Question

1. simplify and state restrictions.

a) \\(\frac{5x^3y^2}{27x} \times \frac{9xy}{25x^2y^2}\\) \\((2\\ marks)\\)
b) \\(\frac{7x^2 + 14x}{5x + 10}\\) \\((2\\ marks)\\)
c) \\(\frac{x^2 - 4x - 5}{x^2 - 3x - 10} \times \frac{9x + 3x^2}{x^2 - 9}\\) \\((3\\ marks)\\)
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Explanation:

Part (a)

Step 1: Multiply numerators and denominators

Multiply the numerators together and the denominators together:
$$\frac{5x^{3}y^{2}\times9xy}{27x\times25x^{2}y^{2}}=\frac{45x^{4}y^{3}}{675x^{3}y^{2}}$$

Step 2: Simplify the fraction

Simplify the coefficients and use the rule of exponents \( \frac{a^{m}}{a^{n}} = a^{m - n}\) for like bases:
For \(x\): \(x^{4-3}=x\)
For \(y\): \(y^{3 - 2}=y\)
For coefficients: \(\frac{45}{675}=\frac{1}{15}\)
So the simplified form is \(\frac{xy}{15}\)

Step 3: State restrictions

The original denominators are \(27x\), \(25x^{2}y^{2}\). So \(x
eq0\) and \(y
eq0\) (since if \(x = 0\) or \(y=0\), the denominators will be zero which is undefined)

Step 1: Factor numerator and denominator

Factor the numerator \(7x^{2}+14x = 7x(x + 2)\)
Factor the denominator \(5x + 10=5(x + 2)\)
So the fraction becomes \(\frac{7x(x + 2)}{5(x + 2)}\)

Step 2: Simplify the fraction

Cancel out the common factor \((x + 2)\) (note that \(x
eq - 2\) because if \(x=-2\), the original denominator \(5(x + 2)=0\))
The simplified form is \(\frac{7x}{5}\)

Step 3: State restrictions

From the original denominator \(5x + 10=5(x + 2)\), so \(x
eq - 2\)

Step 1: Factor all expressions

Factor numerator of first fraction: \(x^{2}-4x - 5=(x - 5)(x+ 1)\)
Factor denominator of first fraction: \(x^{2}-3x - 10=(x - 5)(x + 2)\)
Factor numerator of second fraction: \(9x+3x^{2}=3x(x + 3)\)
Factor denominator of second fraction: \(x^{2}-9=(x - 3)(x + 3)\)
So the expression becomes \(\frac{(x - 5)(x + 1)}{(x - 5)(x + 2)}\times\frac{3x(x + 3)}{(x - 3)(x + 3)}\)

Step 2: Cancel common factors

Cancel out \((x - 5)\) ( \(x
eq5\) ), \((x + 3)\) ( \(x
eq - 3\) )
We get \(\frac{(x + 1)}{(x + 2)}\times\frac{3x}{(x - 3)}=\frac{3x(x + 1)}{(x + 2)(x - 3)}\)

Step 3: State restrictions

From the original denominators:

• \(x^{2}-3x - 10=(x - 5)(x + 2)\) so \(x

eq5,x
eq - 2\)

• \(x^{2}-9=(x - 3)(x + 3)\) so \(x

eq3,x
eq - 3\)

Answer:

Simplified form: \(\boldsymbol{\frac{xy}{15}}\), Restrictions: \(x
eq0,y
eq0\)

Part (b)