Question

solve by cross multiplication method , 2ax+3by=a+2b;3ax+2by=2a+b

Answer 1

May Be This Is Answer

Answer 2

Answer:

Value of x and y are $x=\frac{4ab-b^2}{5ab}\:\:and\:\:y=\frac{+4ab-a^2}{5ab}$

Step-by-step explanation:

Given Equations in 2 variable are

2ax + 3by = a+2b and 3ax + 2by = 2a+b

To find: Solution Of equation by Cross-Multiplication method

Solution by Cross-Multiplication of equation $a_1x+b_1+c_1=0\:\:and\:\:a_2x+b_2+c_2=0$ are give by,

$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

By rewriting given equation in standard form, we get

2ax + 3by - (a+2b) = 0 and 3ax + 2by - (2a+b) = 0

here, $a_1=2a\:,\:b_1=3b\:,\:c_1=-(a+2b)\:\:and\;\:a_2=3a\;,\;b_2=2b\;,\:c_2=-(2a+b)$

So, By Cross-Multiplication Method we get,

$\frac{x}{(3b)(-(2a+b))-(2b)(-(a+2b))}=\frac{y}{(-(a+2b))(3a)-(-(2a+b))(2a)}=\frac{1}{(2a)(2b)-(3a)(3b)}$

$\frac{x}{-3b(2a+b)+2b(a+2b)}=\frac{y}{-3a(a+2b)+2a(2a+b)}=\frac{1}{4ab-9ab)}$

$\frac{x}{-6ab-3b^2+2ab+4b^2}=\frac{y}{-3a^2-6ab+4a^2+2ab}=\frac{1}{-5ab}$

$\frac{x}{-4ab+b^2}=\frac{y}{a^2-4ab}=\frac{1}{-5ab}$

Now, Equating 2 at a time we get,

$\frac{x}{-4ab+b^2}=\frac{y}{a^2-4ab}=\frac{1}{-5ab}$

$\frac{x}{-4ab+b^2}=\frac{1}{-5ab}\:\:and\:\:\frac{y}{a^2-4ab}=\frac{1}{-5ab}$

$x=\frac{-4ab+b^2}{-5ab}\:\:and\:\:y=\frac{a^2-4ab}{-5ab}$

$x=\frac{-(4ab-b^2)}{-5ab}\:\:and\:\:y=\frac{-(-a^2+4ab)}{-5ab}$

$x=\frac{4ab-b^2}{5ab}\:\:and\:\:y=\frac{+4ab-a^2}{5ab}$

Therefore, Value of x and y are $x=\frac{4ab-b^2}{5ab}\:\:and\:\:y=\frac{+4ab-a^2}{5ab}$