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Question

Solve for $$x, y:$$
$$(a+2b)x+ (2a -b) y = 2, (a-2b)x+(2a+b)y=3$$.


A
 x=(5b2a),y=10ab
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B
x=5b2a10ab,y=a+10b10ab
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C
x=5b2a10ab,y=5a+10b10ab
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D
x=5b+2a10ab,y=a+10b10ab
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Solution

The correct option is B $$x=\dfrac{5b-2a}{10ab}, y=\dfrac{a+10b}{10ab}$$
$$(a+2b)x+(2a-b)y=2\\ $$
$$\Rightarrow y=\cfrac { 2-ax-2bx }{ 2a-b } \\ (a-2b)x+(2a+b)y=3\\ $$
$$\Rightarrow (a-2b)x+(2a+b)\times \cfrac { 2-ax-2bx }{ 2a-b } =3\\ $$
$$\Rightarrow \cfrac { (ax-2bx)(2a-b)+(2a+b)(2-ax-2bx) }{ 2a-b } =3\\ $$
$$\Rightarrow 2{ a }^{ 2 }x-4abx-abx+2{ b }^{ 2 }x+4a+2b-2{ a }^{ 2 }x-abx-4abx-2{ b }^{ 2 }x=3(2a-b)\\ $$
$$\Rightarrow 4a+2b-10abx=6a-3b\\ $$
$$\Rightarrow 2a-5b+10abx=0\\ $$
$$\Rightarrow x=\cfrac { 5b-2a }{ 10ab } \\ y=\cfrac { 2-a(\cfrac { 5b-2a }{ 10ab } )-2b(\cfrac { 5b-2a }{ 10ab } ) }{ 2a-b } \\ $$
$$\Rightarrow y=\cfrac { 2(10ab)-a(5b-2a)-2b(5b-2a) }{ 10ab } \times \cfrac { 1 }{ 2a-b } \\ $$
$$\Rightarrow y=\cfrac { 20ab-5ab+2{ a }^{ 2 }-10{ b }^{ 2 }+4ab }{ 10ab\times (2a-b) } \\ $$
$$\Rightarrow y=\cfrac { 19ab+2{ a }^{ 2 }-10{ b }^{ 2 } }{ 10ab\times (2a-b) } \\ $$
$$\Rightarrow y=\cfrac { a+10b }{ 10ab } $$

Mathematics

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