Question

How do solve the following linear system \[5x + y = 6\] and \[12x - 2y = - 16?\]

Answer 1

Hint: In this question, solving the elimination method is generally used to solve the system of equations. In this method, first solve the equation for one variable, and substitution the value of the variable in the other equation. Finally we get values of $x$ and $y$.
A solution to a linear system is an assignment of values to the variables such that all the equations are simultaneously satisfied. A solution to the system below is given by. It makes all two equations valid.

Complete step-by-step solution:
To find $x$ and $y$ value:
First we solve the linear system:
Take,
Equation 1: \[5x + y = 6\]
Equation 2: \[12x - 2y = - 16\]
We can solve the system by elimination:
 Now, multiply Equation 1 by 2 and we get
\[ \Rightarrow 2(5x + y = 6)\]
\[ \Rightarrow 10x + 2y = 12\]
Next we add the above equation and Equation 2 and we get
\[\begin{array}{*{20}{c}}
  {10x + 2\not{y} = 12} \\
  {\underline {12x - 2\not{y} = - 16} } \\
  {\underline {22x{\text{ }} = - 4} }
\end{array}\]
$ \Rightarrow 22x = - 4$
Then, divide both sides by 22 and we get
\[ \Rightarrow x = \dfrac{{ - 4}}{{22}}\]
Simplify the above value:
\[ \Rightarrow x = \dfrac{{ - 2}}{{11}} or \approx - 0.18\]
Substitute the value for $x$ into Equation 1. Solve for $y$
First we take Equation 1,
\[ \Rightarrow 5x + y = 6\]
Put, \[x = \dfrac{{ - 2}}{{11}}\] in the above equation and we get
\[ \Rightarrow 5\left( {\dfrac{{ - 2}}{{11}}} \right) + y = 6\]
Expand the above equation:
\[ \Rightarrow \dfrac{{ - 10}}{{11}} + y = 6\]
Next, add \[\dfrac{{10}}{{11}}\] to both sides in the above equation and we get
\[ \Rightarrow \dfrac{{10}}{{11}} + \dfrac{{ - 10}}{{11}} + y = 6 + \dfrac{{10}}{{11}}\]
\[ \Rightarrow y = 6 + \dfrac{{10}}{{11}}\]
Multiply 6 by \[\dfrac{{11}}{{11}}\] to get an equivalent fraction with 11 as the denominator.
\[ \Rightarrow y = 6 \times \dfrac{{11}}{{11}} + \dfrac{{10}}{{11}}\]
\[ \Rightarrow y = \dfrac{{66}}{{11}} + \dfrac{{10}}{{11}}\]
Adding the numerators and we get
\[\;\;\]\[ \Rightarrow y = \dfrac{{76}}{{11}} or \approx 6.9.\]

The solution to the linear system is \[\left( {\dfrac{{ - 2}}{{11}},\dfrac{{76}}{{11}}} \right)\].

Note: The elimination method allows us to cancel out one of the variables, making it an easy algebra equation to solve for the remaining variable. You'll see.
Let's eliminate the y variable. In order to do that, we need to simply multiply one of the equations by a negative one, let's do it to equation one.
Equation 1: \[ - y = + 5x - 6\]
Equation 2: \[y = 6x + 8\]
Now, we add the equations together, getting one combined equation:
\[\begin{array}{*{20}{c}}
  { - \not{y} = + 5x - 6} \\
  {\underline {\not{y} = 6x + 8} } \\
  {\underline {{\text{ }}0 = 11x + 2} }
\end{array}\]
\[ \Rightarrow 0 = 11x + 2\]
\[ \Rightarrow x = \dfrac{{ - 2}}{{11}} or \approx - 0.18\]
Now, we plug in the $x$ value into one of the original equations to solve for $y$.
Equation1:
\[ \Rightarrow y = - 5\left( {\dfrac{{ - 2}}{{11}}} \right) + 6\]
\[ \Rightarrow y = \dfrac{{76}}{{11}} or \approx 6.9.\]
The solution to the linear system is \[\left( {\dfrac{{ - 2}}{{11}},\dfrac{{76}}{{11}}} \right)\].
The approximate solution is \[( - 0.18,6.9)\].