Question3. $x+5=\frac{14}{x}$
5. $x+\frac{4 x}{x-3}=\frac{12}{x-3}$

Studdy Solution

STEP 1

1. The equation is a rational equation involving a variable in the denominator.
2. We need to eliminate the fraction to solve for $x$ .

STEP 2

1. Eliminate the fraction by multiplying through by $x$ .
2. Rearrange the equation to form a quadratic equation.
3. Solve the quadratic equation.
4. 1. Combine the terms over a common denominator.
2. Simplify and solve for $x$ .
STEP_1: High_Level_Step: 1 Combine the terms over a common denominator, $x - 3$ :
$x + \frac{4x}{x-3} = \frac{12}{x-3}$
Multiply the entire equation by $x-3$ to eliminate the denominators:
$(x)(x-3) + 4x = 12$
High_Level_Step_Completed: TRUE
STEP_2: High_Level_Step: 2 Expand and simplify the equation:
$x^2 - 3x + 4x = 12$ $x^2 + x = 12$
Rearrange to form a quadratic equation:
$x^2 + x - 12 = 0$
High_Level_Step_Completed: TRUE
STEP_3: High_Level_Step: 2 Solve the quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 1$ , $b = 1$ , and $c = -12$ :
$x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1}$ $x = \frac{-1 \pm \sqrt{1 + 48}}{2}$ $x = \frac{-1 \pm \sqrt{49}}{2}$ $x = \frac{-1 \pm 7}{2}$
This gives two possible solutions:
$x = \frac{-1 + 7}{2} = 3$ $x = \frac{-1 - 7}{2} = -4$
High_Level_Step_Completed: TRUE
The solutions for $x$ are:
$x = 3 \quad \text{or} \quad x = -4$

STEP 3

Multiply both sides of the equation by $x$ to eliminate the fraction:
$x(x + 5) = x \cdot \frac{14}{x}$ $x^2 + 5x = 14$

STEP 4

Rearrange the equation to form a quadratic equation:
$x^2 + 5x - 14 = 0$

STEP 5

Solve the quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 1$ , $b = 5$ , and $c = -14$ :
$x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot (-14)}}{2 \cdot 1}$ $x = \frac{-5 \pm \sqrt{25 + 56}}{2}$ $x = \frac{-5 \pm \sqrt{81}}{2}$ $x = \frac{-5 \pm 9}{2}$
This gives two possible solutions:
$x = \frac{-5 + 9}{2} = 2$ $x = \frac{-5 - 9}{2} = -7$
The solutions for $x$ are:
$x = 2 \quad \text{or} \quad x = -7$
### Problem 5: $x + \frac{4x}{x-3} = \frac{12}{x-3}$
_ASSUMPTIONS_:
1. The equation involves rational expressions with a common denominator.
2. We need to combine terms and solve for $x$ .

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