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

Last updated date: 17th Sep 2024
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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 $2 \times 2$ .

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.