Transform each of the following equations to determine whether each has one solution, infinitely many solutions or no solution. Use the result of your transformation to state the number of solutions.

1. 2x + 4 = 3(x – 2) + 1

2. 4(x + 3) = 3x + 17

3. 3x + 4 + 2x = 5(x – 2) + 7

4. 2(1 + 5x) = 5(2x – 1)

5. –18 + 15x = 3(4x – 6) + 3x

1.
we distribute then move placeholders to get
2x+4=4x-6+1
9=x
one solution

2. distribute
4x+12=3x+17
x=5
one solution

3. distribute and combine like terms
5x+4=5x-10+7
4=-3
false
no solution

4. distribute
2+10x=10x-5
2=-5
false
no solution

5. distribute
-18+15x=12x-18+3x
x=x
true
infinite solutions

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v1nce123 v1nce123 · Answer 2 · 2016-10-20T08:19:03+02:00

1. 2x + 4 = 3(x – 2) + 1
     2x +4=3x-6+1
     2x-3x=-5-4
         -x =-9
          x = 9 one solution

2. 4(x + 3) = 3x + 17
    4x +12=3x+17
      4x-3x=17-12
            x = 5 one solution

3. 3x + 4 + 2x = 5(x – 2) + 7
     5x+4=5x-10+7
     5x-5x=-3-4
          0x=-7 no solution

4. 2(1 + 5x) = 5(2x – 1)
      2+10x=10x-5
         10x-10x=-5-2
               0x =-7 no solution

5. –18 + 15x = 3(4x – 6) + 3x
    -18+15x=12x-18+3x
         15x-15x = -18+18
               0    =   0 infinite solutions