1.) Let's solve for x.
2(2x + 5) = 4(x + 3)
4x + 10 = 4x + 12
4x - 4x = 12 - 10
0 = 2
Since we can't solve for x and also the argument is not true that 0 is not equal to 2. Therefore, NO SOLUTION (C).
2.) Let's solve for x.
a + 5 = 1/5 (5a + 25)
a + 5 = a + 5
a - a = 5 - 5
0 = 0
Since we can't solve for x but the argument is true which zero is equal to zero. This case becomes "indeterminate" because 0/1 = 0, 0/2 = 0, 0/3 = 0 and so on. Therefore, the answer is INFINITE SOLUTIONS (B).
3) Let's solve for x.
3(4x - 3) - 7x = 5x - 9
12x - 9 - 7x = 5x - 9
5x - 9 = 5x - 9
5x - 5x = -9 + 9
0 = 0
Therefore, INFINITE SOLUTIONS (B).
4.) Let's test each choices if the solution does not exist.
a.) 8 + 2(8x – 6) = 2(4x - 7)
8 + 16x -12 = 8x - 14
16x - 8x = -14 - 8 + 12
8x = -10
x = -1.25 (one solution)
b.) 8 + 2(8x – 6) = 9x - 19
8 + 16x - 12 = 9x - 19
16x - 9x = -19 - 8 + 12
7x = -15
x = -15/7 (one solution)
c.) 8 + 2(8x – 6) = 16x - 4
8 + 16x - 12 = 16x - 4
16x - 16x = -4 - 8 + 12
0 = 0 (infinite solutions)
d.) 8 + 2(8x – 6) = 4(4x -1)
8 + 16x - 12 = 16x - 4
16x - 16x = -4 - 8 + 12
0 = 0 (infinite solutions)
Therefore, the answers are C and D.
5.) Since we know how to solve for x, I am gonna conclude the answer instead. Identity is another term of "one solution". Contradiction is another term of "no solution". Neither is considered as "infinite solutions."
Number 1 is NEITHER.
Number 2 is CONTRADICTION.
Number 3 is IDENTITY.
Number 4 is NEITHER.
