Question

1\frac{4(6c + 4)}{8} = -2c + 4\
2\frac{6c + 4}{2} = -2c + 4\
\square = -2c + 4

Explanation:

Step1: Simplify the left - hand side fraction

We have the equation \(\frac{6c + 4}{2}=-2c + 4\). First, we can simplify the left - hand side by dividing each term in the numerator by 2. \(\frac{6c}{2}+\frac{4}{2}=3c + 2\). So the equation becomes \(3c+2=-2c + 4\).

Step2: Add \(2c\) to both sides

To get all the \(c\) terms on one side, we add \(2c\) to both sides of the equation \(3c + 2=-2c+4\). \((3c + 2)+2c=(-2c + 4)+2c\). Combining like terms, we have \(3c+2c+2=-2c + 2c+4\), which simplifies to \(5c+2 = 4\).

Step3: Subtract 2 from both sides

Subtract 2 from both sides of the equation \(5c+2 = 4\). \((5c + 2)-2=4 - 2\). This gives us \(5c=2\).

Step4: Divide both sides by 5

To solve for \(c\), we divide both sides of the equation \(5c = 2\) by 5. \(c=\frac{2}{5}\)

Wait, maybe I misread the problem. Looking back, the original problem seems to be about simplifying the left - hand side after canceling. Wait, the first step in the given work has \(\frac{4(6c + 4)}{8}=-2c + 4\), then they simplify \(\frac{4}{8}=\frac{1}{2}\), so \(\frac{6c + 4}{2}=-2c + 4\). Now, we can also multiply both sides by 2 to eliminate the denominator.

Step1: Multiply both sides by 2

Given \(\frac{6c + 4}{2}=-2c + 4\), multiply both sides by 2: \(2\times\frac{6c + 4}{2}=2\times(-2c + 4)\). The left - hand side simplifies to \(6c + 4\), and the right - hand side is \(-4c+8\). So we have the equation \(6c + 4=-4c + 8\).

Step2: Add \(4c\) to both sides

\(6c+4 + 4c=-4c + 8+4c\), which gives \(10c+4 = 8\).

Step3: Subtract 4 from both sides

\(10c+4-4=8 - 4\), so \(10c = 4\).

Step4: Divide by 10

\(c=\frac{4}{10}=\frac{2}{5}\)

But maybe the problem is to simplify \(\frac{6c + 4}{2}\) to get the left - hand side equal to the right - hand side? Wait, no, the box is for the left - hand side after simplifying \(\frac{6c + 4}{2}\). Wait, \(\frac{6c+4}{2}=\frac{6c}{2}+\frac{4}{2}=3c + 2\). Wait, but the right - hand side is \(-2c + 4\). Maybe there is a miscalculation in the original problem's right - hand side? Or maybe I misread. Wait, let's start over.

Wait, the first line: \(\frac{4(6c + 4)}{8}=-2c + 4\). Simplify \(\frac{4}{8}=\frac{1}{2}\), so \(\frac{6c + 4}{2}=-2c + 4\). Now, if we want to write the left - hand side as a simplified expression, \(\frac{6c+4}{2}=3c + 2\). But if we are solving for \(c\):

1. Start with \(\frac{6c + 4}{2}=-2c + 4\)
2. Multiply both sides by 2: \(6c+4=-4c + 8\)
3. Add \(4c\) to both sides: \(6c + 4c+4=-4c+4c + 8\) → \(10c+4 = 8\)
4. Subtract 4: \(10c=4\)
5. Divide by 10: \(c=\frac{4}{10}=\frac{2}{5}\)

But the box is in the third line, where the left - hand side is \(\frac{6c + 4}{2}\) simplified. Wait, \(\frac{6c+4}{2}=3c + 2\). But if we consider the equation \(\frac{6c + 4}{2}=-2c + 4\), and we want to find what the left - hand side is equal to when we solve for the expression. Wait, maybe the problem is to simplify \(\frac{6c + 4}{2}\) correctly. \(\frac{6c+4}{2}=3c + 2\). But if we follow the steps in the given work, after canceling 4 and 8 to get \(\frac{6c + 4}{2}\), then we can also expand the right - hand side? No, maybe the original problem has a typo, but assuming we need to simplify \(\frac{6c + 4}{2}\), we get \(3c + 2\). But if we are solving for \(c\) from \(\frac{6c + 4}{2}=-2c + 4\):

Multiply both sides by 2: \(6c + 4=-4c+8\)

Add \(4c\) to both sides: \(10c+4 = 8\)

Subtract 4: \(10c = 4\)

Divide by 10: \(c=\frac{2}{5}\)

But the box is for the left - hand side expression. Wait, \(\frac{6c + 4}{2}\) simplifies to \(3c + 2\). But maybe the problem is to find the value of the expression in the box such that \(\frac{6c + 4}{2}=\) box \(=-2c + 4\). Wait, no, let's check the arithmetic again.

Wait, if we have \(\frac{4(6c + 4)}{8}=-2c + 4\), simplify \(\frac{4}{8}=\frac{1}{2}\), so \(\frac{6c + 4}{2}=-2c + 4\). Now, multiply both sides by 2: \(6c + 4=-4c + 8\). Then \(6c+4c=8 - 4\), \(10c = 4\), \(c=\frac{2}{5}\). But the box is in the third line, where the left - hand side is \(\frac{6c + 4}{2}\) simplified. Wait, \(\frac{6c+4}{2}=3c + 2\). But if we substitute \(c = \frac{2}{5}\) into \(3c + 2\), we get \(3\times\frac{2}{5}+2=\frac{6}{5}+2=\frac{6 + 10}{5}=\frac{16}{5}\), and substituting \(c=\frac{2}{5}\) into \(-2c + 4\) gives \(-2\times\frac{2}{5}+4=-\frac{4}{5}+4=\frac{-4 + 20}{5}=\frac{16}{5}\). So the left - hand side simplified is \(3c + 2\), but if we are to write the expression in the box as the simplified form of \(\frac{6c + 4}{2}\), it is \(3c + 2\). But maybe the original problem has a mistake in the right - hand side. Alternatively, maybe the problem is to solve for the expression in the box such that \(\frac{6c + 4}{2}=\) box and box \(=-2c + 4\). So we can set \(3c + 2=-2c + 4\) and solve for \(c\), but the box is for the expression, not \(c\). Wait, I think I misread the problem. The box is for the left - hand side after simplifying \(\frac{6c + 4}{2}\). So \(\frac{6c+4}{2}=3c + 2\).

Answer:

If we are simplifying \(\frac{6c + 4}{2}\), the answer is \(3c + 2\). If we are solving for \(c\) from \(\frac{6c + 4}{2}=-2c + 4\), the value of \(c\) is \(\frac{2}{5}\). But based on the box's position, it seems to be the simplified form of \(\frac{6c + 4}{2}\), so the answer is \(3c + 2\).