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Section 2.4 Solutions of Systems of Equations: Elimination

Insight 2.4.1. Elimination Method.

To solve a system of equations via elimination, we add/subtract multiples of an equation to another equation with the objective of eliminating one of variables. This is repeating until there is only one variable left.
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.
Use elimination to solve the following system of equations
\begin{equation*} \begin{array}{ccccc} 3x_1 \amp - \amp 2x_2 \amp = \amp -1\\ x_1 \amp + \amp 4x_2 \amp = \amp 9\\ \end{array} \end{equation*}
Subtract 3 times equation 2 from equation 1,
\begin{equation*} -14 x_2 = -28, \end{equation*}
Solve for \(x_2\text{,}\)
\begin{equation*} x_2 = \frac{-28}{-14} = 2. \end{equation*}
We could either use the value of \(x_2 \) to find \(x_1\) or eliminate \(x_2\) from an equation.
To eliminate \(x_2\) add 2 times equation 2 to equation 1,
\begin{equation*} 7 x_1 = 7 \end{equation*}
Then the solution of the system is \(x_1 =1\) and \(x_2 = 2\text{.}\)
Solve the system using elimination
\begin{equation*} \left\{ \begin{array}{r@{}r@{}r@{}r} 4 x \amp - 7 y \amp = \amp 5 \cr 7 x \amp +8 y \amp = \amp 110 \end{array}\right. \end{equation*}
\(x=\)
\(y=\)
Answer 1.
\(10\)
Answer 2.
\(5\)
Solve the system using the substitution or elimination method
\begin{equation*} \left\{ \begin{array}{l} 5x+2y = 11 \\ 7x+3y = 15 \end{array}\right. \end{equation*}
How many solutions are there to this system?
  • None
  • Exactly 1
  • Exactly 2
  • Exactly 3
  • Infinitely many
  • None of the above
If there is one solution, give its coordinates in the answer spaces below.
If there are infinitely many solutions, enter x in the answer blank for \(x\) and enter a formula for \(y\) in terms of \(x\) in the answer blank for \(y\text{.}\)
If there are no solutions, leave the answer blanks for \(x\) and \(y\) empty.
\(x =\)
\(y =\)
Answer 1.
B
Answer 2.
\(3\)
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