Question

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

Explanation:

Step1: Analyze the first equation

Expand both sides: Left side: \(x - 4(x + 1)=x - 4x - 4=-3x - 4\). Right side: \(-3(x + 1)+1=-3x - 3 + 1=-3x - 2\). Since \(-3x - 4
eq -3x - 2\), not an identity.

Step2: Analyze the second equation

Expand the right side: \(\frac{1}{2}(4x + 2)+2 = 2x + 1 + 2=2x + 3\). Left side is \(2x + 3\). So \(2x + 3 = 2x + 3\), which is an identity.

Step3: Analyze the third equation (optional, but for thoroughness)

Expand both sides: Left side: \(3(x - 1)=3x - 3\). Right side: \(x + 2(x - 1)-1=x + 2x - 2 - 1=3x - 3\). Wait, wait, let's check again. Wait, original third equation: \(3(x - 1)=x + 2(x - 1)-1\)? Wait, expanding right side: \(x + 2x - 2 - 1 = 3x - 3\), left side is \(3x - 3\). Wait, but maybe a typo? Wait, no, let's check the second equation again. Wait, second equation: \(2x + 3=\frac{1}{2}(4x + 2)+2\). \(\frac{1}{2}(4x + 2)=2x + 1\), then \(2x + 1 + 2 = 2x + 3\), so left side equals right side. So second equation is identity. Wait, but let's check third: \(3(x - 1)=3x - 3\), right side: \(x + 2(x - 1)-1=x + 2x - 2 - 1=3x - 3\), so that's also equal? Wait, maybe the third equation has a typo? Wait, the user's image: third equation: \(3(x - 1)=x + 2(x - 1)-1\)? Wait, no, maybe I misread. Wait, no, let's re - check the second equation. The second equation is \(2x + 3=\frac{1}{2}(4x + 2)+2\). Calculating right side: \(\frac{1}{2}\times4x+\frac{1}{2}\times2 + 2=2x + 1+2 = 2x + 3\), which is equal to left side. So second equation is identity.

Answer:

The equation \(2x + 3=\frac{1}{2}(4x + 2)+2\) is an identity.