Question

(2) $(x + y)\div a$
$x + y\div a$
$\dfrac{x + y}{a}$
(4) $6a^{3}\times(-2a)\div3a^{2}$
(2) $12a^{3}\div3a^{2}\div(-2a)$

Explanation:

For \((x + y)\div a\)

Step1: Recall division as fraction

Division of a sum by a number can be written as a fraction, where the dividend is the numerator and the divisor is the denominator. So \((x + y)\div a=\frac{x + y}{a}\)

Step1: Multiply the coefficients and variables

First, multiply the coefficients \(6\times(-2)=- 12\) and multiply the variables using the rule \(a^{m}\times a^{n}=a^{m + n}\), so \(a^{3}\times a=a^{3+1}=a^{4}\). So \(6a^{3}\times(-2a)=-12a^{4}\)

Step2: Divide by \(3a^{2}\)

Now divide \(-12a^{4}\) by \(3a^{2}\). Divide the coefficients \(\frac{-12}{3}=-4\) and divide the variables using the rule \(\frac{a^{m}}{a^{n}}=a^{m - n}\), so \(\frac{a^{4}}{a^{2}}=a^{4-2}=a^{2}\). So \(-12a^{4}\div3a^{2}=-4a^{2}\)

Step1: Divide \(12a^{3}\) by \(3a^{2}\)

Divide the coefficients \(\frac{12}{3} = 4\) and divide the variables \(\frac{a^{3}}{a^{2}}=a^{3 - 2}=a\). So \(12a^{3}\div3a^{2}=4a\)

Step2: Divide \(4a\) by \(-2a\)

Divide the coefficients \(\frac{4}{-2}=-2\) and divide the variables \(\frac{a}{a}=1\) (since \(a
eq0\)). So \(4a\div(-2a)=-2\)

Answer:

\(\frac{x + y}{a}\)

For \(6a^{3}\times(- 2a)\div3a^{2}\)