The solution for the equations c1 x-b1 y-a1-0 and c2 x-b2 y-a2-0 is S 1- x - b 1 a 2- b 2 a 1- c 1 b 2- c 2 b 1S2- y - c 1 a 2- c 2 a 1- a 1 b 2- a 2 b 1A- S1 and S 2 are trueB- S1 and S2 are falseC- S1 is true but S2 is falseD- S1 is false but S 2 is true

The correct option is C S1 is true but S2 is false S1 : x = (b_1a_2 – b_2a_1)/(c_1b_2 c_2b_1) S2 : y = (c_1a_2 – c_2a_1)/(c_2b_1 c_1b_2) ...

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Byju's Answer

Standard X

Mathematics

Cross-Multiplication Method of Finding Solution of a Pair of Linear Equations

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

S1 :x = $\frac{(b_{1} a_{2} - b_{2} a_{1})}{(c_{1} b_{2} - c_{2} b_{1})}$

S2 : 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 C
S1 is true but S2 is false

S1 : x = $\frac{(b_{1} a_{2} - b_{2} a_{1})}{(c_{1} b_{2} - c_{2} b_{1})}$

S2 : y = $\frac{(c_{1} a_{2} - c_{2} a_{1})}{(c_{2} b_{1} - c_{1} b_{2})}$

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Similar questions

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S1 : $△$ APD $≅$ $△$ BQC

S2 : area( $△$ APD) + area(PBCD) = area( $△$ BQC) + area(CDPB)

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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 C S1 is true but S2 is false S1 : x = (b1a2–b2a1)(c1b2−c2b1) S2 : y = (c1a2–c2a1)(c2b1−c1b2)

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

S1 :x = $\frac{(b_{1} a_{2} - b_{2} a_{1})}{(c_{1} b_{2} - c_{2} b_{1})}$

S2 : y = $\frac{(c_{1} a_{2} - c_{2} a_{1})}{(a_{1} b_{2} - a_{2} b_{1})}$

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

S1 : x = $\frac{(b_{1} a_{2} - b_{2} a_{1})}{(c_{1} b_{2} - c_{2} b_{1})}$

S2 : y = $\frac{(c_{1} a_{2} - c_{2} a_{1})}{(c_{2} b_{1} - c_{1} b_{2})}$