The solution for the equations C 1 x - b 1 y - a 1-0 and c 2 x - b 2 y - a 2-0 isS 1- X - b 1 a 2- b 2 a 1- c 1 b 0-S 2- y - a 2 c 1- a 1 c 2- c 2 b 1- c 1 b 2A- S1 and S2 are trueB- S1 and S2 are falseC- S1 is true but S 2 is falseD- S1 is false but S 2 is true

The correct option is C S1 is true but S2 is false Using cross multiplication, x = (b_1a_2 – b_2a_1)/(c_1b_2 c_2b_1) y = (a_1c_2 – a_2c_1)/(c_1b_2 c_2b_1) ...

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Byju's Answer

Standard VIII

Mathematics

Cross-multiplication Method of finding solution of a pair of LE

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

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

Using cross multiplication,

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

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

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Similar questions

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

In the given figure, ABCD is a parallelogram and PD is parallel to QC.

S1 : $△$ APD $≅$ $△$ BQC

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

We are given the equation $x + 2 y = 3$

Statement 1: The degree of the equation is one.

Statement 2 : It is a linear equation in two variable.

The graph for the following system of equations :

$2 x + 3 y = 5$ and $6 x + 9 y = 40$ is shown

$S_{1} : \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

S2 : The two lines intersect each other.

In the given figure, ABCD is a parallelogram and PD is parallel to QC.

S1 : $△$ APD $≅$ $△$ BQC

S2 : ar( $△$ APD) + ar(PBCD) = ar( $△$ BQC) + ar(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=(a2c1–a1c2)(c2b1−c1b2)

The correct option is C S1 is true but S2 is false Using cross multiplication, x=(b1a2–b2a1)(c1b2−c2b1) y=(a1c2–a2c1)(c1b2−c2b1)

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

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

Using cross multiplication,

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

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