To determine the missing values that make the equation \(2(6x + 3) = 6x + ?x + 6\) have zero solutions, we'll start by simplifying both sides and analyzing the equation.
-
Expand the left side: \[ 2(6x + 3) = 12x + 6 \]
-
Rewrite the equation with the left side expanded: \[ 12x + 6 = 6x + ?x + 6 \] Simplifying the right side gives: \[ 12x + 6 = (6 + ? )x + 6 \]
-
Remove the common term (6) from both sides: \[ 12x = (6 + ? )x \]
-
Set the coefficients of \(x\) on both sides equal to each other: \[ 12 = 6 + ? \]
-
Solve for the missing value: \[ ? = 12 - 6 = 6 \]
Now, let's analyze the equation \(12x = (6 + 6)x\) or \(12x = 12x\). This indicates that the equation is true for all values of \(x\).
To find a scenario where there are zero solutions, we want to adjust the right side such that the coefficients are not equal. If we let \(? < 6\) (significantly more negative, or adjust to a value not equal to zero), then:
- If we choose \(? = -1\), we can see what happens: \[ 12x = (6 - 1)x \Rightarrow 12x = 5x \] Which simplifies to: \[ 12x - 5x = 0 \Rightarrow 7x = 0 \] This leads to \(x = 0\) which implies a solution exists.
To derive a case with zero solutions, simply set: \[ ? = -6 \] Substituting back will yield: \[ 12x = (6 - 6)x \Rightarrow 12x = 0 \] Since this states that \(12x\) must equal \(0\), we have determined that there are indeed no possible non-zero solutions from any \(x \ne 0\).
Thus, to achieve the equation yielding zero solutions, set \(\mathbf{? = -6}\).
1 answer