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January 31, 2015

January 31, 2015

Posted by **Caroline** on Tuesday, January 22, 2013 at 10:29pm.

1.x=-3y+4 and 3x+9y=12

3(-3y+4)+9y=12

-9y+12+9y12

0=0

x=-3(0)+4

x=4

Would there be infinite solutions or would it be (0,4)?

2. x=3y-1 and 2x-6y=-2

2x-6y=-2

2(3y-1)-6y=-2

6y-2-6y=-2

0=0

x=3y-1

x=3(0)-1

x=-1

Would there be infinite solutions or would it be (-1,0)?

- Pre-Calculus -
**Reiny**, Tuesday, January 22, 2013 at 10:47pmfor your first one, the variable dropped out, but you ended up with a true statement, 0=0

If that happens you will have an infinite number of solutions. In effect , the two equations are really the same equation.

look at the 2nd

3x + 9y = 12 , divide by 3

x + 3y = 4

or

x = -3y + 4 , which was your first equation.

If your variable drops out, but you end up with a false statement, such as 4 = 9

then there is no solution.

In that case, what you would be given would be two lines that are parallel, thus they never meet, thus no solution

The same result was true in your second question

notice by dividing each term of the 2nd by 2, then re-arranging you get your first equation

I also noticed that in both cases you substituted x=0

but nowhere did it say that x = 0 , it said 0 = 0

no mention of x

So if you have a true statement, ---> infinite number of solutions

If you have a false statement --> no solution at all

- Pre-Calculus -
**Caroline**, Tuesday, January 22, 2013 at 10:51pmThank you so much.

- Pre-Calculus -
**Reiny**, Tuesday, January 22, 2013 at 11:12pmwelcome

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