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14 (a) Simplify \( \frac{x^2 - 16}{2x^2 - 5x - 12} \) (b) Make \( v \) the subject of the formula \( w = \frac{15(u - 2v)}{v} \) - Edexcel - GCSE Maths - Question 14 - 2017 - Paper 3
Question 14
14 (a) Simplify \( \frac{x^2 - 16}{2x^2 - 5x - 12} \) (b) Make \( v \) the subject of the formula \( w = \frac{15(u - 2v)}{v} \)
Worked Solution & Example Answer:14 (a) Simplify \( \frac{x^2 - 16}{2x^2 - 5x - 12} \) (b) Make \( v \) the subject of the formula \( w = \frac{15(u - 2v)}{v} \) - Edexcel - GCSE Maths - Question 14 - 2017 - Paper 3
Step 1
Simplify \( \frac{x^2 - 16}{2x^2 - 5x - 12} \)
Answer
To simplify ( \frac{x^2 - 16}{2x^2 - 5x - 12} ), we first factor both the numerator and the denominator.
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Factor the Numerator: ( x^2 - 16 ) is a difference of squares: [ x^2 - 16 = (x - 4)(x + 4) ]
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Factor the Denominator:
We need to factor ( 2x^2 - 5x - 12 ). We can do this by finding two numbers that multiply to ( 2 \times -12 = -24 ) and add up to ( -5 ), which are ( -8 ) and ( 3 ). Thus: [ 2x^2 - 8x + 3x - 12 = 2x(x - 4) + 3(x - 4) = (2x + 3)(x - 4) ] -
Combine and Cancel:
The expression simplifies to: [ \frac{(x - 4)(x + 4)}{(2x + 3)(x - 4)} = \frac{x + 4}{2x + 3} \text{ (for ( x \neq 4 ))} ]
Step 2
Make \( v \) the subject of the formula \( w = \frac{15(u - 2v)}{v} \)
Answer
To make ( v ) the subject of the formula, we can follow these steps:
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Start with the given equation: [ w = \frac{15(u - 2v)}{v} ]
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Multiply both sides by ( v ) to eliminate the fraction: [ wv = 15(u - 2v) ]
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Expand the right-hand side: [ wv = 15u - 30v ]
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Rearrange to isolate terms involving ( v ): [ wv + 30v = 15u ]
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Factor out ( v ): [ v(w + 30) = 15u ]
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Finally, solve for ( v ): [ v = \frac{15u}{w + 30} ]