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How do you solve the simultaneous equations #x²+y²=5# and #y=3x+1 #?

Answer 1

Nathan Axel

#x = 2/5,quad y=11/5#

#x = -1, y=2#

Substitute #y=3x+1# in #x^2+y^2=5# to obtain

#x^2+(3x+1)^2=5qquad implies qquad 10x^2+6x-4=0#

Hence

#5x^2+3x-2 = 0 qquad implies qquad 5x^2+5x-2x-2 = 0qquad implies# #5x(x+1)-2(x+1) = 0 qquad implies qquad (5x-2)(x+1) = 0#

Hence

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Answer 2

Avery Alexander

To solve the simultaneous equations ( x^2 + y^2 = 5 ) and ( y = 3x + 1 ), substitute the expression for ( y ) from the second equation into the first equation. This gives you a quadratic equation in terms of ( x ). Solve for ( x ) and then use the solution(s) to find the corresponding values of ( y ) using the second equation.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.