solve pairs of simultaneous equation: 3x+y=1 ; x²+y²=5
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Given,
3x +y=1 ....equ(1) and x^2+y^2 = 5 ...equ(2)
y = 1-3x from (1)
substituting equ(2)
x^2 + (1 +3x)^2 = 5
x^2 + 1 +6x+9x^2 = 5
10x^2 + 6x -4 = 0
5x^2 +3x-2=0
5x^2 +5x-2x-2=0
5x(x+1)-2(x+1) = 0
(x+1)(5x-2) = 0
x = -1 and 2/5.
y = 1-3x
y=1+3 = 4
y=1-6/5 = -1/5.
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