Leaving Cert Mathematics Algebra: (x + 5)(3x - 4)

(x + 5)(3x - 4) - 3(x^2 + 2) + 4 = 0 (b) Find the solutions of \[ \frac{5}{x+3} - \frac{1}{x} = \frac{1}{2} \] where $x \neq -3, 0, x \in \mathbb{R}$. - Leaving Cert Mathematics - Question 6 - 2018

Question 6

$(x-+-5)(3x---4)---3(x^2-+-2)-+-4-=-0--(b)-Find-the-solutions-of--\[-\frac{5}{x+3}---\frac{1}{x}-=-\frac{1}{2}-\]-where-$x-\neq--3,-0,-x-\in-\mathbb{R}$.-Leaving Cert Mathematics-Question 6-2018.png$

(x + 5)(3x - 4) - 3(x^2 + 2) + 4 = 0 (b) Find the solutions of \[ \frac{5}{x+3} - \frac{1}{x} = \frac{1}{2} \] where $x \neq -3, 0, x \in \mathbb{R}$.

Worked Solution & Example Answer:(x + 5)(3x - 4) - 3(x^2 + 2) + 4 = 0 (b) Find the solutions of \[ \frac{5}{x+3} - \frac{1}{x} = \frac{1}{2} \] where $x \neq -3, 0, x \in \mathbb{R}$. - Leaving Cert Mathematics - Question 6 - 2018

Step 1

Solve for x.

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Answer

To solve the equation

$(x + 5)(3x - 4) - 3(x^2 + 2) + 4 = 0,$

we first expand the left-hand side:

Expanding: $(x + 5)(3x) + (x + 5)(-4) - 3x^2 - 6 + 4 = 0$ $3x^2 + 15x - 4x - 20 - 3x^2 - 6 + 4 = 0$ $11x - 22 = 0$
Simplifying results in: $11x = 22$
Dividing both sides by 11 gives us: $x = 2.$

Thus, the solution to part (a) is:

$x = 2.$

Step 2

Find the solutions of $ \frac{5}{x+3} - \frac{1}{x} = \frac{1}{2} $.

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104 rated

Answer

To solve the equation

$\frac{5}{x+3} - \frac{1}{x} = \frac{1}{2},$

we start by finding a common denominator.

The common denominator is $2x(x + 3)$ .
Rewriting each term gives:

$10x - 2(x + 3) = x(x + 3)$
Expanding leads to:

$10x - 2x - 6 = x^2 + 3x$
Rearranging yields:

$x^2 - 5x + 6 = 0.$
This factors to:

$(x - 2)(x - 3) = 0$
Thus, we find:

$x = 2$ and $x = 3.$

The solutions for part (b) are:

$x = 2, x = 3.$

(x + 5)(3x - 4) - 3(x^2 + 2) + 4 = 0 (b) Find the solutions of \[ \frac{5}{x+3} - \frac{1}{x} = \frac{1}{2} \] where $x \neq -3, 0, x \in \mathbb{R}$. - Leaving Cert Mathematics - Question 6 - 2018

Worked Solution & Example Answer:(x + 5)(3x - 4) - 3(x^2 + 2) + 4 = 0 (b) Find the solutions of \[ \frac{5}{x+3} - \frac{1}{x} = \frac{1}{2} \] where $x \neq -3, 0, x \in \mathbb{R}$. - Leaving Cert Mathematics - Question 6 - 2018

Solve for x.

Find the solutions of \( \frac{5}{x+3} - \frac{1}{x} = \frac{1}{2} \).

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