Question

• 3a + b = 8
3a² +b² = 28

How to solve this SIMULTANEOUS EQUATION?
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Answer 1

Answer:

(a,b) = (1,5) , (3,-1)

Step-by-step explanation:

Given equations are

3a + b = 8 ..........(1)

3a² + b² = 28 ......(2)

From (1) b = 8 - 3a. Putting this value in (2) we get

3a² + (8-3a)² = 28

⇒ 3a² + 64 -48a + 9a² = 28

⇒ 12a² - 48a + 36 = 0

⇒ a² - 4a + 3 = 0

⇒ (a-1)(a-3) = 0

⇒ a = 1 or 3

Using b = 8 -3a

when a= 1, b = 8-3 = 5

when a= 3, b = 8-3*3 = -1

Hence the solutions are (a,b) = (1,5) , (3,-1)