How do you solve $ x + 2y = 4 $ and $ 2x - y = 3 $ using substitution?

Hint: To solve this question you should know about:
Linear equation of two variables: It is an equation of first order. These equations are defined for lines in the coordinate system. An equation for a straight line is called a linear equation. When there are two variables then this equation will be a linear equation of two variables.
For example: $ax + by = c$
Matrix: It is a collection of numbers in a row and column format.
 \[\left[ {\begin{array}{*{20}{c}}
  2&4 \\
  3&6
\end{array}} \right]\] is a matrix having $2 \times 2$ .

Complete step-by-step answer:
As given in question.
We have, $x + 2y = 4$ and $2x - y = 3$
Here we have to find the value of $x$ and $y$ .
Try it write in matrix form.
 $AX = B$
 $x + 2y = 4$
And
 $2x - y = 3$
We can write it as,
 \[\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  2&{ - 1}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  x \\
  y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  4 \\
  3
\end{array}} \right]\]
To solve we will change this equation into new one,
 $X = {A^{ - 1}}B$
 ${A^{ - 1}}$ is inverse matrix of $A$
We calculate the inverse of $A$ .
 ${A^{ - 1}} = \dfrac{1}{{\det A}}\left[ {\begin{array}{*{20}{c}}
  { - 1}&{ - 2} \\
  { - 2}&1
\end{array}} \right]$
 $\det A = 1 \times \left( { - 1} \right) - 2 \times 2$
 $\det A = \left( { - 1} \right) - 4 = - 5$
Keeping it in above equation. We get,
 \[{A^{ - 1}} = \dfrac{1}{{ - 5}}\left[ {\begin{array}{*{20}{c}}
  { - 1}&{ - 2} \\
  { - 2}&1
\end{array}} \right]\]
 \[ \Rightarrow {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
  {\dfrac{{ - 1}}{{ - 5}}}&{\dfrac{{ - 2}}{{ - 5}}} \\
  {\dfrac{{ - 2}}{{ - 5}}}&{\dfrac{1}{{ - 5}}}
\end{array}} \right]\]
Keeping it back in the equation;
 $X = \left[ {\begin{array}{*{20}{c}}
  {\dfrac{{ - 1}}{{ - 5}}}&{\dfrac{{ - 2}}{{ - 5}}} \\
  {\dfrac{{ - 2}}{{ - 5}}}&{\dfrac{1}{{ - 5}}}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  4 \\
  3
\end{array}} \right]$
By doing multiplication. We get,
 $X = \left[ {\begin{array}{*{20}{c}}
  {\dfrac{1}{5} \times 4 + \,\dfrac{2}{5} \times 3} \\
  {\dfrac{2}{5} \times 4 + \,\dfrac{{ - 1}}{5} \times 3}
\end{array}} \right]$
 $ \Rightarrow X = \left[ {\begin{array}{*{20}{c}}
  {\dfrac{4}{5} + \,\dfrac{6}{5}} \\
  {\dfrac{8}{5} - \,\dfrac{3}{5}}
\end{array}} \right]$
 \[ \Rightarrow X = \left[ {\begin{array}{*{20}{c}}
  {\dfrac{{10}}{5}} \\
  {\dfrac{5}{5}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  2 \\
  1
\end{array}} \right]\]
By keeping value $X$ . We get,
 $\left[ {\begin{array}{*{20}{c}}
  x \\
  y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  2 \\
  1
\end{array}} \right]$
By equating. We will get,
 $x = 2$ and $y = 1$ .
So, the correct answer is “ $ x = 2 $ and $ y = 1 $ ”.

Note: check the value $x = 2$ and $y = 1$ is true or not.
Keep in,
 $
  x + 2y = 4 \\
   \Rightarrow 2 + 2(1) = 4 \;
 $
Hence, for this is true.
Keep in,
 $
  2x - y = 3 \\
   \Rightarrow 2(2) - 1 = 3 \\
   \Rightarrow 4 - 1 = 3 \;
 $
Hence, this value is true for this equation.