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Solve the system of equations: x^2 + y^2 = 1 3x + y = -1

1 Answer

4 votes

Answer:

(0, - 1 ) , (-
(3)/(5),
(4)/(5) )

Explanation:

given the 2 equations to solve simultaneously

x² + y² = 1 → (1)

3x + y = - 1 ( subtract 3x from both sides )

y = - 1 - 3x → (2)

substitute y = - 1 - 3x into (1)

x² + (- 1 - 3x)² = 1 ← expand parenthesis using FOIL

x² + 1 + 6x + 9x² = 1

10x² + 6x + 1 = 1 ( subtract 1 from both sides )

10x² + 6x = 0 ← factor out 2x from each term on the left side

2x(5x + 3) = 0 ← in factored form

equate each factor to zero and solve for x

2x = 0 ⇒ x = 0

5x + 3 = 0 ( subtract 3 from both sides )

5x = - 3 ( divide both sides by 5 )

x = -
(3)/(5)

substitute these values of x into (2) for corresponding y values

x = 0 : y = - 1 - 3(0) = - 1 - 0 = - 1 ⇒ (0, - 1 )

x = -
(3)/(5) : y = - 1 - 3(-
(3)/(5) ) = -
(5)/(5) +
(9)/(5) =
(4)/(5) ⇒ (-
(3)/(5),
(4)/(5) )

solutions are (0, - 1 ) and (-
(3)/(5),
(4)/(5) )

User JAY RAPARKA
by
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