Solve the following equations: x + y + z = a + b + c x a + y b + z c = 1 a x + b y + c z = b c + c a + a b

Given equations are x + y + z = a + b + c . . . ( 1 ) x a + y b + z c = 3 . . . 2 a x + b y + c z = b c + c a + a b . . . 3 Let x = a + h , y = b + k , z = c + l . . . ( 4 ) Then, from 1 and 2 , we have, h + k + l = 0 ⇒ h a + k b + l c = 0 Solving above two equations by cross-multiplication method, h a ( b - c ) = k b ( c - a ) = l c ( a - b ) = λ (say) then, h = a λ ( b - c ) , k = b λ ( c - a ) , l = c λ ( a - b ) . . . 5 Also, from equation 3 , we have a h + b k + c l = a b + b c + a c - a 2 - b 2 - c 2 Putting the values of h , k , l in terms of λ , we have λ a 2 ( b - c ) + b 2 ( c - a ) + c 2 ( a - b ) = a b + b c + a c - a 2 - b 2 - c 2 ∴ λ = a b + b c + a c - a 2 - b 2 - c 2 a 2 ( b - c ) + b 2 ( c - a ) + c 2 ( a - b ) . . . 6 Therefore, by equation 4 and 5 , we have x = a + a λ ( b - c ) y = b + b λ ( c - a ) z = c + c λ ( a - b ) where λ has the same value as given in 6 .

Eliminate x , y , z from the equations x 2 - x y - x z a = y 2 - y z - y x b = z 2 - z x - z y c and a x + b y + c z = 0 .

Let x 2 - x y - x z a = y 2 - y z - y x b = z 2 - z x - z y c = P . . . 1 a x + b y + c z = 0 . . . 2 ∴ From 1 , we have x 2 - x y - x z = a P . . . 3 y 2 - y z - y x = b P . . . 4 z 2 - z x - z y = c P . . . 5 Subtracting 3 from 4 and multiplying z , we get P b - a z = x - y z 2 - z x - z y . . . 6 But by 5 , we have z 2 - z x - z y = c P Therefore, equation 6 becomes P b - a z = x - y c P or c x - c y + a - b z = 0 . . . 7 Now subtracting 5 from 3 , we get P ( a - c ) = x 2 - z 2 + y ( z - x ) = z - x y - x - z or P ( a - c ) y = ( z - x ) y 2 - y x - y z . . . 8 By 4 , we have P ( a - c ) y = ( z - x ) b P or b z - b x - a - c y = 0 or b x + a - c y - b z = 0 . . . 9 Now solving 7 and 9 by cross multiplication method, we have x c b - a - b a - c = y a - b b + c b = z c a - c + c b or x a b + c - a = y b c + a - b = z c a + b - c = k Putting the value of x , y , z in 2 we have, a · a k b + c - a + b · b k c + a - b + c · c k a + b - c = 0 or a 2 b + c + b 2 c + a + c 2 a + b = a 3 + b 3 + c 3 which is the required condition.

E M B I B E

Higher Algebra>Chapter 36 - Miscellaneous Examples>Exercise 1>Q 217

HARD

JEE Main

IMPORTANT

Earn 100

In a shooting competition, a man can score $5, 4, 3, 2$ or $0$ points for each shot. Find the number of different ways in which he can score $30$ in $7$ shots.

Important Questions on Miscellaneous Examples

HARD

JEE Main

IMPORTANT

Higher Algebra>Chapter 36 - Miscellaneous Examples>Exercise 1>Q 219

A bag contains

6

black balls and an unknown number (not greater than

6

) of white balls. Three are drawn successively and not replaced and are all found to be white. Prove that the chance that a black ball will be drawn next is

\frac{477}{909}

.

MEDIUM

JEE Main

IMPORTANT

Higher Algebra>Chapter 36 - Miscellaneous Examples>Exercise 1>Q 223

Solve the equations

$x^{2} + 2 y z = y^{2} + 2 z x = z^{2} + 2 x y + 3 = 76$

HARD

JEE Main

IMPORTANT

Higher Algebra>Chapter 36 - Miscellaneous Examples>Exercise 1>Q 223

Solve the following equations:

$x + y + z = a + b + c$

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

$a x + b y + c z = b c + c a + a b$

HARD

JEE Main

IMPORTANT

Higher Algebra>Chapter 36 - Miscellaneous Examples>Exercise 1>Q 224

Prove that if each of

m

points in one straight line be joined to each of

n

in another by straight lines terminated by the points, then excluding the given points, the line will intersect

\frac{1}{4} m n (m - 1) (n - 1)

times.

MEDIUM

JEE Main

IMPORTANT

Higher Algebra>Chapter 36 - Miscellaneous Examples>Exercise 1>Q 226

A farmer spent three equal sums of money in buying hens, pigeons and parrots. Each hen cost Rs.

1

more than a pigeon and Rs.

2

more than a parrot, altogether he bought

47

birds. The number of pigeons exceeded that of the hens by as many parrots as he could have bought for

9

. Find the number of birds of each kind.

HARD

JEE Main

IMPORTANT

Higher Algebra>Chapter 36 - Miscellaneous Examples>Exercise 1>Q 228

In a certain examination six papers are set and to each are assigned $100$ marks as a maximum. Show that the number of ways in which a candidate may obtain forty percent of the whole numbers of marks is $\frac{1}{5!} \{\frac{245!}{240!} - 6 \frac{144!}{139!} + 15 \frac{43!}{38!}\}$ .

HARD

JEE Main

IMPORTANT

Higher Algebra>Chapter 36 - Miscellaneous Examples>Exercise 1>Q 233

Eliminate $x, y, z$ from the equations $\frac{x^{2} - x y - x z}{a} = \frac{y^{2} - y z - y x}{b} = \frac{z^{2} - z x - z y}{c}$ and $a x + b y + c z = 0$ .