How do you solve the system of equations : $6x - 7y = 19$ and $3x + 4y = 2$?

Hint: Here we are given two linear equations in two variables x and y. We will use elimination method to find the value of unknown variables x and y. Firstly, we will try to make the coefficient of x or y to have the same values in both the given equation. For this we multiply by a suitable number to both the equations and make the terms of x or y to contain the same value. Then we add or subtract them to obtain the value of one unknown. If we get the value of one variable, we substitute it back in any one of the equations and get the value of another unknown variable.

Complete step by step answer:
Given two linear equations,
$6x - 7y = 19$ …… (1)
$3x + 4y = 2$ …… (2)
If we carefully observe the equations, there are no terms of x or y containing the same coefficient.
So, to make the unknown variables x and y to have the same coefficient, we multiply them with a suitable number and then we try to eliminate the terms with the same coefficient using addition or subtraction.
In this problem, we multiply equation (2) by -2, so that the x terms will have the same coefficient in both the equations.
Multiplying the equation (2) by -2, we get,
$ \Rightarrow - 2(3x + 4y) = - 2(2)$
$ \Rightarrow - 6x - 8y = - 4$ …… (3)
Now adding the equations (1) and (3), we get,
$ \Rightarrow 6x - 7y - 6x - 8y = 19 - 4$
Rearranging the terms we get,
$ \Rightarrow 6x - 6x - 7y - 8y = 19 - 4$
Combining the like terms $6x - 6x = 0$ and $ - 7y - 8y = - 15y$
Hence the above equation becomes,
$ \Rightarrow 0 - 15y = 15$
$ \Rightarrow - 15y = 15$
Taking -15 to the right hand side we get,
$ \Rightarrow y = \dfrac{{15}}{{ - 15}}$
$ \Rightarrow y = - 1$.
Now to get the value of x we substitute back $y = - 1$in equation (1) or (2).
Substituting $y = - 1$ in the equation (1), we get,
 $ \Rightarrow 6x - 7( - 1) = 19$
$ \Rightarrow 6x + 7 = 19$
Taking 7 to the right hand side we get,
$ \Rightarrow 6x = 19 - 7$
$ \Rightarrow 6x = 12$
Dividing by 6 on both sides we get,
$ \Rightarrow \dfrac{{6x}}{6} = \dfrac{{12}}{6}$
$ \Rightarrow x = 2$

Hence the values of unknown are given by $x = 2$ and $y = - 1$.

Note: We must choose a suitable number to multiply the given linear equations to eliminate any one of the variables by making them to have the same value.
We can verify whether the obtained value of the variable x and y are correct, by substituting them any one of the equations given. If the equation satisfies, then they are the required values.
We need to be careful while taking the terms to the other side. When transferring any variable or number to the other side, the sign of the same will be changed to its opposite sign.
It is important to know the following basic facts.
An equation remains unchanged or undisturbed if it satisfies the following conditions.
(1) If L.H.S. and R.H.S. are interchanged.
(2) If the same number is added on both sides of the equation.
(3) If the same number is subtracted on both sides of the equation.
(4) When both L.H.S. and R.H.S. are multiplied by the same number.
(5) When both L.H.S. and R.H.S. are divided by the same number.