Find x & y if [ x 3 x; y 4 y ][ 2; 1 ]=[ 5; 12 ]

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Finding Solution for Consistent Pair of Linear Equations

Find x & y if...

Question

= 1, y = 2

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= 2, y = -1

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= 2, y = 1

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= 3, y = 2

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= -2, y = 2

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Solution

The correct option is A
= 1, y = 2

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

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

(2 a - 3)^{2} - 2 (2 a - 3) (a - 1) + (a - 1)^{2}

Factorise:

$(2 a - 3)^{2} - 2 (2 a - 3) (a - 1) + (a - 1)^{2}$

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.

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