Given that the equations in x : 3 x - 2 x + a 3 = 2 x and 3 x + a 3 - 1 + 4 x 6 = 0 have a common solution. Find the common solution

From the first equation we have 3 x - 6 x + a 3 = 2 x ⇒ 3 x - 6 x - 2 a = 2 x ⇒ 5 x = - 2 a ⇒ x = - 2 a 5 From the second equation we have 2 ( 3 x + a ) - ( 1 + 4 x ) = 0 ⇒ 2 x = 1 - 2 a ⇒ x = 1 - 2 a 2 For the common root between the given equations ⇒ - 2 a 5 = 1 - 2 a 2 ⇒ - 4 a = 5 ( 1 - 2 a ) ⇒ - 4 a = 5 - 10 a ⇒ 6 a = 5 ⇒ a = 5 6 Thus, x = - 2 a 5 = - 2 5 · 5 6 = - 1 3

If positive numbers a , b , c satisfy a b c = 1 , solve the equation in x 2 a x a b + a + 1 + 2 b x b c + b + 1 + 2 c x c a + c + 1 = 1

Given, a , b , c are positive numbers and a b c = 1 Also given, 2 a x a b + a + 1 + 2 b x b c + b + 1 + 2 c x c a + c + 1 = 1 ⇒ 2 a x a b + a + 1 × b c b c + 2 b x b c + b + 1 + 2 c x c a + c + 1 × b b = 1 ⇒ 2 a b c x a b c · b + a b c + b c + 2 b x b c + b + 1 + 2 b c x a b c + b c + b = 1 ⇒ 2 x 1 · b + 1 + b c + 2 b x b c + b + 1 + 2 b c x 1 + b c + b = 1 ⇒ 2 x 1 1 + b + b c + b 1 + b + b c + b c 1 + b + b c = 1 ⇒ 2 x 1 + b + b c 1 + b + b c = 1 ⇒ 2 x = 1 ⇒ x = 1 2

If a n + 1 = 1 1 + 1 a n n = 1 , 2 , … , 2008 and a 1 = 1 , find the value of a 1 a 2 + a 2 a 3 + a 3 a 4 + … + a 2008 a 2009

For n = 1 , 2 … . . , 2008 , we have a n + 1 = 1 1 + 1 a n ⇒ a n + 1 = a n a n + 1 ⇒ a n + 1 a n + a n + 1 = a n ⇒ a n + 1 a n = a n - a n + 1 Therefore, a 1 a 2 + a 2 a 3 + a 3 a 4 + … + a 2008 a 2009 = a 1 - a 2 + a 2 - a 3 + … + a 2008 - a 2009 = a 1 - a 2009 Now, we have a n + 1 = a n a n + 1 . . . i Put n = 1 , then a 2 = a 1 1 + a 1 ⇒ a 2 = 1 2 Put n = 2 , then a 3 = a 2 1 + a 2 ⇒ a 3 = 1 3 Put n = 3 , then a 4 = a 3 1 + a 3 ⇒ a 4 = 1 4 Hence, a n - 1 = 1 n - 1 Then, a n = 1 n - 1 1 n - 1 + 1 = 1 n ⇒ a 2009 = 1 2009 Then, a 1 a 2 + a 2 a 3 + a 3 a 4 + … + a 2008 a 2009 = 1 - 1 2009 = 2008 2009

Solve the system of equations x - y + z = 1 y - z + u = 2 z - u + v = 3 u - v + x = 4 v - x + y = 5

Given five equations x - y + z = 1 . . . i y - z + u = 2 . . . ii z - u + v = 3 . . . iii u - v + x = 4 . . . iv v - x + y = 5 . . . v Apply the addition of all the given equations ⇒ ( x - y + z ) + ( y - z + u ) + ( z - u + v ) + ( u - v + x ) + ( v - x + y ) = 1 + 2 + 3 + 4 + 5 ⇒ x + y + z + u + v = 15 . . . vi Apply vi - 2 × ii - i - iii ⇒ x + y + z + u + v - 2 ( y - z + u ) - ( x - y + z ) - ( z - u + v ) = 15 - 2 × 2 - 1 - 3 ⇒ x + y + z + u + v + ( - 2 y + 2 z - 2 u ) + ( - x + y - z ) + ( - z + u - v ) = 15 - 4 - 4 ⇒ z = 7 Apply vi - 2 × iii - ii - iv ⇒ x + y + z + u + v - 2 ( z - u + v ) - ( y - z + u ) - u - v + x = 15 - 2 × 3 - 2 - 4 ⇒ u = 3 Apply vi - 2 × iv - iii - v ⇒ x + y + z + u + v - 2 u - v + x - ( z - u + v ) - v - x + y = 15 - 2 × 4 - 3 - 5 ⇒ v = - 1 Apply vi - 2 × v - iv - i ⇒ x + y + z + u + v - 2 v - x + y - u - v + x - ( x - y + z ) = 15 - 2 × 5 - 4 - 1 ⇒ x = 0 Apply vi - 2 × i - ii - v ⇒ x + y + z + u + v - 2 x - y + z - y - z + u - v - x + y = 15 - 2 × 1 - 2 - 5 ⇒ y = 6

Solve the system of equations 1 x + 1 y + z = 1 2 1 y + 1 z + x = 1 3 1 z + 1 x + y = 1 4

Given, 1 x + 1 y + z = 1 2 , 1 y + 1 z + x = 1 3 & 1 z + 1 x + y = 1 4 ⇒ y + z + x x y + z x = 1 2 , z + x + y y z + x y = 1 3 & x + y + z z x + y z = 1 4 ⇒ x y + z x = 2 ( x + y + z ) , y z + x y = 3 ( x + y + z ) & z x + y z = 4 ( x + y + z ) Let, x y + z x = 2 ( x + y + z ) . . . ( i ) y z + x y = 3 ( x + y + z ) . . . ( ii ) And z x + y z = 4 ( x + y + z ) . . . ( iii ) Apply ( i ) + ( ii ) + ( iii ) ⇒ x y + z x + y z + x y + z x + y z = 2 ( x + y + z ) + 3 ( x + y + z ) + 4 ( x + y + z ) ⇒ 2 ( x y + y z + z x ) = 9 ( x + y + z ) ⇒ x y + y z + z x = 9 2 ( x + y + z ) . . . ( iv ) Apply ( iv ) - ( iii ) , ( iv ) - ( ii ) & ( iv ) - ( i ) ⇒ x y = 1 2 ( x + y + z ) , y z = 3 2 ( x + y + z ) & z x = 5 2 ( x + y + z ) ⇒ x y = 1 2 ( x + y + z ) , y z = 3 2 ( x + y + z ) , z x = 5 2 ( x + y + z ) & x y × y z × z x = 15 8 ( x + y + z ) 3 ⇒ x y = 1 2 ( x + y + z ) , y z = 3 2 ( x + y + z ) , z x = 5 2 ( x + y + z ) & ( x y z ) 2 = 15 8 ( x + y + z ) 3 ⇒ 1 4 ( x + y + z ) 2 z 2 = 15 8 ( x + y + z ) 3 , 9 4 ( x + y + z ) 2 x 2 = 15 8 ( x + y + z ) 3 & 25 4 ( x + y + z ) 2 y 2 = 15 8 ( x + y + z ) 3 ⇒ z = 15 2 ( x + y + z ) , x = 5 6 ( x + y + z ) & y = 3 10 ( x + y + z ) ⇒ x = 5 6 ( x + y + z ) , y = 3 10 ( x + y + z ) , z = 15 2 ( x + y + z ) & x + y + z = 5 6 ( x + y + z ) + 3 10 ( x + y + z ) + 15 2 ( x + y + z ) ⇒ x = 5 6 x + y + z , y = 3 10 x + y + z , z = 15 2 x + y + z & x + y + z = 5 6 + 3 10 + 15 2 ⇒ x = 5 6 5 6 + 3 10 + 15 2 , y = 3 10 5 6 + 3 10 + 15 2 & z = 15 2 5 6 + 3 10 + 15 2 ⇒ x = 5 6 + 1 2 + 5 2 , y = 1 2 + 3 10 + 3 2 & z = 5 2 + 3 2 + 15 2 ⇒ x = 23 6 , y = 23 10 & z = 23 2 .

E M B I B E

Non-Routine Mathematics Resource Book-1 for PRMO>Chapter 1 - Algebra>EXERCISE - 1.4>Q 5

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IOQM - PRMO and RMO

IMPORTANT

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If $a, b, c$ are real numbers such that $a + (\frac{1}{b}) = \frac{7}{3}; b + (\frac{1}{c}) = 4; c + (\frac{1}{a}) = 1$ , find $a b c$

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Important Questions on Algebra

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IOQM - PRMO and RMO

IMPORTANT

Non-Routine Mathematics Resource Book-1 for PRMO>Chapter 1 - Algebra>EXERCISE - 1.4>Q 6

If

a, b, c, d

, satisfy the equations

a + 7 b + 3 c + 5 d = 0, 8 a + 4 b + 6 c + 2 d = - 16, 2 a + 6 b + 4 c + 8 d = 16

and

5 a + 3 b + 7 c + d = - 16

, then the value of

|(a + d) (b + c)| =

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IOQM - PRMO and RMO

IMPORTANT

Non-Routine Mathematics Resource Book-1 for PRMO>Chapter 1 - Algebra>EXERCISE - 1.4>Q 7

If

x + y = 5 x y, y + z = 6 y z, z + x = 7 z x

, find the value of

x + y + z

.

EASY

IOQM - PRMO and RMO

IMPORTANT

Non-Routine Mathematics Resource Book-1 for PRMO>Chapter 1 - Algebra>EXERCISE - 1.4>Q 8

Given that the equations in $x$ :

$3 [x - 2 (x + \frac{a}{3})] = 2 x$ and $\frac{3 x + a}{3} - \frac{1 + 4 x}{6} = 0$ have a common solution. Find the common solution

EASY

IOQM - PRMO and RMO

IMPORTANT

Non-Routine Mathematics Resource Book-1 for PRMO>Chapter 1 - Algebra>EXERCISE - 1.4>Q 9

If positive numbers $a, b, c$ satisfy $a b c = 1$ , solve the equation in $x$

$\frac{2 a x}{a b + a + 1} + \frac{2 b x}{b c + b + 1} + \frac{2 c x}{c a + c + 1} = 1$

HARD

IOQM - PRMO and RMO

IMPORTANT

Non-Routine Mathematics Resource Book-1 for PRMO>Chapter 1 - Algebra>EXERCISE - 1.4>Q 10

If $a_{n + 1} = \frac{1}{1 + \frac{1}{a_{n}}} (n = 1, 2, \dots, 2008)$ and $a_{1} = 1$ , find the value of $a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{4} + \dots + a_{2008} a_{2009}$

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IOQM - PRMO and RMO

IMPORTANT

Non-Routine Mathematics Resource Book-1 for PRMO>Chapter 1 - Algebra>EXERCISE - 1.4>Q 12

Solve the system of equations

$\{\begin{cases} x - y + z = 1 \\ y - z + u = 2 \\ z - u + v = 3 \\ u - v + x = 4 \\ v - x + y = 5 \end{cases}$

HARD

IOQM - PRMO and RMO

IMPORTANT

Non-Routine Mathematics Resource Book-1 for PRMO>Chapter 1 - Algebra>EXERCISE - 1.4>Q 13

Given

$\frac{1}{x} + \frac{2}{y} + \frac{3}{z} = 0 . . . (i)$

$\frac{1}{x} - \frac{6}{y} - \frac{5}{z} = 0 . . . (ii)$

Find the absolute value of $\frac{x}{y} + \frac{y}{z} + \frac{z}{x}$

HARD

IOQM - PRMO and RMO

IMPORTANT

Non-Routine Mathematics Resource Book-1 for PRMO>Chapter 1 - Algebra>EXERCISE - 1.4>Q 14

Solve the system of equations

$\{\begin{cases} \frac{1}{x} + \frac{1}{y + z} = \frac{1}{2} \\ \frac{1}{y} + \frac{1}{z + x} = \frac{1}{3} \\ \frac{1}{z} + \frac{1}{x + y} = \frac{1}{4} \end{cases}$