GCSE Edexcel Maths Inequalities: (a) Express \( \frac{x}{x + 2} +

(a) Express $ \frac{x}{x + 2} + \frac{2x}{x - 4} $ as a single fraction in its simplest form - Edexcel - GCSE Maths - Question 13 - 2020 - Paper 3

Question 13

$(a)-Express-$-\frac{x}{x-+-2}-+-\frac{2x}{x---4}-$-as-a-single-fraction-in-its-simplest-form-Edexcel-GCSE Maths-Question 13-2020-Paper 3.png$

(a) Express $ \frac{x}{x + 2} + \frac{2x}{x - 4} $ as a single fraction in its simplest form. (b) Expand and simplify $ (x - 3)(2x + 3)(4x + 5) $

Worked Solution & Example Answer:(a) Express $ \frac{x}{x + 2} + \frac{2x}{x - 4} $ as a single fraction in its simplest form - Edexcel - GCSE Maths - Question 13 - 2020 - Paper 3

Step 1

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Answer

To add the two fractions, we first need to find a common denominator.

The common denominator is ((x + 2)(x - 4)).

Rewriting each fraction with the common denominator:

$\frac{x(x - 4)}{(x + 2)(x - 4)} + \frac{2x(x + 2)}{(x + 2)(x - 4)}$

Now, this gives:

$\frac{x(x - 4) + 2x(x + 2)}{(x + 2)(x - 4)}$

Now we can simplify the numerator:

$x(x - 4) + 2x(x + 2) = x^2 - 4x + 2x^2 + 4x = 3x^2$

Thus, the single fraction becomes:

$\frac{3x^2}{(x + 2)(x - 4)}$

Step 2

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Answer

To expand ( (x - 3)(2x + 3)(4x + 5) ), we can start by expanding two of the factors first:

Expanding ( (x - 3)(2x + 3) ):

$= x(2x) + x(3) - 3(2x) - 3(3) = 2x^2 + 3x - 6x - 9 = 2x^2 - 3x - 9$
Now, we multiply this result by ( (4x + 5) ):

$= (2x^2 - 3x - 9)(4x + 5)$

Expanding this:

$= 2x^2(4x) + 2x^2(5) - 3x(4x) - 3x(5) - 9(4x) - 9(5)$

$= 8x^3 + 10x^2 - 12x^2 - 15x - 36x - 45$

Combine like terms:

$= 8x^3 + (-12x^2 + 10x^2) + (-15x - 36x) - 45$

$= 8x^3 - 2x^2 - 51x - 45$