The solution for the equations c1 x-b1 y-a1-0 and c2 x-b2 y-a2-0 isS 1- x - b 1 a 2- b 2 a 1- c 1 h 3-S 2- y - c 1 a 2- c 2 a 1- a 1 b 2- a 2 b 1A- S 1 and S 2 are trueB- S 1 and S 2 are falseC- S 1 is true but S 2 is falseD- S 1 is false but S 2 is true

The correct option is D S1 is false but S2 is true S_1: x = (b_1c_2 – b_2c_1)/(a_1b_2 a_2b_1) S_2 : y = (c_1a_2 – c_2a_1)/(a_1b_2 a_2b_1) ...

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Byju's Answer

Standard X

Mathematics

Equations Reducible to a Pair of Linear Equations in Two Variables

The solution ...

Question

The solution for the equations $c_{1}$ x + $b_{1}$ y + $a_{1}$ = 0 and $c_{2}$ x + $b_{2}$ y + $a_{2}$ = 0 is

$S_{1}$ : x = $\frac{(b_{1} a_{2} - b_{2} a_{1})}{(c_{1} b_{2} - c_{2} b_{1})}$

$S_{2}$ : y = $\frac{(c_{1} a_{2} - c_{2} a_{1})}{(a_{1} b_{2} - a_{2} b_{1})}$

S₁ and S₂ are true

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S₁ and S₂ are false

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S₁ is true but S₂ is false

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S₁ is false but S₂ is true

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Solution

The correct option is D
S₁ is false but S₂ is true

$S_{1}$ : x = $\frac{(b_{1} c_{2} - b_{2} c_{1})}{(a_{1} b_{2} - a_{2} b_{1})}$

$S_{2}$ : y = $\frac{(c_{1} a_{2} - c_{2} a_{1})}{(a_{1} b_{2} - a_{2} b_{1})}$

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Similar questions

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The solution for the equations c1x + b1y + a1 = 0 and c2x + b2y + a2 = 0 is S1 : x = (b1a2–b2a1)(c1b2−c2b1) S2 : y = (c1a2–c2a1)(a1b2−a2b1)

The correct option is D S1 is false but S2 is true S1: x = (b1c2–b2c1)(a1b2−a2b1) S2 : y = (c1a2–c2a1)(a1b2−a2b1)

The solution for the equations $c_{1}$ x + $b_{1}$ y + $a_{1}$ = 0 and $c_{2}$ x + $b_{2}$ y + $a_{2}$ = 0 is

$S_{1}$ : x = $\frac{(b_{1} a_{2} - b_{2} a_{1})}{(c_{1} b_{2} - c_{2} b_{1})}$

$S_{2}$ : y = $\frac{(c_{1} a_{2} - c_{2} a_{1})}{(a_{1} b_{2} - a_{2} b_{1})}$

The correct option is D
S₁ is false but S₂ is true

$S_{1}$ : x = $\frac{(b_{1} c_{2} - b_{2} c_{1})}{(a_{1} b_{2} - a_{2} b_{1})}$

$S_{2}$ : y = $\frac{(c_{1} a_{2} - c_{2} a_{1})}{(a_{1} b_{2} - a_{2} b_{1})}$