To determine which equations have infinitely many solutions, we need to find out if the equations simplify to a true statement (e.g., 0=0) or if they isolate to a contradiction (e.g., 0=5).
Let's analyze each equation:
A. \( 6(x-9)=6x-54+x \)
Expanding both sides:
- Left side: \( 6x - 54 \)
- Right side: \( 6x - 54 + x \) simplifies to \( 7x - 54 \)
Setting them equal: \( 6x - 54 = 7x - 54 \)
Subtracting \( 6x \) and adding \( 54 \) gives: \( 0 = x \)
This has a unique solution (x = 0). Not infinitely many solutions.
B. \( 5(x-8)+10=5(x+2) \)
Expanding both sides:
- Left side: \( 5x - 40 + 10 = 5x - 30 \)
- Right side: \( 5x + 10 \)
Setting them equal: \( 5x - 30 = 5x + 10 \)
Subtracting \( 5x \) gives: \( -30 = 10 \)
This is a contradiction, so it has no solutions.
C. \( 20-8x-6=2(7-4x) \)
Simplifying both sides:
- Left side: \( 14 - 8x \)
- Right side: \( 14 - 8x \)
Setting them equal: \( 14 - 8x = 14 - 8x \)
This is a true statement, indicating infinitely many solutions.
D. \( 4(x+4)=2(2x+5)+6 \)
Expanding both sides:
- Left side: \( 4x + 16 \)
- Right side: \( 4x + 10 + 6 = 4x + 16 \)
Setting them equal: \( 4x + 16 = 4x + 16 \)
This is also a true statement, indicating infinitely many solutions.
E. \( 2x+9=18+2x \)
Subtracting \( 2x \) from both sides: \( 9 = 18 \)
This is a contradiction, hence no solutions.
Conclusion:
The equations that have infinitely many solutions are C and D.
1 answer