GCSE Edexcel Maths Estimating Gradients & Areas under Graphs: Show that 6 + \[ \frac{(x + 5) +

Show that 6 + \[ \frac{(x + 5) + \frac{x^2 + 3x - 10}{x - 1}}{x - 1} \] simplifies to $ \frac{ax - h}{cx - d} $ where a, b, c and d are integers. - Edexcel - GCSE Maths - Question 14 - 2019 - Paper 2

Question 14

$Show-that-6-+-\[-\frac{(x-+-5)-+-\frac{x^2-+-3x---10}{x---1}}{x---1}-\]-simplifies-to-$-\frac{ax---h}{cx---d}-$-where-a,-b,-c-and-d-are-integers.-Edexcel-GCSE Maths-Question 14-2019-Paper 2.png$

Show that 6 + \[ \frac{(x + 5) + \frac{x^2 + 3x - 10}{x - 1}}{x - 1} \] simplifies to $ \frac{ax - h}{cx - d} $ where a, b, c and d are integers.

Worked Solution & Example Answer:Show that 6 + \[ \frac{(x + 5) + \frac{x^2 + 3x - 10}{x - 1}}{x - 1} \] simplifies to $ \frac{ax - h}{cx - d} $ where a, b, c and d are integers. - Edexcel - GCSE Maths - Question 14 - 2019 - Paper 2

Step 1

Step 1: Simplify the Expression Inside the Bracket

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Answer

Start with the expression inside the brackets:

[ (x + 5) + \frac{x^2 + 3x - 10}{x - 1} ]

To combine these terms, we need a common denominator. The common denominator is ( x - 1 ), so we rewrite the first term:

[ (x + 5) = \frac{(x + 5)(x - 1)}{x - 1} = \frac{x^2 + 4x - 5}{x - 1} ]

Now substitute this back:

[ \frac{x^2 + 4x - 5}{x - 1} + \frac{x^2 + 3x - 10}{x - 1} = \frac{(x^2 + 4x - 5) + (x^2 + 3x - 10)}{x - 1} ]

Step 2

Step 2: Combine the Numerators

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Answer

Now we simplify the numerator:

[ (x^2 + 4x - 5) + (x^2 + 3x - 10) = 2x^2 + 7x - 15 ]

Thus, we have:

[ \frac{2x^2 + 7x - 15}{x - 1} ]

Step 3

Step 3: Add 6 to the Expression

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Answer

Next, we add 6 to the entire expression:

[ 6 + \frac{2x^2 + 7x - 15}{x - 1} ]

Again, we need a common denominator, which will be ( x - 1 ):

[ 6 = \frac{6(x - 1)}{x - 1} = \frac{6x - 6}{x - 1} ]

Therefore, we can write:

[ \frac{6x - 6 + 2x^2 + 7x - 15}{x - 1} = \frac{2x^2 + 13x - 21}{x - 1} ]

Step 4

Step 4: Factor the Numerator

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Answer

The next step is to factor the numerator, ( 2x^2 + 13x - 21 ). This factors to:

[ 2x^2 + 14x - x - 21 = (2x - 3)(x + 7) ]

Thus, the expression becomes:

[ \frac{(2x - 3)(x + 7)}{x - 1} ]

Step 5

Step 5: Express in the Required Form

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Answer

Rearranging this gives us:

[ \frac{ax - h}{cx - d} = \frac{2x - 3}{x - 1}, ]

Where:

a = 2,
h = 3,
c = 1,
d = 1.

The integers a, b, c, and d are confirmed.

Show that 6 + \[ \frac{(x + 5) + \frac{x^2 + 3x - 10}{x - 1}}{x - 1} \] simplifies to \( \frac{ax - h}{cx - d} \) where a, b, c and d are integers. - Edexcel - GCSE Maths - Question 14 - 2019 - Paper 2

Worked Solution & Example Answer:Show that 6 + \[ \frac{(x + 5) + \frac{x^2 + 3x - 10}{x - 1}}{x - 1} \] simplifies to \( \frac{ax - h}{cx - d} \) where a, b, c and d are integers. - Edexcel - GCSE Maths - Question 14 - 2019 - Paper 2

Step 1: Simplify the Expression Inside the Bracket

Step 2: Combine the Numerators

Step 3: Add 6 to the Expression

Step 4: Factor the Numerator

Step 5: Express in the Required Form

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