Using factorisation method, solve the following quadratic equations.`(2x)/(x-4)+(2x-5)/(x-3)=25/3; x!=4,3`

Options

A. `x=6orx=(45)/(13)`
B. `x=0orx=(40)/(13)`
C. `x=4orx=(40)/(13)`
D. `x=6orx=(40)/(13)`

Explanation

To solve the quadratic equation \(\frac{2x}{x-4} + \frac{2x-5}{x-3} = \frac{25}{3}\) using the factorization method, we will follow these steps:

### Step 1: Find a common denominator and rewrite the equation

The common denominator for the left side of the equation is \((x - 4)(x - 3)\). We rewrite the equation:

\[ \frac{2x(x - 3) + (2x - 5)(x - 4)}{(x - 4)(x - 3)} = \frac{25}{3} \]

### Step 2: Expand the numerator

Now, we expand the numerator:

\[ 2x(x - 3) + (2x - 5)(x - 4) = 2x^2 - 6x + (2x^2 - 8x - 5x + 20) \]

Combining like terms:

\[ 2x^2 - 6x + 2x^2 - 8x - 5x + 20 = 4x^2 - 19x + 20 \]

### Step 3: Set up the equation

Now we have:

\[ \frac{4x^2 - 19x + 20}{(x - 4)(x - 3)} = \frac{25}{3} \]

Cross-multiplying gives us:

\[ 3(4x^2 - 19x + 20) = 25(x^2 - 7x + 12) \]

### Step 4: Expand both sides

Expanding both sides:

\[ 12x^2 - 57x + 60 = 25x^2 - 175x + 300 \]

### Step 5: Move all terms to one side

Rearranging gives us:

\[ 12x^2 - 57x + 60 - 25x^2 + 175x - 300 = 0 \]

Combining like terms:

\[ -13x^2 + 118x - 240 = 0 \]

Multiplying through by -1 to simplify:

\[ 13x^2 - 118x + 240 = 0 \]

### Step 6: Factor the quadratic equation

Now we will factor the quadratic:

We look for two numbers that multiply to \(13 \times 240 = 3120\) and add to \(-118\). The numbers are \(-78\) and \(-40\).

So we can split the middle term:

\[ 13x^2 - 78x - 40x + 240 = 0 \]

Grouping gives us:

\[ 13x(x - 6) - 40(x - 6) = 0 \]

Factoring out \((x - 6)\):

\[ (13x - 40)(x - 6) = 0 \]

### Step 7: Solve for x

Setting each factor to zero gives:

1. \(13x - 40 = 0 \Rightarrow x = \frac{40}{13}\)
2. \(x - 6 = 0 \Rightarrow x = 6\)

### Final Solutions

The solutions to the equation are:

\[ x = \frac{40}{13} \quad \text{and} \quad x = 6 \]