If x x a + y b + z c = y y a + z b + x c = z z a + x b + y c & x + y + z ≠ 0 , then show that each ratio can be equal to 1 a + b + c .

Given: x x a + y b + z c = y y a + z b + x c = z z a + x b + y c Add the antecedents and consequents. By the property of equal ratio: ⇒ x + y + z x a + y b + z c + y a + z b + x c + z a + x b + y c ⇒ x + y + z a x + b x + c x + a y + b y + c y + a z + b z + c z ⇒ x + y + z x a + b + c + y a + b + c + z a + b + c ⇒ x + y + z a + b + c x + y + z ⇒ 1 a + b + c ∴ x x a + y b + z c = y y a + z b + x c = z z a + x b + y c = 1 a + b + c Therefore, each ratio can be equals to 1 a + b + c by the property of equal ratio. Hence, proved.

If b - c b z + c y = c - a c x + a z = a - b a y + b x = 1 then show that a + b + c x + y + z = a x + b y + c z .

Given, b - c b z + c y = c - a c x + a z = a - b a y + b x = 1 We need to show that a + b + c x + y + z = a x + b y + c z Let's take L.H.S. side, ⇒ a + b + c x + y + z ⇒ a x + a y + a z + b x + b y + b z + c x + c y + c z ⇒ a x + b y + c z + b z + c y + c x + a z + a y + b x ⇒ a x + b y + c z + b z + c y + b z + c y c - a b - c + b z + c y a - b b - c (from the given data) ⇒ a x + b y + c z + 1 b - c b z + c y b - c + b z + c y c - a + b z + c y a - b ⇒ a x + b y + c z + 1 b - c b z b - b z c + c y b - c y c + b z c - b z a + c y c - c y a + b z a - b z b + c y a - c y b ⇒ a x + b y + c z + 1 b - c 0 ⇒ a x + b y + c z = R.H.S Hence, Proved.

If x = 4 a b a + b then show that x + 2 a x - 2 a + x + 2 b x - 2 b = 2 .

Given, x = 4 a b a + b We have to show that x + 2 a x - 2 a + x + 2 b x - 2 b = 2 Put the value x in the above equation we get, ⇒ 4 a b a + b + 2 a 4 a b a + b - 2 a + 4 a b a + b + 2 b 4 a b a + b - 2 b = 2 ⇒ 4 a b + 2 a 2 + 2 a b a + b 4 a b - 2 a 2 - 2 a b a + b + 4 a b + 2 a 2 + 2 a b a + b 4 a b - 2 a 2 - 2 a b a + b = 2 ⇒ 6 a b + 2 a 2 2 a b - 2 a 2 + 6 a b + 2 b 2 2 a b - 2 b 2 = 2 Now take the common terms out from both numerator and denominator, we get, ⇒ 2 a 3 b + a 2 a b - a + 2 b 3 a + b 2 b a - b = 2 ⇒ 3 b + a b - a + 3 a + b a - b = 2 ⇒ 3 b + a b - a - 3 a + b b - a = 2 ⇒ 3 b + a - 3 a - b b - a = 2 ⇒ b - a 3 - 1 b - a = 2 ⇒ 2 = 2 L.H.S. = R.H.S Hence, Proved.

If x 3 + 3 x y 2 3 x 2 y + y 3 = a 3 + 3 a b 2 3 a 2 b + b 3 then show that x a = y b .

Given, x 3 + 3 x y 2 3 x 2 y + y 3 = a 3 + 3 a b 2 3 a 2 b + b 3 We have to show that x a = y b or we can write it as x y = a b Add 1 on both sides of the given equation we get, x 3 + 3 x y 2 3 x 2 y + y 3 + 1 = a 3 + 3 a b 2 3 a 2 b + b 3 + 1 ⇒ x 3 + 3 x y 2 + 3 x 2 y + y 3 3 x 2 y + y 3 = a 3 + 3 a b 2 + 3 a 2 b + b 3 3 a 2 b + b 3 ⇒ x + y 3 3 x 2 y + y 3 = a + b 3 3 a 2 b + b 3 a 3 + 3 a b 2 + 3 a 2 b + b 3 = a + b 3 x y + 1 3 3 x y 2 + 1 = a b + 1 3 3 a b 2 + 1 We can see that the above equation looks the same as if x y = a b Hence, Proved.

Given a 3 + 3 a b 2 b 3 + 3 a 2 b = 63 62 then find a : b .

Given, a 3 + 3 a b 2 b 3 + 3 a 2 b = 63 62 By componendo and dividendo we get, ⇒ a 3 + 3 a b 2 + b 3 + 3 a 2 b a 3 + 3 a b 2 - b 3 - 3 a 2 b = 63 + 62 63 - 62 ⇒ a + b 3 a - b 3 = 125 1 x + y 3 = x 3 + 3 x 2 y + 3 y 2 x + y 3 ⇒ a + b a - b 3 = 5 1 3 ⇒ a + b a - b = 5 1 ⇒ a + b = 5 a - b ⇒ a + b = 5 a - 5 b ⇒ 5 b + b = 5 a - a ⇒ 6 b = 4 a ⇒ a b = 6 4 ⇒ a b = 3 2 Hence, a : b = 3 : 2

If x : a = y : b = z : c then show that, x 3 a 2 + y 3 b 2 + z 3 c 2 = x + y + z 3 a + b + c 2 .

Given, x : a = y : b = z : c ⇒ x a = y b = z c = k ∴ x = ka , y = kb , z = kc x 3 a 2 + y 3 b 2 + z 3 c 2 = x + y + z 3 a + b + c 2 L . H . S = x 3 a 2 + y 3 b 2 + z 3 c 2 ⇒ ka 3 a 2 + kb 3 b 2 + kc 3 c 2 ⇒ ak 3 + bk 3 + ck 3 ⇒ k 3 a + b + c R . H . S = x + y + z 3 a + b + c 2 ⇒ ka + kb + kc 3 a + b + c 2 ∵ x = ka , y = kb , z = kc ⇒ k 3 a + b + c 3 a + b + c 2 ⇒ k 3 a + b + c ∴ L . H . S = R . H . S . Hence, it is proved.

If a b = b c , then show that a 2 + b 2 b 2 + c 2 = a b + b c 2 .

Given: a b = b c So, b 2 = a c Consider L.H.S: a 2 + b 2 b 2 + c 2 ⇒ a 2 + a c a c + c 2 ⇒ a a + c c a + c ⇒ a c a + c 2 ⇒ b 2 a + c 2 ⇒ a + c b 2 ⇒ a b + c b 2 L . H . S = R . H . S ∴ a 2 + b 2 b 2 + c 2 = a b + b c 2 Hence, proved.

E M B I B E

SCHOOL MATHEMATICS Class 10>Miscellaneous Exercise : Algebra>Chapter 27 - Miscellaneous Exercise : Algebra>Miscellaneous Exercise : Algebra

Bidyut Guha Thakurta and Amitava Mitra Solutions for Exercise 1: Miscellaneous Exercise : Algebra

Author: Bidyut Guha Thakurta & Amitava Mitra

Bidyut Guha Thakurta Mathematics Solutions for Exercise - Bidyut Guha Thakurta and Amitava Mitra Solutions for Exercise 1: Miscellaneous Exercise : Algebra

Attempt the practice questions from Exercise 1: Miscellaneous Exercise : Algebra with hints and solutions to strengthen your understanding. SCHOOL MATHEMATICS Class 10 solutions are prepared by Experienced Embibe Experts.

Questions from Bidyut Guha Thakurta and Amitava Mitra Solutions for Exercise 1: Miscellaneous Exercise : Algebra with Hints & Solutions

MEDIUM

10th West Bengal Board

IMPORTANT

SCHOOL MATHEMATICS Class 10>Chapter 27 - Miscellaneous Exercise : Algebra>Miscellaneous Exercise : Algebra>Q 81

If $\frac{x}{x a + y b + z c} = \frac{y}{y a + z b + x c} = \frac{z}{z a + x b + y c} & x + y + z \neq 0$ , then show that each ratio can be equal to $\Rightarrow \frac{x + y + z}{(a + b + c) (x + y + z)}$ .

MEDIUM

10th West Bengal Board

IMPORTANT

SCHOOL MATHEMATICS Class 10>Chapter 27 - Miscellaneous Exercise : Algebra>Miscellaneous Exercise : Algebra>Q 82

If $\frac{b - c}{b z + c y} = \frac{c - a}{c x + a z} = \frac{a - b}{a y + b x} = 1$ then show that $(a + b + c) (x + y + z) = a x + b y + c z$ .

MEDIUM

10th West Bengal Board

IMPORTANT

SCHOOL MATHEMATICS Class 10>Chapter 27 - Miscellaneous Exercise : Algebra>Miscellaneous Exercise : Algebra>Q 82

If $x = \frac{4 a b}{a + b}$ then show that $\frac{x + 2 a}{x - 2 a} + \frac{x + 2 b}{x - 2 b} = 2$ .

HARD

10th West Bengal Board

IMPORTANT

SCHOOL MATHEMATICS Class 10>Chapter 27 - Miscellaneous Exercise : Algebra>Miscellaneous Exercise : Algebra>Q 82

If $\frac{x^{3} + 3 x y^{2}}{3 x^{2} y + y^{3}} = \frac{a^{3} + 3 a b^{2}}{3 a^{2} b + b^{3}}$ then show that $\frac{x}{a} = \frac{y}{b}$ .

MEDIUM

10th West Bengal Board

IMPORTANT

SCHOOL MATHEMATICS Class 10>Chapter 27 - Miscellaneous Exercise : Algebra>Miscellaneous Exercise : Algebra>Q 83

Given $\frac{a^{3} + 3 a b^{2}}{b^{3} + 3 a^{2} b} = \frac{63}{62}$ then find $a : b$ .

MEDIUM

10th West Bengal Board

IMPORTANT

SCHOOL MATHEMATICS Class 10>Chapter 27 - Miscellaneous Exercise : Algebra>Miscellaneous Exercise : Algebra>Q 83

If $x : a = y : b = z : c$ then show that, $\frac{x^{3}}{a^{2}} + \frac{y^{3}}{b^{2}} + \frac{z^{3}}{c^{2}} = \frac{{(x + y + z)}^{3}}{{(a + b + c)}^{2}}$ .