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Solve the given linear equations:$\begin{array}{l}x + 2y + z = 7\\x + 3z = 11\\2x - 3y = 11\end{array}$

Last updated date: 16th Sep 2024
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Hint: We are required to solve the given system of linear equations. To solve this question, we will use the concept of substitution. We will first convert the linear system of three variables into two variables, by eliminating anyone unknown term. We will then find the value of a variable in terms of the other, and then back-substitute it in the previous equation to obtain the final values of the unknown variables.

We are provided with the following linear equations. We will number each of the equations to ease the process of solving.
$x + 2y + z = 7$………………..$\left( 1 \right)$
$x + 3z = 11$…………………….$\left( 2 \right)$
$2x - 3y = 11$……………………$\left( 3 \right)$
Now we will consider equation $\left( 2 \right)$ and equation $\left( 3 \right)$. We will find the value of the unknown variable $x$ from equation $\left( 2 \right)$ and substitute it in equation $\left( 3 \right)$. On doing so we will get –
$x + 3z = 11$
$\Rightarrow x = 11 - 3z$……………$\left( 4 \right)$
On substituting the above value of $x$ from equation $\left( 4 \right)$ to equation $\left( 3 \right)$, we get,
$\begin{array}{l}2x - 3y = 11\\ \Rightarrow 2\left( {11 - 3z} \right) - 3y = 11\end{array}$
Now we will multiply the terms further. After that, we will rearrange the equation such that the like terms are together. On doing so we obtain,
$\begin{array}{l}22 - 6z - 3y = 11\\ \Rightarrow - 6z - 3y = 11 - 22\end{array}$
Simplifying the equation further, we get
$- 6z - 3y = - 11$
$\Rightarrow 6z + 3y = 11$………$\left( 5 \right)$
Let us now substitute the value of $x$ from equation $\left( 4 \right)$ to equation $\left( 1 \right)$. On doing so, we get
$\begin{array}{l}x + 2y + z = 7\\ \Rightarrow 11 - 3z + 2y + z = 7\end{array}$
Rearranging and adding the like terms, we get
$\begin{array}{l} - 3z + z + 2y = 7 - 11\\ \Rightarrow - 2z + 2y = - 4\end{array}$
Dividing both side by $- 2$, we get
$\dfrac{{ - 2z}}{{ - 2}} + \dfrac{{2y}}{{\left( { - 2} \right)}} = \dfrac{{ - 4}}{{ - 2}}$
$\Rightarrow z - y = 2$……………$\left( 6 \right)$
We have now obtained equation $\left( 5 \right)$ and equation $\left( 6 \right)$ in two variables. So, we will solve these now.
We will multiply equation $\left( 6 \right)$ with 3 on both sides of the equation. We will thus obtain
$\left( {z - y} \right) \times 3 = 2 \times 3$
$\Rightarrow 3z - 3y = 6$………..$\left( 7 \right)$
Now we will add equation $\left( 5 \right)$ and equation $\left( 7 \right)$ together. We will then obtain
$\begin{array}{l}6z + 3y + 3z - 3y = 11 + 6\\ \Rightarrow 9z = 17\\ \Rightarrow z = \dfrac{{17}}{9}\end{array}$
Substituting the value of $z$ in equation (2), we get
$\begin{array}{l}x + 3\left( {\dfrac{{17}}{9}} \right) = 11\\ \Rightarrow x + \dfrac{{17}}{3} = 11\end{array}$
Gathering and subtracting all the like terms, we get
$\begin{array}{l}x = 11 - \dfrac{{17}}{3}\\ \Rightarrow x = \dfrac{{33}}{3} - \dfrac{{17}}{3}\\ \Rightarrow x = \dfrac{{16}}{3}\end{array}$
We will now back substitute the value of $x$ and $z$ in equation (1). On doing so we will obtain the value of $y$ as follows –
$\begin{array}{l}x + 2y + z = 7\\ \Rightarrow \dfrac{{16}}{3} + 2y + \dfrac{{17}}{9} = 7\end{array}$
Gathering all the like terms, we get
$\Rightarrow 2y = 7 - \dfrac{{16}}{3} - \dfrac{{17}}{9}$
Taking LCM of all the like terms, we get
$\Rightarrow 2y = \dfrac{{7 \times 9}}{9} - \dfrac{{16 \times 3}}{3} - \dfrac{{17}}{9}$
Simplifying the terms, we get
$\begin{array}{l} \Rightarrow 2y = \dfrac{{63}}{9} - \dfrac{{48}}{9} - \dfrac{{17}}{9}\\ \Rightarrow 2y = \dfrac{{63}}{9} - \dfrac{{65}}{9}\\ \Rightarrow 2y = \dfrac{{ - 2}}{9}\end{array}$
Dividing both sides by 2, we get
$\begin{array}{l}\dfrac{{2y}}{2} = \dfrac{{ - 2}}{{9 \times 2}}\\ \Rightarrow y = \dfrac{{ - 1}}{9}\end{array}$
Thus, the values obtained are as follows –
$\begin{array}{l}x = \dfrac{{16}}{3}\\y = \dfrac{{ - 1}}{9}\\z = \dfrac{{17}}{9}\end{array}$.

Note: To check whether the values of the unknown variables obtained is correct or not, we always substitute our values in the equations provided. If the equations are satisfied then it means that our solution is correct. If the equation is not satisfied then our solutions become incorrect.
Let us check our solutions for the given equations.
On substituting the value of $x = \dfrac{{16}}{3}$,$y = \dfrac{{ - 1}}{9}$ and $z = \dfrac{{17}}{9}$ in equation $\left( 1 \right)$, we get
$\begin{array}{l}\dfrac{{16}}{3} + 2\left( {\dfrac{{ - 1}}{9}} \right) + \dfrac{{17}}{9} = 7\\ \Rightarrow \dfrac{{16}}{3} - \dfrac{2}{9} + \dfrac{{17}}{9} = 7\end{array}$
Adding the like terms, we get
$\dfrac{{16}}{3} + \dfrac{{15}}{9} = 7$
Taking LCM on left hand side, we get
$\Rightarrow \dfrac{{16 \times 3}}{{3 \times 3}} + \dfrac{{15}}{9} = 7$
Simplifying the terms, we get
$\begin{array}{l}\dfrac{{48}}{9} + \dfrac{{15}}{9} = 7\\ \Rightarrow \dfrac{{63}}{9} = 7\\ \Rightarrow 7 = 7\end{array}$
Hence, the above equation is satisfied.
On substituting the value of $x = \dfrac{{16}}{3}$,$y = \dfrac{{ - 1}}{9}$ and $z = \dfrac{{17}}{9}$ in equation $\left( 2 \right)$, we get
$\begin{array}{l}\dfrac{{16}}{3} + 3\left( {\dfrac{{17}}{9}} \right) = 11\\ \Rightarrow \dfrac{{16}}{3} + \dfrac{{17}}{3} = 11\end{array}$
Adding the terms on left hand side, we get
$\begin{array}{l}\dfrac{{33}}{3} = 11\\ \Rightarrow 11 = 11\end{array}$
Hence, the above equation is satisfied.
On substituting the value of $x = \dfrac{{16}}{3}$,$y = \dfrac{{ - 1}}{9}$ and $z = \dfrac{{17}}{9}$ in equation $\left( 3 \right)$, we get
$\begin{array}{l}2\left( {\dfrac{{16}}{3}} \right) - 3\left( {\dfrac{{ - 1}}{9}} \right) = 11\\ \Rightarrow \dfrac{{32}}{3} + \dfrac{1}{9} = 11\end{array}$
Adding the terms on left hand side, we get
$\begin{array}{l}\dfrac{{33}}{3} = 11\\ \Rightarrow 11 = 11\end{array}$
Hence, the above equation is satisfied.
Thus, as all the above equations are satisfied, so our solution is correct.