Question

How do you solve the system of linear equations \[ - 8x + 3y = 7\] and \[13x - 3y = - 1\] ?

Answer 1

Hint: We use the substitution method to solve two linear equations given in the question. We find the value of 3y from the first equation in terms of x and substitute in the second equation which becomes an equation in x entirely. Solve for the value of x and substitute back the value of y to obtain the value of y.

Complete step-by-step answer:
We have two linear equations \[ - 8x + 3y = 7\] and \[13x - 3y = - 1\]
Let us solve the first equation to obtain a value of 3y in terms of x.
We have \[ - 8x + 3y = 7\]
Shift the value of x to RHS of the equation.
\[ \Rightarrow 3y = 7 + 8x\] … (1)
Now we substitute the value of \[3y = 8x + 7\]from equation (1) in the second linear equation.
Substitute \[3y = 8x + 7\]in \[13x - 3y = - 1\]
\[ \Rightarrow 13x - (8x + 7) = - 1\]
Open the bracket in LHS of the equation
\[ \Rightarrow 13x - 8x - 7 = - 1\]
Add like terms in numerator of LHS of the equation
\[ \Rightarrow 5x - 7 = - 1\]
Shift the constant values to one side of the equation.
\[ \Rightarrow 5x = - 1 + 7\]
\[ \Rightarrow 5x = 6\]
Divide both sides by 5
\[ \Rightarrow \dfrac{{5x}}{5} = \dfrac{6}{5}\]
Cancel same terms from numerator and denominator.
\[ \Rightarrow x = \dfrac{6}{5}\]
Now substitute the value of \[x = \dfrac{6}{5}\]in equation (1) to get the value of y
\[ \Rightarrow 3y = 7 + 8 \times \left( {\dfrac{6}{5}} \right)\]
Open the bracket in RHS of the equation.
\[ \Rightarrow 3y = 7 + \dfrac{{48}}{5}\]
Take LCM in RHS of the equation
\[ \Rightarrow 3y = \dfrac{{35 + 48}}{5}\]
\[ \Rightarrow 3y = \dfrac{{83}}{5}\]
Cross multiply 3 from numerator of LHS to RHS
\[ \Rightarrow y = \dfrac{{83}}{{15}}\]

\[\therefore \]Solution of the system of linear equations is \[x = \dfrac{6}{5};y = \dfrac{{83}}{{15}}\]

Note:
We can use combination method to solve the system of linear equations as there are exact same values in both linear equations with opposite signs.
We have equations \[ - 8x + 3y = 7\] and \[13x - 3y = - 1\]
Multiply second equation by -1
Then equations become \[ - 8x + 3y = 7\] and \[ - 13x + 3y = 1\]
Now we subtract second equation from first equation
\[\begin{gathered}
   - 8x + 3y = 7 \\
  \underline { - 13x + 3y = 1} \\
  5x = 6 \\
\end{gathered} \]
Cross multiply 5 from LHS to RHS
\[ \Rightarrow x = \dfrac{6}{5}\]
Now substitute the value of \[x = \dfrac{6}{5}\]in equation (1) to get the value of y
\[ \Rightarrow 3y = 7 + 8 \times \left( {\dfrac{6}{5}} \right)\]
Open the bracket in RHS of the equation.
\[ \Rightarrow 3y = 7 + \dfrac{{48}}{5}\]
Take LCM in RHS of the equation
\[ \Rightarrow 3y = \dfrac{{35 + 48}}{5}\]
\[ \Rightarrow 3y = \dfrac{{83}}{5}\]
Cross multiply 3 from numerator of LHS to RHS
\[ \Rightarrow y = \dfrac{{83}}{{15}}\]
\[\therefore \]Solution of the system of linear equations is \[x = \dfrac{6}{5};y = \dfrac{{83}}{{15}}\]