Question

Q 1. Solve the following equations by elimination method 3x+4y= -6 3x-2y= 4​

Answer 1

Step-by-step explanation:

Eq. no. 1 = 3x + 4y = -6

Eq. no. 2 = 3x - 2y = 4

Applying elimination on eqs. 1 and 2

3x + 4y = -6

3x - 2y = +4

- + -

0x + 6y = -10

Therefore, y = -10/6

y = -5/3

Putting value of y in eq. no. 1

3x + 4 x -5/3 = -6

3x - 20/3 = -6

3x = -6 + 20/3

3x = -18 + 20 / 3

3x = 2/3

x = 2 / 3 x 3

x = 2 / 9

Therefore the values of x and y are 2/9 and -5/3

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Answer 2

Answer:

=> The value of $x$ is $\frac{2}{9}$ and $y$is $\frac{-5}{3}.$

Step-by-step explanation:

Given that:-

$3x + 4y = -6$    $...(i)$

$3x - 2y = 4$      $...(ii)$

By using the method of elimination:

   $3x + 4y = -6$

    $3x - 2y = +4$

   ---------------------

           $6y = -10$

Therefore,  

=> $6y = -10$

$=> y =\frac{-10}{6}$

By reducing to its lowest terms,

$=> y=\frac{-5}{3}$

Now, substitute the value of $y$ in the equation ( $i$ ), we get  

=> $3x + 4 \times \frac{-5}{3} = -6$

By further calculation, we get

$=> 3x - \frac{20}{3} = -6$

$=> 3x = -6+ \frac{20}{3}$

 $=> 3x = \frac{2}{3}$

 $=> x = \frac{2}{9}$

Therefore the values of $x$ and $y$ are $\frac{2}{9}$ and $\frac{-5}{3}.$