Name: Writing Equivalent Expressions
Uploaded: 2019-07-19
Duration: 2 min 40 s
Description: In this video, Let us learn to write equivalent expressions by using the additive identity property of zero and the multiplicative identity property of one.

Question

If

p = x^{m + n} . y^{l}, q = x^{n + l} . y^{m}

and

r = x^{l + m} . y^{n}

, prove that

p^{m - n} . q^{n - l} . r^{l - m} = 1

Accepted Answer

Given, $p = x^{m + n} . y^{l}, q = x^{n + l} . y^{m}$ and $r = x^{l + m} . y^{n}$

We have to prove that $p^{m - n} . q^{n - l} . r^{l - m} = 1$

$p^{m - n} . q^{n - l} . r^{l - m} - - - - (1)$

Substitute the values of $p, q and r$ in equation $(1)$ , we get

${[x^{m + n} . y^{l}]}^{m - n} \times {[x^{n + l} . y^{m}]}^{n - l} \times {[x^{l + m} . y^{n}]}^{l - m}$

$\Rightarrow [x^{(m + n) (m - n)} . y^{l (m - n)}] \times [x^{(n + l) (n - l)} . y^{m (n - l)}] \times [x^{(l + m) (l - m)} . y^{n (l - m)}]$

$\Rightarrow [x^{(m^{2} - n^{2})} . y^{m l - n l}] \times [x^{(n^{2} - l^{2})} . y^{m n - m l}] \times [x^{(l^{2} - m^{2})} . y^{n l - m n}]$

$\Rightarrow [x^{(m^{2} - n^{2}) + (n^{2} - l^{2}) + (l^{2} - m^{2})} . y^{(m l - n l) + (m n - m l) + (n l - m n)}]$

$\Rightarrow [x^{m^{2} - n^{2} + n^{2} - l^{2} + l^{2} - m^{2}} . y^{m l - n l + m n - m l + n l - m n}]$

$\Rightarrow [x^{0} . y^{0}]$

$\Rightarrow 1 \times 1 = 1$

So, $p^{m - n} . q^{n - l} . r^{l - m} = 1$

Hence, proved.

If $p = x^{m + n} . y^{l}, q = x^{n + l} . y^{m}$ and $r = x^{l + m} . y^{n}$ , prove that $p^{m - n} . q^{n - l} . r^{l - m} = 1$

Important Questions on Number System

Learn and Prepare for any exam you want !

If p=xm+n.yl, q=xn+l.ym and r=xl+m.yn, prove that pm-n.qn-l.rl-m=1

Important Questions on Number System

Learn and Prepare for any exam you want !

If $p = x^{m + n} . y^{l}, q = x^{n + l} . y^{m}$ and $r = x^{l + m} . y^{n}$ , prove that $p^{m - n} . q^{n - l} . r^{l - m} = 1$