Question

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

Answer 1

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.

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.