# AlgebraII-Reiny could you please check-I'm stuck

posted by on .

Sikve the following by using substitution method

x=3y -5
6x-18y = -30

I got this far and I'm stuck
6(3y-5) - 18y = -30
18y - 30 - 18y = -30
-30 = -30

So this doesn't make sense anymore

• AlgebraII-Reiny could you please check-I'm stuck - ,

you did nothing wrong.
Notice that if you divide your second equation by 6 , you get
x - 3y = -5, which is really the same as your first equation.

So you were actually only given one equation twice.

In general, if you solve using proper algebra and your variables disappear, one of two things can happen
1. the resulting statement is TRUE.
in that case, there will be an infinite number of correct solutions. This is your case

2. the resulting statement in FALSE.
in that case there is no solution.

So your pair of equations have an infinite number of solutions, just form a table of values for
x = 3y - 5 to get a few of them if you have to

• AlgebraII-Please recheckReiny or anyone available to help could you please check-I'm stuck - ,

So some possible answers would be:
(13,6)
(22,9)
etc:

is that correct then?