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1 vote
Solve the simultaneous equation:
3y=2(x^2-3)
2x-y=2

User Gowrath
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2 Answers

21 votes
21 votes
Solve the simultaneous equation
User HoangHieu
by
2.7k points
10 votes
10 votes

Answer:

(0, - 2 ) and (3, 4 )

Explanation:

Given the equations

3y = 2(x² - 3) , that is

3y = 2x² - 6 → (1)

2x - y = 2 (add y to both sides )

2x = 2 + y ( subtract 2 from both sides )

2x - 2 = y → (2)

Substitute y = 2x - 2 into (1)

3(2x - 2) = 2x² - 6

6x - 6 = 2x² - 6 ( subtract 6x - 6 from bpth sides )

0 = 2x² - 6x ( factor out 2x from each term )

0 = 2x(x - 3)

Equate each factor to zero and solve for x

2x = 0 ⇒ x = 0

x - 3 = 0 ⇒ x = 3

Substitute these values into (2) for corresponding values of y

x = 0 : y = 2(0) - 2 = 0 - 2 = - 2 ⇒ (0, - 2 )

x = 3 : y = 2(3) - 2 = 6 - 2 = 4 ⇒ (3, 4 )

User Pattie
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2.9k points