Solve the system of equations 3x^2-3y^2=27 and 2x-2y=2 .

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The system of equations to be solved is

3x^2-3y^2=27 ...(1)

2x-2y=2 ...(2)

(2)

=> x - y = 1

=> y = x - 1

substitute in (1)

3(x^2) - 3(x - 1)^2 = 27

=> x^2 - (x - 1)^2 = 9

=> x^2 - x^2 - 1 + 2x = 9

=> 2x = 10

=> x = 5

y = x - 1 = 4

The solution of the equations is (5 , 4)

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll re-write the first equation factorizing by 3:

3(x^2 - y^2) = 27

We'll divide by 3:

x^2 - y^2  = 9 (1)

x - y = 1 (2)

We notice that we can re-write (1) as a product, being a difference of squares:

(x-y)(x+y) = 9

But x-y = 1 => x + y = 9 (3)

We'll add (3) to (2):

x + y + x – y = 9 + 1

We'll eliminate like terms:

2x = 10

x = 5

We'll substitute x = 5 into (2):

5 – y = 1

-y = -5 + 1

y = 4

The solution of the system is represented by the pair {5 ; 4}.

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