = 0 and x + 2 = 0
x = {13/5, -2}
Substitute those for either equation and solve for x.
y = 3(13/5) - 1
=> 39/5 - 5/5
=> 34/5
y = 3(-2) - 1
=> -6 - 1
=> -7
Solutions: (13/5, 34/5) and (-2,-7)
2. You are good with the first steps. Keep going with the computation. Follow the steps as I did for the first problem.
5x² + 4x + 1 = 2
5x² + 4x - 1 = 0
(5x - 1)(x + 1) = 0
5x - 1 = 0 and x + 1 = 0
x = {1/5, -1}
I hope this helps!
Source(s): Knowledge - 1 decade ago
x^2+y^2=53
y= 3x-1
x^2+(3x-1)^2=53
x^2+9x^2-6x+1=53
10x^2-6x-52=0
a=10 b=-6 c=-52
6+/- sq rt 6^2-4(10)(-52)/20
6+/- sq rt 36+2080/20
6+/- sq rt 2116/20
6+/- 46/20
x=-2 or 13/5
4+y^2=53
y^2=49
y=+/-7 -2,13/5 7,-7