Question

3x+6y=19 ; x+2y=21 by elimination method​

Answer 1

3x + 6y =19 × 1

X + 2y = 21 ×3

then equation becomes

3x+6y = 19

3x + 6 y = 63

subtracting both

0 + 0 = -44

False Statement

having no solution

thanks mate

Answer 2

Answer:

It has no solution.

Step by Step Explanation:

Since $\frac{a_{1} }{a_{2} } =\frac{b_{1} }{b_{2} } \neq \frac{c_{1} }{c_{2} }$

        $\frac{3}{1} =\frac{6}{2} \neq \frac{19}{21}$

        $\frac{3}{1} =\frac{3}{1} \neq \frac{19}{21}$

∴It has no solution.

             or

When we subtract 3x+6y=19 and x+2y=21 by elimination method, we get 0+0= -44.

∴It has no solution.

Hope this helps :)

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