Question

solving linear equations: distributive property
which of these equations have no solution? choose three correct answers.
$square$ $2x + 3(x + 5) = 5(x - 3)$
$square$ $4 - (2x + 6) = \frac{1}{2}(-4x - 2)$
$square$ $4(x + 3) = x + 12$
$square$ $3(x - 2) + 2 = 3(x + 3) + 1$
$square$ $5(x + 1) - x = 4(x + 5) - 1$

Explanation:

Step1: Analyze \(2x + 3(x + 5)=5(x - 3)\)

Expand both sides: \(2x + 3x + 15 = 5x - 15\)
Simplify: \(5x + 15 = 5x - 15\)
Subtract \(5x\): \(15=-15\) (False, no solution)

Step2: Analyze \(4-(2x + 6)=\frac{1}{2}(-4x - 4)\)

Simplify left: \(4 - 2x - 6=-2x - 2\)
Simplify right: \(-2x - 2\)
So \(-2x - 2=-2x - 2\) (True, infinite solutions)

Step3: Analyze \(4(x + 3)=x + 12\)

Expand: \(4x + 12 = x + 12\)
Subtract \(x\) and \(12\): \(3x = 0\Rightarrow x = 0\) (Has solution)

Step4: Analyze \(3(x - 2)+2 = 3(x + 3)+1\)

Expand: \(3x - 6 + 2 = 3x + 9 + 1\)
Simplify: \(3x - 4 = 3x + 10\)
Subtract \(3x\): \(-4 = 10\) (False, no solution)

Step5: Analyze \(5(x + 1)-x = 4(x + 5)-1\)

Expand: \(5x + 5 - x = 4x + 20 - 1\)
Simplify: \(4x + 5 = 4x + 19\)
Subtract \(4x\): \(5 = 19\) (False, no solution)

Answer:

The equations with no solution are \(2x + 3(x + 5)=5(x - 3)\), \(3(x - 2)+2 = 3(x + 3)+1\), and \(5(x + 1)-x = 4(x + 5)-1\) (select their corresponding checkboxes).