Answer

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**Hint:**Substitution Method:

In this method from the given two equations of two variables, we have to substitute the equation of any one variable from one of the equations and then substitute it in the other one such that the second equation becomes an equation of one variable and thereby we can solve for that one variable. So by using any of the above definitions we can solve the given pair of linear equations.

**Complete step by step solution:**

Given

\[

3x - y = 4.............\left( i \right) \\

2x - 3y = - 9.............\left( {ii} \right) \\

\]

Now we are using the substitution method which is one of the algebraic methods, as given above to solve the question.

So by using the above definition, let’s substitute x from equation (i) to (ii):

$

\Rightarrow 3x - y = 4 \\

\Rightarrow x = \dfrac{{4 + y}}{3}..................\left( {iii} \right) \\

$

Now substituting (iii) in (ii):

$ \Rightarrow x = \dfrac{{4 + y}}{3}$

Now substituting (iii), we get:

\[

\Rightarrow 2x - 3y = - 9 \\

\Rightarrow 2\left( {\dfrac{{4 + y}}{3}} \right) - 3y = - 9 \\

\Rightarrow 2\left( {\dfrac{{4 + y}}{3}} \right) = 3y - 9 \\

\Rightarrow 2\left( {4 + y} \right) = 3\left( {3y - 9} \right) \\

\Rightarrow 8 + 2y = 9y - 27 \\

\Rightarrow 9y - 2y = 8 + 27 \\

\Rightarrow 7y = 8 + 27..................\left( {iv} \right) \\

\]

On observing (iv) we get that it’s an equation of only ‘y’ such that we can solve for the variable ‘y’.

$

\Rightarrow 7y = 8 + 27 \\

\Rightarrow 7y = 35 \\

\Rightarrow y = \dfrac{{35}}{7} \\

\Rightarrow y = 5......................\left( v \right) \\

$

So now we get: $y = 5$

Now substituting (v) in (iii) to get the value of ‘x’:

$

\Rightarrow x = \dfrac{{4 + y}}{3} \\

\Rightarrow x = \dfrac{{4 + 5}}{3} \\

\Rightarrow x = \dfrac{9}{3} \\

\Rightarrow x = 3.........................\left( {vi} \right) \\

$

Therefore on solving $3x - y = 4\;{\text{and}}\;2x - 3y = - 9$ by substitution method we get $x =

3\;{\text{and}}\;{\text{y = 5}}{\text{.}}$

Additional Information:

A given pair of linear equations can be solved either by graphical or algebraic method. The algebraic

method is of three types:

1. Substitution Method

2. Elimination Method

3. Cross Multiplication Method

**Note:**

While solving a pair of linear equations one should take care of following things:

We need to express the two linear equations in two different variables, we can solve them either by substitution, elimination, cross multiplication method or by graphical method. We can also check the validation of the ‘x’ and ‘y’ values by putting them in the given equations and checking whether it satisfies mathematically or not.

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