# 4x+y=5 , 27x*2+21xy+2y*2=0add math question (simultaneous equation)

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### 1 Answer

Given:

4x + y = 5 ... (1)

And

27x^2 + 21 xy + 2y^2 = 0 ... (2)

We factorise the left hand side of this equation as follows:

27x^2 + 18 xy + 3xy + 2y^2 = 0

9x(3x + 2y) + y(3x + 2y) = 0

(3x + 2y)(9x + y) = 0

Therefore:

3x + 2y = 0 ... (3)

9x + y = 0 ... (4)

Equation (3) and equation (4) representing two factors of equation (2) will give us two different sets of value for x an y. To find these two sets of values we form two separate sets of equation, which are (1) with (3), and (1) with (4).

First we solve equations (1) and (3) as follows.

Multiplying equation (1) by 2 we get:

8x + 2y = 10 ... (5)

Subtracting equation (3) from equation (5) we get:

8x - 3x + 2y - 2y = 10

5x = 10

x = 2

Substituting this value of x in equation (5)we get:

8*2 + 2y = 10

16 + 2y = 10

2y = 10 - 16 = - 6

Therefore:

y = - 6/2 = - 3

This gives first set of values as x = 2, and y = -3

Now we solve equation (1) and (4) as follows:

Subtracting equation (1) from (4) we get

9x - 4x + y - y = 0 - 5

5x = - 5

Therefore x = - 5/5 = - 1

Substituting this value of x in equation (1) we get

4(-1) + y = 5

- 4 + y = 5

y = 5 +4 = 9

This gives second set of values as x = -1, and y = 9