Question

Given the system of equations: $5{{x}_{1}}+3{{x}_{2}}+2{{x}_{3}}=4$, $3{{x}_{1}}+3{{x}_{2}}+2{{x}_{3}}=2$, ${{x}_{2}}+{{x}_{3}}=5$. How to find ${{x}_{1}},{{x}_{2}},{{x}_{3}}$?

Answer 1

Hint: To solve the given equations first we will consider two equations. Then by adding or subtracting the equations we will eliminate one variable and get the value of another variable. Then again substituting the obtained value in the original equation we get the value of other variables.

Complete step by step solution:
We have been given the system of equations $5{{x}_{1}}+3{{x}_{2}}+2{{x}_{3}}=4$, $3{{x}_{1}}+3{{x}_{2}}+2{{x}_{3}}=2$, ${{x}_{2}}+{{x}_{3}}=5$.
We have to find the values of ${{x}_{1}},{{x}_{2}},{{x}_{3}}$.
Now, we have three equations as
$5{{x}_{1}}+3{{x}_{2}}+2{{x}_{3}}=4........(i)$
$3{{x}_{1}}+3{{x}_{2}}+2{{x}_{3}}=2........(ii)$
${{x}_{2}}+{{x}_{3}}=5........(iii)$
Now, let us consider equation (i) and equation (ii). We observe that the coefficients of ${{x}_{2}}$ and ${{x}_{3}}$ are same, so by subtracting the equations we get the value of ${{x}_{1}}$. Then by subtracting equation (ii) from equation (i) we will get
\[\Rightarrow 5{{x}_{1}}+3{{x}_{2}}+2{{x}_{3}}-\left( 3{{x}_{1}}+3{{x}_{2}}+2{{x}_{3}} \right)=4-2\]
Now, simplifying the above obtained equation we will get
\[\begin{align}
  & \Rightarrow 5{{x}_{1}}+3{{x}_{2}}+2{{x}_{3}}-3{{x}_{1}}-3{{x}_{2}}-2{{x}_{3}}=4-2 \\
 & \Rightarrow 2{{x}_{1}}=2 \\
 & \Rightarrow {{x}_{1}}=1 \\
\end{align}\]
Now, substituting the value of ${{x}_{1}}$ in equation (i) we will get
$\Rightarrow 5\times 1+3{{x}_{2}}+2{{x}_{3}}=4$
Now, simplifying the above obtained equation we will get
\[\begin{align}
  & \Rightarrow 5+3{{x}_{2}}+2{{x}_{3}}=4 \\
 & \Rightarrow 3{{x}_{2}}+2{{x}_{3}}=4-5 \\
 & \Rightarrow 3{{x}_{2}}+2{{x}_{3}}=-1........(iv) \\
\end{align}\]
Now, multiplying the equation (iii) by 3 and subtracting the obtained equation from equation (iv) we will get
\[\Rightarrow 3{{x}_{2}}+2{{x}_{3}}-\left( 3{{x}_{2}}+3{{x}_{3}} \right)=-1-15\]
Now, simplifying the above obtained equation we will get
\[\begin{align}
  & \Rightarrow 3{{x}_{2}}+2{{x}_{3}}-3{{x}_{2}}-3{{x}_{3}}=-16 \\
 & \Rightarrow -{{x}_{3}}=-16 \\
 & \Rightarrow {{x}_{3}}=16 \\
\end{align}\]
Now, substituting the value of ${{x}_{3}}$ in equation (iii) we will get
$\begin{align}
  & \Rightarrow {{x}_{2}}+16=5 \\
 & \Rightarrow {{x}_{2}}=5-16 \\
 & \Rightarrow {{x}_{2}}=-11 \\
\end{align}$
Hence on solving the given equations we get the values of ${{x}_{1}},{{x}_{2}},{{x}_{3}}$ as 1, $-11$, 16 respectively.

Note: To eliminate the variable from the equations we have to make the coefficients of a variable equal by dividing or multiplying the equation by some number. We can also solve the system of linear equations by using elimination method, substitution method or graphing method.