Dividing the first one by two we obtain x+2y=5 whence x=5-2y Now we substitute this expression into (2): 3(5-2y)+6y=15 Opening the braces we obtain: 15-6y+6y=15 which gives us 15=15 - right equality Basicaly, such outcome means that our two equations are not independent. We can clearly see that if we divide (2) by 3: we obtain x+2y=5 which is teh same as (1) divided by 2. Actually, we have one distinct equation. Such system is called indefinite. We cannot uniquely calculate the solution. Instead such system has infinit number of solutions for which the equality x=5-2y holds.

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Assignment Expert24.10.13, 19:312x+4y=10 (1)

3x+6y=15 (2)

Dividing the first one by two we obtain x+2y=5 whence x=5-2y

Now we substitute this expression into (2):

3(5-2y)+6y=15

Opening the braces we obtain:

15-6y+6y=15 which gives us 15=15 - right equality

Basicaly, such outcome means that our two equations are not independent. We can clearly see that if we divide (2) by 3: we obtain x+2y=5 which is teh same as (1) divided by 2. Actually, we have one distinct equation. Such system is called indefinite. We cannot uniquely calculate the solution. Instead such system has infinit number of solutions for which the equality x=5-2y holds.

abel02.10.13, 12:27Solve the linear equations 2x+4y=10 and 3x+6y=15

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