Solve for X: Rational Equation 7/(6+3x-5(x+2)) = 1/(4(2-x))

Solve for X:

$\frac{7}{6+3x-5(x+2)}=\frac{1}{4(2-x)}$

To solve the given equation:

$\frac{7}{6 + 3x - 5(x + 2)} = \frac{1}{4(2-x)}$

we will follow these steps:

• Simplify the expression inside the denominators.
• Cross-multiply to eliminate the fractions.
• Solve the resulting linear equation for $x$.
• Check the solution in the original equation to ensure there are no extraneous solutions.

Let's go through each step:

Step 1: Simplify the denominators
The first step is to simplify the expression in the denominator on the left-hand side: $6 + 3x - 5(x + 2)$.

Distribute the $-5$ in the expression:

$6 + 3x - 5(x + 2) \implies 6 + 3x - 5x - 10$

Combine like terms:

$6 - 10 + 3x - 5x \implies -4 - 2x$

So, the equation becomes:

$\frac{7}{-4 - 2x} = \frac{1}{4(2-x)}$

Now, simplify $4(2-x)$:

$4(2-x) = 8 - 4x$

So the equation is:

$\frac{7}{-4 - 2x} = \frac{1}{8 - 4x}$

Step 2: Cross-multiply to eliminate fractions
Cross-multiply to get rid of the fractions:

$7 \times (8 - 4x) = 1 \times (-4 - 2x)$

Distribute on both sides:

$56 - 28x = -4 - 2x$

Step 3: Solve the linear equation for $x$
Rearrange the equation to bring like terms together:

$56 + 4 = 28x - 2x$

Simplify:

$60 = 26x$

Divide both sides by 26 to solve for $x$:

$x = \frac{60}{26} = \frac{30}{13}$

Step 4: Verify the solution
We need to ensure that our solution satisfies the original equation and doesn't create a situation where the denominator is zero:

We found $x = \frac{30}{13}$, so check that:

$6 + 3x - 5(x + 2) \neq 0$

Substitute $x = \frac{30}{13}$ back into the simplified denominator:

$-4 - 2 \left(\frac{30}{13}\right) \neq 0$

Calculate:

$-4 - \frac{60}{13} = \frac{-52 - 60}{13} = \frac{-112}{13} \neq 0$

Thus, the solution is valid.

Therefore, the solution to the problem is $x = \frac{30}{13}$.