The Elimination Method (Reduction Method)

The elimination method, also called the reduction method, is a method for algebraically solving a system of equations when the two equations have the form |ax + by = c.|

The elimination method can be performed with any variable, dependent or independent.

The elimination can be done either by adding, or subtracting the equations term by term. The coefficients should be added when a variable’s coefficients are opposite and subtracted when the coefficients of one of the variables are the same.

To algebraically solve of a system of equations using the reduction method, follow these steps.

1. In a word problem, define the variables and rewrite it as a system of equations.

2. If necessary, form an equivalent system of equations where the coefficients of one variable are opposite (or equal) to each other.

3. Eliminate by adding (or subtracting) the two equations term-by-term to form a first-degree equation with one variable.

4. Solve the equation.

5. Substitute the value found in step 4 into one of the two initial equations to find the value of the second variable.

6. Verify the answer by substituting the values obtained for the variables into each of the system’s equations.

Elimination method — Subtracting equations

Consider the following system of linear equations. ||\begin{cases}15y = 9x + 6 \\ -5y = 2x - 10\end{cases}||

1. For a word problem, define the variables and translate the situation into a system of equations
   Since the system is already provided, this step is not necessary.

2. If necessary, form an equivalent system of equations where the coefficients of one variable are equal to each other
   Choose the variable |y:| find the operation that makes the coefficients of |y| equal in both equations. In other words, multiply all the terms in one of the two equations by the appropriate constant. Here, multiply the second equation by the constant |(-3).| ||\begin{align}-3 \times (-5y &= 2x - 10)\\ \Rightarrow 15y &= -6x + 30\end{align}||

3. Eliminate by subtracting term by term the two equations to form a first-degree equation with one variable
   ||\begin{align}15y&=\ \ \ 9x+\ \, 6 \\ -(15y&=-6x+30)\\ \hline \color{red}{0y}&=\ \, 15x-24\end{align}||

4. Solve the equation
   ||\begin{align} 0 &= -15x + 24\\
   0 \color{red}{+ 15x} &= -15x\color{red}{+ 15x}+ 24\\
   15x &= 24\\
   15x\color{red}{\div 15} &= 24\color{red}{\div 15}\\
   x &= {\dfrac{24}{15}} = {\dfrac{8}{5}}\end{align}||

5. Substitute the value found in step 4 into one of the two initial equations to find the value of the second variable
   ||\begin{align} -5y &= 2x - 10\\
   
   -5y &= 2\left({\dfrac{8}{5}}\right) - 10\\
   
   -5y &= {\dfrac{16}{5}} - 10\\
   
   -5y &= {\dfrac{-34}{5}}\\
   
   y &= {\dfrac{34}{25}}\end{align}||

6. Verify the answer by substituting the values obtained for the variables into each of the system’s equations
   ||\begin{align} 15y &= 9x + 6 & -5y &= 2x - 10\\
   15\left({\dfrac{34}{25}}\right)&=9\left({\dfrac{8}{5}}\right)+6 & -5\left({\dfrac{34}{25}}\right)&=2\left({\dfrac{8}{5}}\right)-10\\
   {\dfrac{102}{5}}&=\dfrac{72}{5}+6 & {\dfrac{-34}{5}}&={\dfrac{16}{5}}-10\\
   {\dfrac{102}{5}}&=\dfrac{102}{5} & {\dfrac{-34}{5}}&={\dfrac{-34}{5}}\end{align}||
   Since the values confirm the two equations, the solution is the following coordinate|\left(\dfrac{8}{5},\dfrac{34}{25}\right).|

Elimination method — Adding equations

In the previous example, instead of multiplying the second equation by |(-3)|, it would have been possible to multiply the second equation by |(3)|. In this case, instead of subtracting, add the two equations to perform the elimination. The final solution is the same.
||\begin{align}3\times (-5y &= 2x - 10)\\ \Rightarrow\ -15y &=6x-30\end{align}||
Add the two equations term by term to remove the variable |y.|
||\begin{align}15y&=\ \ 9x+\ \, 6\\ ^{\Large+}\ -15y&=\ \ 6x-30\\ \overline{\phantom{24} \color{red}{0y}} &\overline{\,\, =15x-24}\end{align}||
Solve the single-variable equation obtained by isolating the variable |x.|
||\begin{align} 0 &= 15x - 24\\
0 \color{red}{- 15x} &= 15x\color{red}{- 15x}- 24\\
-15x &= -24\\
15x\color{red}{\div -15} &= 24\color{red}{\div -15}\\
x &= \dfrac{24}{15} = \dfrac{8}{5}\end{align}||
Note that the same result is obtained for both ways of carrying out the elimination. The defining feature of the elimination method is the removal of one of the two variables from the system of equations to obtain a single-variable equation.

Elimination method — Manipulating the two equations

Consider the following system of linear equations. ||\begin{cases} 2x+2y=12 \\ 3x-5y = 26\end{cases}||

1. For a word problem, define the variables and rewrite it as a system of equations
   Since the system is already provided, this step is not necessary.

2. If necessary, form an equivalent system of equations where the coefficients of one variable are equal to each other
   Choose the variable |x:| find the operation that makes the coefficients of |x| equal in both equations. In other words, multiply all the terms of one of the two equations by the appropriate constant. Here, multiply the first equation by |3| and the second by |2|.
   
   First equation
   ||\begin{align}3\times (2x + 2y &= 12)\\
   \Rightarrow\ 6x + 6y &= 36\end{align}||
   
   Second equation
   ||\begin{align}2\times (3x - 5y &= 26)\\
   \Rightarrow\ 6x - 10y &= 52\end{align}||

3. Eliminate by subtracting term by term the two equations to form a first-degree equation with one variable
   ||\begin{align}6x+\phantom{1}6y\ &=\ \ \ 36\\ -(6x-10y\ &=\ \ \ 52)\\ \hline \color{red}{0x}+16y\ &=-16\end{align}||

4. Solve the equation
   ||\begin{align}0 + 16y &= -16\\
   \dfrac{16y}{\color{red}{16}} &= \dfrac{-16}{\color{red}{16}}\\
   y &= -1\end{align}||

5. Substitute the value found in step 4 into one of the two initial equations to find the value of the second variable
   ||\begin{align}6x + 6y &= 36\\
   6x + 6(-1) &= 36\\
   6x - 6 &= 36\\
   6x - 6 \color{red}{+ 6} &= 36 \color{red}{+ 6}\\
   6x &= 42\\
   6x\color{red}{\div 6} &= 42\color{red}{\div 6}\\
   x &= 7\end{align}||

6. Verify the answer by substituting the values obtained for the variables into each of the system’s equations
   ||\begin{align} 2x + 2y &= 12 & 3x - 5y &= 26\\
   2(7)+2(-1)&=12 & 3(7)-5(-1)&=26\\
   14-2&=12 & 21+5&=26\\
   12&=12 & 26&=26\end{align}||

Since the values confirm the two equations, the solution is the point |(7,-1).|