Question

Solve 2bx +ay= 2ab and bx-ay=4ab by cross multiplication method

Answer 1

Answer:

The solution for the given pair of the equation is   $x=2a, y=-2b$ .

Step-by-step explanation:

We'll use the cross-multiplication method to solve linear equations with two variables.

                 $\frac{x}{b_{1}c_{2}\hspace{0.1cm} - \hspace{0.1cm}b_{2}c_{1} } = \frac{-y}{a_{1}c_{2}\hspace{0.1cm} - \hspace{0.1cm}a_{2}c_{1} } = \frac{1}{a_{1}b_{2}\hspace{0.1cm} - \hspace{0.1cm}a_{2}b_{1} }$

So we conclude, $x = \frac{b_{1}c_{2}\hspace{0.1cm} - \hspace{0.1cm}b_{2}c_{1} }{a_{1}b_{2}\hspace{0.1cm} - \hspace{0.1cm}a_{2}b_{1} }$

                            $y= \frac{a_{2}c_{1}\hspace{0.1cm} - \hspace{0.1cm}a_{1}c_{2}}{a_{1}b_{2}\hspace{0.1cm} - \hspace{0.1cm}a_{2}b_{1} }$

Step 1:

From the given equation we have,

$a_1=2b \hspace{1cm} b_1=a \hspace{1cm} c_1=-2ab\\a_2=b \hspace{1cm} b_2=-a \hspace{1cm} c_2=-4ab$

Step 2:

Putting in the above conclusion, the complete solution is

$\frac{x}{-4a^{2} b \hspace{0.1cm} - \hspace{0.1cm} (2a^{2}b) } =\frac{-y}{-8ab^2-(-2ab^2)} = \frac{1}{-2ab-ab}$

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

Step 3:

Solving,

$\frac{x}{-6a^{2}b} =\frac{1}{-3ab} \hspace{2cm} \implies x=2a$

$\frac{-y}{-6ab^2}=\frac{1}{-3ab} \hspace{2cm} \implies y=-2b$

Therefore, the solution for the given pair of the equation is $x=2a, y=-2b$ .

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