Question
Solve the following equation and verify the answer:
$\frac{\text{2x}}{5}-\frac{3}{2}=\frac{\text{x}}{2}+1$
$\frac{\text{2x}}{5}-\frac{3}{2}=\frac{\text{x}}{2}+1$
✓
Answer
$\frac{\text{2x}}{5}-\frac{3}{2}=\frac{\text{x}}{2}+1$ Multiplying each term by 10, the L.C.M. of 5 and 2, we get $\frac{\text{2x}}{5}\times10-\frac{3}{2}\times10$ $=\frac{\text{x}}{2}\times10+1\times10$⇒ 4x - 15 = 5x + 10
⇒ 4x - 5x = 10 + 15
(transposing 5x to L.H.S.)
⇒ -x = 25
⇒ x = -25
(multiplying both sides by -1)
So, x = -25 is a solution of the given equation.
Check: Substituting x = -25 in the given equation, we get
$\text{L.H.S.}=\frac{2\times(-25)}{5}-\frac{3}{2}=-10-\frac{3}{2}$
$=\frac{-20-3}{2}=\frac{-23}{2}$
$\text{R.H.S.}=\frac{-25}{2}+1=\frac{-25+2}{2}=\frac{-23}{2}$
$\therefore$ When x = -25, we have
L.H.S. = R.H.S.
⇒ 4x - 5x = 10 + 15
(transposing 5x to L.H.S.)
⇒ -x = 25
⇒ x = -25
(multiplying both sides by -1)
So, x = -25 is a solution of the given equation.
Check: Substituting x = -25 in the given equation, we get
$\text{L.H.S.}=\frac{2\times(-25)}{5}-\frac{3}{2}=-10-\frac{3}{2}$
$=\frac{-20-3}{2}=\frac{-23}{2}$
$\text{R.H.S.}=\frac{-25}{2}+1=\frac{-25+2}{2}=\frac{-23}{2}$
$\therefore$ When x = -25, we have
L.H.S. = R.H.S.
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Kaprekar Number.
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iii. Obtain another 4-digit number by arranging the digits of the new number in ascending order.
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i. Take any 4-digit number in which all the digits are not the same.
ii. Obtain a new 4-digit number by arranging the digits in descending order.
iii. Obtain another 4-digit number by arranging the digits of the new number in ascending order.
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