Question

algebra 1 (25-26) practice makes progress #6
which equation has a solution of all real numbers?
-4(1 - 4x) = -28 + 28x
3(-x - 1) = -6(x - 1)
-2(x + 1) = -(2x - 3) - 5
16x - 32 = -4 + 16x + 2

Explanation:

Step1: Simplify Option A

Expand left side: $-4(1-4x) = -4 + 16x$
Right side: $-28 + 28x$
Equation becomes: $-4 + 16x = -28 + 28x$

Step2: Solve Option A

Rearrange terms: $16x - 28x = -28 + 4$
$\Rightarrow -12x = -24$
$\Rightarrow x=2$ (single solution)

Step3: Simplify Option B

Expand left side: $3(-x-1) = -3x -3$
Expand right side: $-6(x-1) = -6x +6$
Equation becomes: $-3x -3 = -6x +6$

Step4: Solve Option B

Rearrange terms: $-3x +6x = 6 +3$
$\Rightarrow 3x=9$
$\Rightarrow x=3$ (single solution)

Step5: Simplify Option C

Expand left side: $-2(x+1) = -2x -2$
Expand right side: $-(2x-3)-5 = -2x +3 -5 = -2x -2$
Equation becomes: $-2x -2 = -2x -2$

Step6: Analyze Option C

Add $2x+2$ to both sides: $0=0$ (always true)

Step7: Simplify Option D

Right side: $-4 +16x +2 = 16x -2$
Equation becomes: $16x -32 = 16x -2$

Step8: Analyze Option D

Subtract $16x$ from both sides: $-32=-2$ (false)

Answer:

C. $3(-x - 1) = -6(x - 1)$