Simultaneous Linear Equations were never this complicated

Algebra Level 3

a 1 x + b 1 y + c 1 = 0 a 2 x + b 2 y + c 2 = 0 \large a_1x+b_1y+c_1=0 \\ \large a_2x+b_2y+c_2=0

Given the set of simultaneous linear equations above which are known to have 2 unique real roots, what is the value of x y \dfrac{x}{y} ?

b 1 c 2 − b 2 c 1 c 1 a 2 − c 2 a 1 \dfrac{b_1c_2-b_2c_1}{c_1a_2-c_2a_1} c 1 a 2 − c 2 a 1 a 1 b 2 − a 2 b 1 \dfrac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1} a 1 b 2 − a 2 b 1 b 1 c 2 − b 2 c 1 \dfrac{a_1b_2-a_2b_1}{b_1c_2-b_2c_1} None of the above.

Sharky Kesa by Sharky Kesa , 19, United Kingdom

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1 solution

Nihar Mahajan
May 31, 2015

This question can be easily solved using Cramer's Rule.

a 1 x + b 1 y = − c 1 a 2 x + b 2 y = − c 2 a_1x+b_1y=-c_1 \\ a_2x+b_2y=-c_2

So by Cramer's rule we have:

x = [ ( − c 1 b 1 − c 2 b 2 ] [ ( a 1 b 1 a 2 b 2 ] , y = [ a 1 − c 1 a 2 − c 2 ] [ ( a 1 b 1 a 2 b 2 ] w h e r e [ ( a 1 b 1 a 2 b 2 ] ≠ 0 x=\dfrac{\left[\begin{array}{c}(-c_1 & b_1 \\ -c_2 & b_2 \end{array}\right]}{\left[\begin{array}{c}(a_1 & b_1 \\ a_2 & b_2 \end{array}\right]} \ , \ y=\dfrac{\left[\begin{array}{c}a_1 & -c_1 \\ a_2 & -c_2 \end{array}\right]}{\left[\begin{array}{c}(a_1 & b_1 \\ a_2 & b_2 \end{array}\right]} \ where \ \left[\begin{array}{c}(a_1 & b_1 \\ a_2 & b_2 \end{array}\right] \neq 0

So ,

x y = [ ( − c 1 b 1 − c 2 b 2 ] [ ( a 1 b 1 a 2 b 2 ] [ ( a 1 − c 1 a 2 − c 2 ] [ ( a 1 b 1 a 2 b 2 ] = [ ( − c 1 b 1 − c 2 b 2 ] [ ( a 1 − c 1 a 2 − c 2 ] = b 1 c 2 − b 2 c 1 c 1 a 2 − c 2 a 1 \dfrac{x}{y} = \dfrac{\dfrac{\left[\begin{array}{c}(-c_1 & b_1 \\ -c_2 & b_2 \end{array}\right]}{\left[\begin{array}{c}(a_1 & b_1 \\ a_2 & b_2 \end{array}\right]}}{\dfrac{\left[\begin{array}{c}(a_1 & -c_1 \\ a_2 & -c_2 \end{array}\right]}{\left[\begin{array}{c}(a_1 & b_1 \\ a_2 & b_2 \end{array}\right]}} = \dfrac{\left[\begin{array}{c}(-c_1 & b_1 \\ -c_2 & b_2 \end{array}\right]}{\left[\begin{array}{c}(a_1 & -c_1 \\ a_2 & -c_2 \end{array}\right]} = \boxed{\dfrac{b_1c_2-b_2c_1}{c_1a_2-c_2a_1}}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

: ) +1 from me. I have solved by elimination. First writing x as an expression in y from 1 st equation. Then finding y from 2nd equation therefore finding x. Done.

Mayank Chaturvedi - 6 years ago

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Thanks. Yeah thats the standard method. You can post it as a separate solution.

Nihar Mahajan - 6 years ago

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Instead I would like monopoly of your great and innovative solution over here.

Mayank Chaturvedi - 6 years ago

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