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Section 2.3 Solutions of Systems of Equations: Substitution

Insight 2.3.1. Substitution Method.

To solve a system of equations via substitution, solve for one of the variables in one of the equations and substitute the result in the other equations. Repeat until only one variable is left unknown.
Use the last equation to find the value of the last variable and then substitute that value back to find the values of the other variables.
Solve the following system of equations,
\begin{equation*} \begin{array}{rcrcrcr} 3 x_1 \amp + \amp 2x_2 \amp \amp \amp = \amp 0\\ x_1 \amp + \amp x_2 \amp + \amp x_3 \amp = \amp -1\\ x_1 \amp \amp \amp - \amp 2x_3 \amp = \amp 2 \end{array} \end{equation*}
We first solve for \(x_1\) using the third equation,
\begin{equation*} x_1 = 2 + 2x_3. \end{equation*}
Next, solve for \(x_2\) in the first equations and substitute the value for \(x_1\text{,}\)
\begin{equation*} 3x_1 + 2x_2 = 0 \hspace{0.2cm} \Rightarrow \hspace{0.2cm} x_2 = - \frac{3}{2}x_1 = -3 - 3 x_3. \end{equation*}
Next, we substitute the expressions for \(x_1\) and \(x_2\) into the second equation and solve for \(x_3\)
\begin{equation*} \begin{array}{rcrcrcr} x_1 \amp + \amp x_2 \amp + \amp x_3 \amp = \amp -1\\ 2 + 2x_3 \amp + \amp ( -3 - 3 x_3 )\amp + \amp x_3\amp = \amp -1\\ \amp \amp \amp \amp 0 \amp = \amp 0\\ \end{array} \end{equation*}
Since the equation is satisfied independently of the value of \(x_3\) we say that \(x_3\) is a free variable. Denote that variable by \(s\text{,}\) then the solution to the system of equations is,
\begin{equation*} \begin{array}{ccr} x_1 \amp= \amp 2 + 2 s\\ x_2 \amp = \amp -3 - 3s\\ x_3 \amp = \amp s. \end{array} \end{equation*}
Use the substitution method to solve the system
\begin{equation*} \begin{array}{l} -x+y = -2, \\ 4x-3y = 6. \\ \end{array} \end{equation*}
Your answer is
\(x=\)
\(y=\)
Answer 1.
\(0\)
Answer 2.
\(-2\)
For the following system to be consistent,
\begin{equation*} \begin{array}{rcrcrcr} 6x \amp -\amp 3 y\amp + \amp 5 z\amp = \amp 3\\ -20x \amp +\amp 17 y\amp +\amp k\, z\amp = \amp 0\\ -7x \amp +\amp 7 y\amp +\amp 6 z\amp = \amp 3 \end{array}, \end{equation*}
we must have, \(k\neq\) .
Answer.
\(7\)
Solution.
Solve for \(x\) using the first equation and substitute in the last two equations,
\begin{equation*} \begin{array}{ccc} x \amp = \amp\displaystyle \frac{3 + 3 y -5 z }{6}\\ \amp = \amp {0.5} +{0.5} y - {0.833333} z\\ \end{array} \end{equation*}
The second and third equation become,
\begin{equation*} \begin{array}{ccc} -20\left({0.5} +{0.5} y - {0.833333} z\right) + 17 y + k\, z\amp = \amp 0\\ {7} y + (k + {16.6667})\, z\amp = \amp {10}\\ \end{array} \end{equation*}
\begin{equation*} \begin{array}{ccc} -7\left({0.5} +{0.5} y - {0.833333} z\right) + 7 y + 6 z\amp = \amp 3\\ {3.5} y + {11.8333} z\amp = \amp {6.5}\\ \end{array} \end{equation*}
Solve for \(y\) using the last equation,
\begin{equation*} \begin{array}{ccc} y \amp =\amp {1.85714}- {3.38095} z \end{array} \end{equation*}
Substituting this into the second equation,
\begin{equation*} \begin{array}{ccc} {7} y + (k + {16.6667})\, z\amp = \amp {10}\\ {7} \left( {1.85714} - {3.38095} z \right) + (k + {16.6667})\, z\amp = \amp {10}\\ (k - {7})\, z \amp = \amp {-3}\\ z \amp = \amp\displaystyle \frac{{-3}}{k - {7}}. \end{array} \end{equation*}
For the system to have a solution, \(k \ne {7}\) (division by zero).
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