Question

Solve by elimination method :- (a+2b)x + (2a -b) y = 2 and (a-2b)x + (2a+b)y = 3

Answer 1

Answer:  The required solution is

$x=\dfrac{-2a+5b}{10ab},~~~~~y=\dfrac{a+10b}{10ab}.$

Step-by-step explanation:  We are given to solve the following system of equations by elimination method :

$(a+2b)x+(2a-b)y=2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\(a-2b)x+(2a+b)y=3~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

Multiplying equation equation (i) by (a - 2b) and equation (ii) by (a + 2b), we have

$(a+2b)(a-2b)x+(2a-b)(a-2b)y=2(a-2b)\\\\\Rightarrow (a^2-4b^2)x+(2a^2-5ab+2b^2)y=2(a-2b)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(iii)$

and

$(a-2b)(a+2b)x+(2a+b)(a+2b)y=3(a+2b)\\\\\Rightarrow (a^2-4b^2)x+(2a^2+5ab+2b^2)y=3(a+2b)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(iv)$

Subtracting equation (iii) from equation (iv), we get

$(a^2-4b^2)x+(2a^2+5ab+2b^2)y-(a^2-4b^2)x-(2a^2-5ab+2b^2)y=3(a+2b)-2(a-2b)\\\\\Rightarrow (5ab+5ab)y=3a+6b-2a+4b\\\\\Rightarrow 10aby=a+10b\\\\\\\Rightarrow y=\dfrac{a+10b}{10ab}.$

Substituting the value of y in equation (i), we get

$(a+2b)x+(2a-b)\times\dfrac{a+10b}{10ab}=2\\\\\\\Rightarrow (a+2b)x+\dfrac{2a^2-ab+20ab-10b^2}{10ab}=2\\\\\\\Rightarrow (a+2b)x+\dfrac{2a^2+19ab-10b^2}{10ab}=2\\\\\Rightarrow 10ab(a+2b)x+2a^2+19ab-10b^2=20ab\\\\\Rightarrow 10ab(a+2b)x=20ab-2a^2-19ab+10b^2\\\\\Rightarrow 10ab(a+2b)x=-2a^2+ab+10b^2\\\\\Rightarrow 10ab(a+2b)x=-2a^2+5ab-4ab+10b^2\\\\\Rightarrow 10ab(a+2b)x=a(-2a+5b)+2b(-2a+5b)\\\\\Rightarrow 10ab(a+2b0x=(a+2b)(-2a+5b)\\\\\Rightarrow 10abx=-2a+5b\\\\\\\Rightarrow x=\dfrac{-2a+5b}{10ab}.$

Thus, the required solution is

$x=\dfrac{-2a+5b}{10ab},~~~~~y=\dfrac{a+10b}{10ab}.$

Answer 2

Step-by-step explanation:

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