Question

determining when an equation has infinite solutions\which equation is an identity?\\$\boldsymbol{3(x - 1)=x + 2(x + 1)+1}$\\$\boldsymbol{x - 4(x + 1)=-3(x + 1)+1}$\\$\boldsymbol{2x + 3=\frac{1}{2}(4x + 2)+2}$\\$\boldsymbol{\frac{1}{3}(6x - 3)=3(x + 1)-x - 2}$

Explanation:

Step1: Simplify Option A

Expand both sides:
Left: $3(x-1)=3x-3$
Right: $(x+2)(x+1)+1=x^2+3x+2+1=x^2+3x+3$
Result: $3x-3=x^2+3x+3$ → $-3=x^2+3$, not an identity.

Step2: Simplify Option B

Expand both sides:
Left: $x-4(x+1)=x-4x-4=-3x-4$
Right: $-3(x+1)+1=-3x-3+1=-3x-2$
Result: $-3x-4=-3x-2$ → $-4=-2$, not an identity.

Step3: Simplify Option C

Expand both sides:
Left: $2x+3$
Right: $\frac{1}{2}(4x+2)+2=2x+1+2=2x+3$
Result: $2x+3=2x+3$, which is an identity.

Step4: Verify Option D (optional)

Expand both sides:
Left: $\frac{1}{3}(6x-3)=2x-1$
Right: $3(x+1)-x-2=3x+3-x-2=2x+1$
Result: $2x-1=2x+1$ → $-1=1$, not an identity.

Answer:

C. $2x+3 = \frac{1}{2}(4x+2)+2$