Class 10 Algebraic Fraction Exercise 8 Solutions | Maths Links Readmore Publishers & Distributors

Given,

$\rm \frac{x^2}{y ( x- y)} + \frac{y^2}{x (y - x)}$

We know, $\rm (y - x) = - (x - y)$

$\rm = \frac{x^2}{y (x - y)} + \frac{y^2}{x \{ - (x-y) \}}$

$\rm = \frac{x^2}{y (x-y)} - \frac{y^2}{x (x -y)}$

Taking LCM and simplifying the equation, we get,

$\rm = \frac{ x^2 \cdot x - y^2 \cdot y}{xy (x - y)}$

$\rm = \frac{x^3 - y^3}{xy (x - y)}$

Using the factorization formula for $\rm (a^3 - b^3) = (a-b)(a^2 + ab + b^2)$, we get,

$\rm = \frac{ (x - y)(x^2 + xy + y^2)}{xy (x -y)}$

$\rm = \frac{ x^2 + xy + y^2}{xy}$

$$\rm \therefore \frac{x^2}{y (x - y)} + \frac{y^2}{x (y -x)} = \frac{x^2 + xy + y^2}{xy}$$

Given

$\rm \frac{x^2}{y ( x  + y)} + \frac{y^2}{x ( x + y)}$

$\rm = \frac{x^2}{y (x + y)} \cdot \frac{x}{x} + \frac{y^2}{x (x + y)} \cdot \frac{y}{y}$

$\rm = \frac{x^2 \cdot x}{xy (x + y)} + \frac{y^2 \cdot y}{xy (x + y)}$

$\rm = \frac{x^3}{xy ( x + y)} + \frac{y^3}{xy ( x + y)}$

$\rm = \frac{x^3 + y^3}{xy (x + y)}$

Using the factor formula for $\rm (x^3 + y^3 = (x + y)(x^2 - xy + y^2)$

$\rm = \frac{ (x + y)(x^2 - xy + y^2)}{xy ( x + y)}$

$\rm = \frac{x^2 - xy + y^2}{xy}$

$\rm \therefore \frac{x^2}{y ( x  + y)} + \frac{y^2}{x ( x + y)}= \frac{x^2 - xy + y^2}{xy}$

Hence, the required simplified form of the above expression is found.

Given

$\rm \frac{a^{2} + b^{2}}{ab} - \frac{b^{2}}{a(a+b)} - \frac{a^{2}}{b (a+b)}$

Let us identify the terms as first, second, and third from the left to the right. Then, multiply and divide the first term by \( \rm (a+b) \), the second term by \( \rm b \), and the third term by \( \rm a \).

$\rm = \frac{a^{2} + b^{2}}{ab} \cdot \frac{a +b}{a+b} -  \frac{b^{2}}{a(a+b)} \cdot \frac{b}{b} - \frac{a^{2}}{b (a+b)} \cdot \frac{a}{a}$

$\rm =  \frac{(a^{2} + b^{2})(a +b)}{ab(a+b)}  - \frac{b^{3}}{ab(a+b)} - \frac{a^{3}}{b(a+b)}$

The denominators of the first, second, and third terms are common. So, we simplify the expression as shown below.

$\rm = \frac{ (a+b)(a^{2} + b^{2}) - b^{3} - a^{3}}{ab (a +b)}$

$\rm = \frac{a (a^{2} + b^{2}) + b(a^{2} + b^{2}) - b^{3} -  a^{3}}{ab ( a + b)}$

$\rm = \frac{a^{3} + ab^{2} + a^{2}b + b^{3} - b^{3} - a^{3}}{ab ( a + b)}$

$\rm = \frac{ ab^{2} + a^{2} b}{ab ( a + b)}$

$\rm = \frac{ ab (a + b) }{ab ( a + b)}$

$\rm = 1$

Hence, $$\rm \frac{a^{2} + b^{2}}{ab} - \frac{b^{2}}{a(a+b)} - \frac{a^{2}}{b (a+b)} = 1$$

Given

$\rm \frac{a - b}{a + b} - \frac{a + b}{a - b} + \frac{2ab}{a^2 - b^2}$

$\rm = \frac{a - b}{a + b} \cdot \frac{a - b}{a - b} - \frac{a + b}{a - b} \cdot \frac{a + b}{a + b} + \frac{2ab}{a^2 - b^2}$

$\rm = \frac{ (a-b)(a-b)}{(a+b)(a-b)} - \frac{(a+b)(a+b)}{(a-b)(a+b)} + \frac{2ab}{a^2 - b^2}$

Using the formula for $\rm (a - b)(a + b) = a^2 - b^2$, we get,

$\rm = \frac{ (a - b)^2}{ a^2 - b^2} - \frac{ (a + b)^2}{a^2 - b^2} + \frac{2ab}{a^2 - b^2}$

$\rm = \frac{ (a - b)^2 - (a + b)^2 + 2ab}{a^2 - b^2}$

$\rm = \frac{ a^2 - 2ab + b^2 - (a^2 + 2ab + b^2) + 2ab}{a^2 - b^2}$

$\rm = \frac{ a^2 + b^2 - a^2 - 2ab - b^2}{ a^2 - b^2}$

$\rm = - \frac{ 2ab}{a^2 - b^2}$

Hence, $\rm \frac{a - b}{a + b} - \frac{a + b}{a - b} + \frac{2ab}{a^2 - b^2} = - \frac{ 2ab}{a^2 - b^2}$

Given

$\rm \frac{a}{x - y} - \frac{x + y}{x^{2} - y^{2}}$

Using the factorization formula for $\rm (x^{2} - y^{2}) = (x + y)(x - y)$, we get,

$\rm = \frac{a}{x - y} - \frac{ x + y}{ (x + y)(x - y)}$

$\rm = \frac{a}{x-y} - \frac{1}{ x - y}$

$\rm = \frac{a - 1}{x - y}$

$\rm \therefore \frac{a}{x - y} - \frac{x + y}{x^{2} - y^{2}}= \frac{a - 1}{x - y}$

Given

$\rm \frac{b}{x - 2} + \frac{x + 3}{2 - x} = \frac{1 - x}{x - 2}$

$\rm or, \frac{b}{x - 2} = \frac{1 - x}{x - 2} - \frac{x + 3}{2 - x}$

$\rm or, \frac{b}{x - 2} = \frac{1 - x}{x - 2} - \frac{x + 3}{ - (x - 2)}$

$\rm or, \frac{b}{x - 2} = \frac{1 - x}{x - 2} + \frac{x + 3}{x - 2}$

$\rm or, \frac{b}{x - 2} = \frac{ (1 - x) + (x + 3)}{x - 2}$

$\rm or, \frac{b}{x - 2} = \frac{1 - x + x + 3}{x - 2}$

$\rm or, \frac{b}{x - 2} = \frac{4}{x - 2}$

The expressions in the denominators of both sides of the equation are equal. So, the numerators of these expressions must be equal to hold the equality.

$\rm \therefore b = 4$

Hence, the required value of b is 4.

Given

$\rm \frac{b}{x^2 - 5x} - \frac{x}{5x - 25} = - \frac{x + 5}{5x}$

$\rm or, \frac{b}{x(x - 5)} - \frac{x}{5(x - 5)} = - \frac{x + 5}{5x}$

$\rm or, \frac{b}{x (x - 5)} = - \frac{x + 5}{5x} + \frac{x}{5 (x - 5)}$

$\rm or, \frac{b}{x (x - 5)} = - \frac{x + 5}{5x} \cdot \frac{(x - 5)}{(x - 5)} + \frac{x}{5 (x - 5)} \cdot \frac{x}{x}$

$\rm or, \frac{b}{x (x - 5)} = - \frac{ (x - 5)(x + 5) }{5x ( x - 5)} + \frac{x \cdot x}{5 (x - 5) \cdot x}$

$\rm or, \frac{b}{x (x - 5)} = - \frac{ x^2 - 25 }{5x ( x - 5)} + \frac{ x^2 }{5x (x - 5)}$

$\rm or, \frac{b}{x (x - 5)} = \frac{ - (x^2 - 25) + x^2 }{5x ( x - 5) }$

$\rm or, \frac{b}{x (x - 5)} = \frac{ - x^2 + 25 + x^2}{5x (x - 5)}$

$\rm or, \frac{b}{x (x - 5)} = \frac{ 25}{5x (x - 5)}$

$\rm or, \frac{b}{x ( x - 5)} = \frac{ 5 \cdot 5}{5x (x - 5)}$

$\rm or, \frac{b}{x (x - 5)} = \frac{5}{x (x - 5)}$

The expressions in the denominators of both sides of the equation are equal. So, the numerators of these expressions must be equal to hold the equality.

$\rm \therefore b = 5$

Hence, the required value of b is 5.

Given

$\rm \frac{a}{2x - 3} - \frac{b}{3x + 4} = \frac{x + 7}{6x^2 - x - 12}$

Solving the Left-Hand Side of the equation first, we get,

$\rm \frac{a}{2x- 3} - \frac{b}{3x + 4}$

$\rm = \frac{a}{2x - 3} \cdot \frac{3x + 4}{3x + 4} - \frac{b}{3x + 4} \cdot \frac{2x - 3}{2x - 3}$

Simplifying the expression

$\rm = \frac{a (3x + 4)}{(2x-3)(3x + 4)} - \frac{ b(2x - 3)}{(3x + 4)(2x - 3)}$

$\rm = \frac{ a(3x + 4) - b(2x - 3)}{ (2x -3)(3x + 4) }$

$\rm = \frac{ 3ax + 4a - 2bx + 3b }{ (2x - 3)(3x  + 4)}$

$\rm = \frac{ 3ax - 2bx + 4a + 3b }{ 2x (3x + 4) - 3(3x + 4)}$

$\rm = \frac{ (3a - 2b)x + (4a + 3b) }{ 6x^2 + 8x - 9x - 12}$

$\rm = \frac{ (3a - 2b)x + (4a + 3b)}{6x^2 - x - 12}$

From the given, the expression obtained above for the LHS is equal to the RHS. So,

$\rm or, \frac{ (3a - 2b)x + (4a + 3b)}{6x^2  - x - 12} = \frac{x + 7}{6x^2 - x - 12}$

The denominators on both sides of the equation are the same, so we equate their numerators.

$\rm or, (3a - 2b) x + (4a + 3b) = x + 7$

For the above equation to hold, the coefficient of like terms on both sides of the equation must be the same.

We equate the coefficients of x, we get,

$\rm or, 3a - 2b = 1$

$\rm or, 3a = 1 + 2b$

$\rm or, a = \frac{1 + 2b}{3}$ – (1)

We equate the coefficients of constant terms, and we get,

$\rm or, 4a + 3b = 7$

$\rm or, 4a = 7 - 3b$

$\rm or, a = \frac{7 - 3b}{4}$ – (2)

From equations (1) and (2), we get,

$\rm or, \frac{1 + 2b}{3} = \frac{7 - 3b}{4}$

Multiplying both sides of the equation by 12, we get,

$\rm or, \frac{1 + 2b}{3} \cdot 12 = \frac{7 - 3b}{4} \cdot 12$

$\rm or, (1 + 2b) \cdot 4 = (7 - 3b) \cdot 3$

$\rm or, 4 + 8b = 21 - 9b$

$\rm or, 8b + 9b = 21 - 4$

$\rm or, 17b = 17$

$\rm \therefore b = 1$

We put the value of b = 1 in equation (1) to find the value of a, we get,

$\rm a = \frac{1 + 2 \cdot 1}{3} = \frac{1 + 2}{3} = \frac{3}{3}$

$\rm \therefore a = 1$

Hence, the required solution is (a,b) = (1,1).