Asked by mizuki akiyama
                decide whether each equation has one solution, no solution, or infinitely many solutions.
1. 2(x-3) = 2x
2. 3(y-3) = 2y-9+y
3. 10x-2-6x=3x-2+x
4. 4(x+3) = 2x= x-8
please be quick!!
 
            
            
        1. 2(x-3) = 2x
2. 3(y-3) = 2y-9+y
3. 10x-2-6x=3x-2+x
4. 4(x+3) = 2x= x-8
please be quick!!
Answers
                    Answered by
            mathhelper
            
    simplify each equation, each of which is linear.
If your variable drops out, and
you get a true statement ----> an infinite number of solutions
you get a false statement ---> no solution
If you end up with the variable remaining in the equation, you will have
one solution
e.g. the first one:
2(x-3) = 2x
2x - 6 = 2x
subtract 2x from both sides:
-6 = 0 , which is false, so.... no solution
do the rest in the same way.
btw, you have a typo in the last, two equal signs.
    
If your variable drops out, and
you get a true statement ----> an infinite number of solutions
you get a false statement ---> no solution
If you end up with the variable remaining in the equation, you will have
one solution
e.g. the first one:
2(x-3) = 2x
2x - 6 = 2x
subtract 2x from both sides:
-6 = 0 , which is false, so.... no solution
do the rest in the same way.
btw, you have a typo in the last, two equal signs.
                    Answered by
            mizuki akiyama
            
    4. 4(x+3) + 2x = x-8
    
                    Answered by
            mizuki akiyama
            
    thx mathhelper
    
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