Question

solve the following pair of linear equations by cross-multiplication method: 8x+2y=9
3x+2y=4.

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Answer 1

S O L U T I O N :

$\bf{\large{\underline{\bf{Given\::}}}}}$

• 8x + 2y = 9
• 3x + 2y = 4

$\bf{\large{\underline{\bf{Explanation\::}}}}}$

We have;

$\mapsto\sf{8x+2y=9}\\\\\mapsto\sf{8x+2y-9=0....................(1)}$

&

$\mapsto\sf{3x+2y=4}\\\\\mapsto\sf{3x+2y-4=0....................(2)}$

So;

$\bullet\:\tt{a_1=8}\\\bullet\tt{a_2=3}\\\bullet\tt{b_1=2}\\\bullet\tt{b_2=2}\\\bullet\tt{c_1=-9}\\\bullet\tt{c_2=-4}$

$\bigstar$ By using cross - multiplication :

$\mapsto\tt{\bigg[\dfrac{x}{b_1c_2-b_2c_1} =\dfrac{y}{c_1a_2-c_2a_1} =\dfrac{1}{a_1b_2-a_2b_1} \bigg]}\\\\\\\mapsto\tt{\dfrac{x}{2\times (-4)-2\times (-9)} =\dfrac{y}{-9\times 3-(-4)\times 8} =\dfrac{1}{8\times 2-3\times 2} }\\\\\\\mapsto\tt{\dfrac{x}{-8-(-18)} =\dfrac{y}{-27-(-32)} =\dfrac{1}{16-6} }\\\\\\\mapsto\tt{\dfrac{x}{-8+18} =\dfrac{y}{-27+32} =\dfrac{1}{10} }\\\\\\\mapsto\tt{\dfrac{x}{10} =\dfrac{y}{5} =\dfrac{1}{10} }$

$\mapsto\tt{\dfrac{x}{10} =\dfrac{1}{10} \:\:\:Or\:\:\:\dfrac{y}{5} =\dfrac{1}{10} }\\\\\\\mapsto\tt{10x=10\:\:\:Or\:\:\:10y=5}\\\\\\\mapsto\tt{x=\cancel{10/10} \:\:\:Or\:\:\:y=\cancel{5/10}}\\\\\\\mapsto\bf{x=1\:\:\:Or\:\:\:y=1/2}$

Thus;

The value of x and y will be 1 & 1/2 .

Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

$\dagger \: \: {\sf{ { \green{8x + 2y = 9}}}}$

$\dagger \: \: {\sf{ { \green{3x + 2y = 4}}}}$

${\bf{\blue{\underline{Now:}}}}$

$: \implies{\sf{ \frac{x}{ \frac{2}{2} \times \frac{9}{4} } = \frac{y}{ \frac{9}{4} \times \frac{8}{3} } = \frac{ - 1}{ \frac{8}{3} \times \frac{2}{2} } }}$

$: \implies{\sf{ \frac{x}{(4)(2) - (2)(9)} = \frac{y}{(9)(3) - (8)(4)} = \frac{ - 1}{(8)(2) - (3)(2)} }}$

$: \implies{\sf{ \frac{x}{8 - 18} = \frac{y}{27 - 32} = \frac{ - 1}{16 - 6} }}$

$: \implies{\sf{ \frac{x}{ - 10} = \frac{y}{ - 5} = \frac{ - 1}{10} }}$

Now,

$: \implies{\sf{ \frac{x}{ - 10} = \frac{ -1}{10} \: \: \: \: \: \: , \: \: \frac{y}{ - 5} = \frac{ - 1}{10} }}$

$: \implies{\sf{ x = \frac{ 10}{10} \: \: \: \ \: \: \: \:, \: \: \: y = \frac{ 5}{10} }}$

$: \implies{\sf{ x = \frac{ 1}{1} \: \: \: \ \: \: \: \: \: \: \: y = \frac{ 1}{2} }}$

$\dagger \: \: \: \boxed{\sf { \purple {hence \: the \: value \: of \: x = 1 \: and \: y = \frac{1}{2} }}}$