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Solve the equations:
x + 3y = 13
x2 + 3y2 = 43

User Tstirrat
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1 Answer

5 votes

Answer:

(4, 3 ) and (2.5, 3.5 )

Explanation:

Given the 2 equations

x + 3y = 13 → (1)

x² + 3y² = 43 → (2)

Rearrange (1) expressing x in terms of y, that is

x = 13 - 3y → (3)

Substitute x = 13 - 3y into (2)

(13 - 3y)² + 3y² = 43 ← expand factor using FOIL

169 - 78y + 9y² + 3y² = 43

12y² - 78y + 169 = 43 ( subtract 43 from both sides )

12y² - 78y + 126 = 0 ( divide through by 6 )

2y² - 13y + 21 = 0 ← in standard form

(y - 3)(2y - 7) = 0 ← in factored form

Equate each factor to zero and solve for y

y - 3 = 0 ⇒ y = 3

2y - 7 = 0 ⇒ 2y = 7 ⇒ y = 3.5

Substitute these values into (3) for corresponding values of x

y = 3 → x = 13 - 3(3) = 13 - 9 = 4 ⇒ (4, 3 )

y = 3.5 → x = 13 - 3(3.5) = 13 - 10.5 = 2.5 ⇒ (2.5, 3.5 )

User Sushant Aryal
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