Question

\frac{2}{4} - \frac{6}{-2x + 8}
2(10 - 3x) = 6x + 5 - 12x
\frac{1}{2}(18x + 100) = 20 - x

Explanation:

Let's solve each equation one by one.

First Equation: \(2(10 - 3x)=6x + 5 - 12x\)

Step 1: Expand the left side and simplify the right side

Expand \(2(10 - 3x)\) to get \(20-6x\).
Simplify the right side: \(6x + 5 - 12x=5 - 6x\).
So the equation becomes \(20-6x = 5 - 6x\).

Step 2: Add \(6x\) to both sides

\(20-6x+6x = 5 - 6x+6x\)
\(20=5\)
This is a contradiction, so the first equation has no solution.

Second Equation: \(\frac{1}{2}(18x + 100)=20 - x\)

Step 1: Multiply both sides by 2 to eliminate the fraction

\(2\times\frac{1}{2}(18x + 100)=2\times(20 - x)\)
\(18x + 100 = 40 - 2x\)

Step 2: Add \(2x\) to both sides

\(18x+2x + 100 = 40 - 2x+2x\)
\(20x + 100 = 40\)

Step 3: Subtract 100 from both sides

\(20x + 100 - 100 = 40 - 100\)
\(20x=-60\)

Step 4: Divide both sides by 20

\(x=\frac{-60}{20}=-3\)

For the first fractional expression \(\frac{2}{4}-\frac{6}{-2x + 8}\) (assuming we need to simplify or solve for \(x\), but it's not an equation. If we assume it's an equation set to zero, let's do that:

Let \(\frac{2}{4}-\frac{6}{-2x + 8}=0\)

Step 1: Simplify \(\frac{2}{4}=\frac{1}{2}\)

The equation becomes \(\frac{1}{2}-\frac{6}{-2x + 8}=0\)

Step 2: Add \(\frac{6}{-2x + 8}\) to both sides

\(\frac{1}{2}=\frac{6}{-2x + 8}\)

Step 3: Cross - multiply

\(1\times(-2x + 8)=2\times6\)
\(-2x + 8 = 12\)

Step 4: Subtract 8 from both sides

\(-2x+8 - 8=12 - 8\)
\(-2x = 4\)

Step 5: Divide by - 2

\(x=-2\)

Final Answers:

• For \(2(10 - 3x)=6x + 5 - 12x\): No solution.
• For \(\frac{1}{2}(18x + 100)=20 - x\): \(x = - 3\)
• For \(\frac{2}{4}-\frac{6}{-2x + 8}=0\) (assumed equation): \(x=-2\)

Answer:

Let's solve each equation one by one.

First Equation: \(2(10 - 3x)=6x + 5 - 12x\)

Step 1: Expand the left side and simplify the right side

Expand \(2(10 - 3x)\) to get \(20-6x\).
Simplify the right side: \(6x + 5 - 12x=5 - 6x\).
So the equation becomes \(20-6x = 5 - 6x\).

Step 2: Add \(6x\) to both sides

\(20-6x+6x = 5 - 6x+6x\)
\(20=5\)
This is a contradiction, so the first equation has no solution.

Second Equation: \(\frac{1}{2}(18x + 100)=20 - x\)

Step 1: Multiply both sides by 2 to eliminate the fraction

\(2\times\frac{1}{2}(18x + 100)=2\times(20 - x)\)
\(18x + 100 = 40 - 2x\)

Step 2: Add \(2x\) to both sides

\(18x+2x + 100 = 40 - 2x+2x\)
\(20x + 100 = 40\)

Step 3: Subtract 100 from both sides

\(20x + 100 - 100 = 40 - 100\)
\(20x=-60\)

Step 4: Divide both sides by 20

\(x=\frac{-60}{20}=-3\)

For the first fractional expression \(\frac{2}{4}-\frac{6}{-2x + 8}\) (assuming we need to simplify or solve for \(x\), but it's not an equation. If we assume it's an equation set to zero, let's do that:

Let \(\frac{2}{4}-\frac{6}{-2x + 8}=0\)

Step 1: Simplify \(\frac{2}{4}=\frac{1}{2}\)

The equation becomes \(\frac{1}{2}-\frac{6}{-2x + 8}=0\)

Step 2: Add \(\frac{6}{-2x + 8}\) to both sides

\(\frac{1}{2}=\frac{6}{-2x + 8}\)

Step 3: Cross - multiply

\(1\times(-2x + 8)=2\times6\)
\(-2x + 8 = 12\)

Step 4: Subtract 8 from both sides

\(-2x+8 - 8=12 - 8\)
\(-2x = 4\)

Step 5: Divide by - 2

\(x=-2\)

Final Answers:

• For \(2(10 - 3x)=6x + 5 - 12x\): No solution.
• For \(\frac{1}{2}(18x + 100)=20 - x\): \(x = - 3\)
• For \(\frac{2}{4}-\frac{6}{-2x + 8}=0\) (assumed equation): \(x=-2\)