Question

Prove that : | x × y | = |x|×|y| ,if x= -4/5 and y= 3/-7

solve with explanation​

Answer 1

$\huge\underline\bold\red{Question}$

Prove that: |x × y| = |x|×|y|, if

x=$\frac{-4}{5} and\: \frac{3}{-7}$

$\huge\underline\bold\purple{Solution}$

We have to prove | x × y | = |x|×|y|

${\boxed{\green{ x=\frac{-4}{5} and\: y=\frac{-3}{7}}}}$

$LHS = |x \times y|$

${\boxed{\green{x=\frac{-4}{5}and\: y=\frac{-3}{7}}}}$

$= | (\frac{ - 4}{5}) \times (\frac{ - 3}{7} )|$

$= | \frac{12}{35} |$

$= \dfrac{12}{35}$

$RHS = |x| \times |y|$

${\boxed{\green{put \:the \:value \:of\: \:x \:and\:y}}}}$

$= \frac{ - 4}{5} ×\frac{ - 3}{7}$

=$\frac{4}{5} × \frac{3}{7}$

$= \frac{12}{35}$

➛${LHS=RHS$

➤${Hence \:proved}$

Answer 2

Question :

Prove that | x × y | = |x|×|y|

if x=$\frac{-4}{5}$ and y=$\frac{-3}{7}$

Modulus Function :

The function f(x) defined by f(x)= |x|

$\bf{f(x)}\begin{cases}\sf{x,when\:x \geqslant\:0}\\ \sf{-x,when\:x< 0}\end{cases}}$

Solution :

We have to prove | x × y | = |x|×|y|

if x=$\frac{-4}{5}$ and y=$\frac{-3}{7}$

LHS

$= |x \times y|$

put x=$\frac{-4}{5}$ and y=$\frac{-3}{7}$

$= | (\frac{ - 4}{5}) \times (\frac{ - 3}{7} )|$

$= | \frac{12}{35} |$

$= \dfrac{12}{35}$

RHS

$= |x| \times |y|$

put the value of x & y

$= | \frac{ - 4}{5} | \times | \frac{ - 3}{7} |$

Here |x| , x>0 i.e |-x| = -(-x)

$= \frac{4}{5} \times \frac{3}{7}$

$= \frac{12}{35}$

⇒LHS = RHS

Hence proved __________________________

More About Modulus Function :

1)For any real number x, √ x² = |x|

2) Modulus Function : It is also called absolute value function.