which equations have infinitely many solutions select two that apply
a. 6 ( x - 9 ) = 6x - 54 + x
b. 5 ( x - 8 ) + 10 = 5 ( x + 2 )
c. 20 - 8x - 6 = 2 ( 7 - 4x )
d. 4 ( x + 4 ) = 2 ( 2x + 5 ) + 6
e. 2x + 9 = 18 + 2x
a. 6 ( x - 9 ) = 6x - 54 + x
b. 5 ( x - 8 ) + 10 = 5 ( x + 2 )
Both equations a and b have infinitely many solutions.
To determine which equations have infinitely many solutions, we need to check if the variables cancel out on both sides of the equation. Let's go through each option:
a. 6(x - 9) = 6x - 54 + x
Expanding the equation:
6x - 54 = 6x - 54 + x
Combining like terms:
6x - 54 = 7x - 54
The variable x cancels out on both sides, which means it has infinitely many solutions.
b. 5(x - 8) + 10 = 5(x + 2)
Expanding the equation:
5x - 40 + 10 = 5x + 10
Combining like terms:
5x - 30 = 5x + 10
The variable x cancels out on both sides, which means it has infinitely many solutions.
c. 20 - 8x - 6 = 2(7 - 4x)
Simplifying the equation:
14 - 8x = 14 - 8x
The variable x cancels out on both sides, which means it has infinitely many solutions.
d. 4(x + 4) = 2(2x + 5) + 6
Expanding the equation:
4x + 16 = 4x + 10 + 6
Combining like terms:
4x + 16 = 4x + 16
The variable x cancels out on both sides, which means it has infinitely many solutions.
e. 2x + 9 = 18 + 2x
Subtracting 2x from both sides:
9 = 18
The variable x cancels out, leaving a false statement. This equation has no solutions.
Therefore, the two equations that have infinitely many solutions are:
a. 6(x - 9) = 6x - 54 + x
b. 5(x - 8) + 10 = 5(x + 2)
To determine which equations have infinitely many solutions, we need to solve each equation and check if they lead to a consistent statement. Let's go through each option.
a. 6(x - 9) = 6x - 54 + x
First, distribute the 6 on the left side of the equation:
6x - 54 = 6x - 54 + x
Now, combine like terms on the right side:
6x - 54 = 7x - 54
At this point, we can observe that the equation is true for all values of x. Since the variables cancel out, the statement becomes 0 = 0, which is always true. Thus, option (a) has infinitely many solutions.
b. 5(x - 8) + 10 = 5(x + 2)
Distribute the 5 on the left side of the equation:
5x - 40 + 10 = 5x + 10
Combine like terms on both sides:
5x - 30 = 5x + 10
In this case, we can see that the x terms will cancel out if we subtract 5x from both sides. This would lead to -30 = 10, which is not a true statement. Therefore, option (b) does not have infinitely many solutions.
c. 20 - 8x - 6 = 2(7 - 4x)
Combine like terms on both sides:
14 - 8x = 14 - 8x
Here, the x terms cancel out, and we obtain 14 = 14, which is always true. Hence, option (c) has infinitely many solutions.
d. 4(x + 4) = 2(2x + 5) + 6
Distribute on both sides:
4x + 16 = 4x + 10 + 6
Combine like terms:
4x + 16 = 4x + 16
In this case, the x terms cancel out, resulting in 16 = 16, which is always true. Therefore, option (d) has infinitely many solutions.
e. 2x + 9 = 18 + 2x
Move the x terms to one side:
2x - 2x = 18 - 9
After subtracting 2x from both sides, we get 0 = 9, which is not a true statement. Therefore, option (e) does not have infinitely many solutions.
In summary, options (a), (c), and (d) have infinitely many solutions.