Fractions

Algebra Level 2

What is the value of x x that satisfies the following equation: x x − 2 + x − 21 x − 19 = x + 1 x − 1 + x − 20 x − 18 ? \frac{x}{x-2}+\frac{x-21}{x-19}=\frac{x+1}{x-1}+\frac{x-20}{x-18}?

10 12 13 11

Ajay Dutta by Ajay Dutta , 24, India

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2 solutions

Ahmed Luqman
Oct 11, 2014

Add the fractions on both sides: x ( x − 19 ) + ( x − 21 ) ( x − 2 ) ( x − 2 ) ( x − 19 ) = ( x + 1 ) ( x − 18 ) + ( x − 20 ) ( x − 1 ) ( x − 1 ) ( x − 18 ) \frac{x(x - 19) + (x - 21)(x - 2)}{(x - 2)(x - 19)} = \frac{(x + 1)(x - 18) + (x - 20)(x - 1)}{(x - 1)(x - 18)}

Expand: 2 x 2 − 42 x + 42 x 2 − 21 x + 38 = 2 x 2 − 38 x + 2 x 2 − 19 x + 18 \frac{2x^{2} - 42x + 42}{x^{2} - 21x + 38} = \frac{2x^{2} - 38x + 2}{x^{2} - 19x + 18}

Factorise the numerators: 2 ( x 2 − 21 x + 21 ) x 2 − 21 x + 38 = 2 ( x 2 − 19 x + 1 ) x 2 − 19 x + 18 \frac{2(x^{2} - 21x + 21)}{x^{2} - 21x + 38} = \frac{2(x^{2} - 19x + 1)}{x^{2} - 19x + 18}

We can cancel both sides by 2: x 2 − 21 x + 21 x 2 − 21 x + 38 = x 2 − 19 x + 1 x 2 − 19 x + 18 \frac{x^{2} - 21x + 21}{x^{2} - 21x + 38} = \frac{x^{2} - 19x + 1}{x^{2} - 19x + 18}

Which can also be written as: 1 − 17 x 2 − 21 x + 38 = 1 − 17 x 2 − 19 x + 18 1 - \frac{17}{x^{2} - 21x + 38} = 1 - \frac{17}{x^{2} - 19x + 18}

Therefore: x 2 − 21 x + 38 = x 2 − 19 x + 18 x^{2} - 21x + 38 = x^{2} - 19x + 18 2 x = 20 2x = 20 x = 10 x = 10

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Simply plug in the values of x x and verify

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