Method of Elimination


Solving a System of Equations Using Elimination

Another algebraic method that can be used to solve a system of equations is called elimination.  

Using the elimination method results in one of the variables being "eliminated" or cancelled out so that only one variable remains in an equation.

Before we look at using this method, it is important to remember the following two key ideas:

  • When we add or subtract the same value to both sides of an equation of a line, the equation still represents the same line.
    • For example: \(3x+2y=5\) is equivalent to the equation \(3x+2y-4=5-4\)
  • When we multiply or divide every term in an equation of a line by the same value, the equation still represents the same line.
    • For example: \(3x+2y=5\) is equivalent to the equation \(12x+8y=20\)

In the following examples, we will use the method of elimination to solve a system of two equations with two unknowns.


Slide Notes

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Example 4

Solve the system of equations using elimination:

\(\begin{align*} 3x+4y&=5 \tag{1}\\ 3x-2y&=11 \tag{2} \end{align*}\)

 

Example 4 Continued

Solve the system of equations using elimination:

\(\begin{align*} 3x+4y&=5 \tag{1}\\ 3x-2y&=11 \tag{2} \end{align*}\)

Solution

Step 3: Solve for the variable

\(6y\)

\(\;=-6\)

 

 

Example 4 Continued

Solve the system of equations using elimination:

\(\begin{align*} 3x+4y&=5 \tag{1}\\ 3x-2y&=11 \tag{2} \end{align*}\)

Solution

Step 5: Check the solution \((x,y)=(3,-1)\) in the equation not used in Step 4

 

Example 4 Continued

Solve the system of equations using elimination:

\(\begin{align*} 3x+4y&=5 \tag{1}\\ 3x-2y&=11 \tag{2} \end{align*}\)

Solution

We can see the solution as the point of intersection on the graph of the two lines.

The lines 3x+4y=5 and 3x minus 2y=11 are plotted on a graph. They intersect at the point (3,negative 1)

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Slide Notes

Glossary

All Slides

 

Example 5

Solve the system of equations using elimination:

\(\begin{align*} 3x-8y&=-30 \tag{1}\\ 7x+4y&=-2 \tag{2} \end{align*}\)

 

Example 5 Continued

Solve the system of equations using elimination:

\(\begin{align*} 3x-8y&=-30 \tag{1}\\ 7x+4y&=-2 \tag{2} \end{align*}\)

Solution

Step 2: Eliminate one variable by adding or subtracting

Equation \((1)\) \(+\) Equation \((3)\): 

\(\begin{align*} \class{timed add1-hl2 remove2-hl2}{3x}\class{timed add2-hl2 remove3-hl2}{-8y}&=\class{timed add3-hl2 remove4-hl2}{-30} \tag{1}\\ \class{timed add1-hl2 remove2-hl2}{14x}\class{timed add2-hl2 remove3-hl2}{+8y}&=\class{timed add3-hl2 remove4-hl2}{-4} \tag{3} \\ \hline \class{timed in5}{\class{timed add1-hl2 remove2-hl2}{17x}\class{timed add2-hl2 remove3-hl2}{+0y}}&\;\class{timed in5}{=\class{timed add3-hl2 remove4-hl2}{-34}} \tag{1)+(3}\\ \class{timed in6}{17x}&\;\class{timed in6}{=-34} \end{align*}\)

 

Example 5 Continued

Solve the system of equations using elimination:

\(\begin{align*} 3x-8y&=-30 \tag{1}\\ 7x+4y&=-2 \tag{2} \end{align*}\)

Solution

Step 3: Solve for the variable

 

 

Example 5 Continued

Solve the system of equations using elimination:

\(\begin{align*} 3x-8y&=-30 \tag{1}\\ 7x+4y&=-2 \tag{2} \end{align*}\)

Solution

Step 5: Check the solution \((x,y)=(-2,3)\) in the equation not used in Step 4

 

Example 5 Continued

Solve the system of equations using elimination:

\(\begin{align*} 3x-8y&=-30 \tag{1}\\ 7x+4y&=-2 \tag{2} \end{align*}\)

Solution

We can see the solution as the point of intersection on the graph of the two lines.

The lines 3x+30=8y and 3x minus 2y=11 are plotted on a graph. They intersect at the point (negative 2, 3)

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Summary of Elimination Steps

  • Step 1: If required, multiply one or both equations by a factor to ensure that both equations have either an \(x\) or \(y\) coefficient that is either equal or opposite to each other.
  • Step 2: Eliminate one variable by adding or subtracting
    • Remember if the coefficients from Step 1 have the same sign we will subtract, and if they have opposite signs, we will add.
  • Step 3: Solve for the variable
  • Step 4: Substitute the value to solve for the second variable
    • In this step it does not matter which equation you choose to work with.
  • Step 5: Check the solution in the equation not used in Step 4

Check Your Understanding 3



Example 6

Solve the system of equations using elimination:

\(\begin{align*} 2x-100&=-9y \tag{1}\\ -4y-35&=-15x \tag{2} \end{align*}\)

Solution

Before we start with solving using elimination, we will first rearrange the equations so they are both in the form \(Ax+By=C\). This will allow us to easily compare the coefficients of \(x\) and \(y\) between the two equations.

\(\begin{align*} 2x+9y&=100 \tag{1}\\ 15x-4y&=35 \tag{2} \end{align*}\)

Step 1: 

In this example, we will need to multiply by a factor to get the same or opposite coefficient for \(x\) or \(y\). If we wish to not use factors that are fractions we will need to multiply both equations by their own factor. 
In our example, we could:

  • choose to multiply so both equations have a coefficient of \(30\) (the lowest common multiple of \(2\) and \(15\)) in front of \(x\) \(\rightarrow\) Equation \((1) \times 15\) and Equation \((2) \times 2\) 

or we could

  • choose to multiply so both equations have a coefficient of \(36\) but with opposite signs (the lowest common multiple of \(9\) and \(4\)) in front of \(y\rightarrow\) Equation \((1)\times 4\) and Equation \((2)\times9\)

For our solution, we will choose the second option. This will give us two equations with opposite coefficients in front of their \(y\) variable.

  • Equation \((1) \times 4\rightarrow \) Equation \((3)\)
  • Equation \((2) \times 9\rightarrow\) Equation \((4)\)

\(\begin{align*} 8x+36y&=400 \tag{3}\\ 135x-36y&=315 \tag{4} \end{align*}\)

Steps 2 and 3:

We now have opposite coefficients for \(y\) so we will add the equations together to eliminate \(y\) and then we can solve for \(x\).

  • Equation \((3)\) \(+\) Equation \((4)\):

\(\begin{align*} 8x+36y&=400 \tag{3}\\ 135x-36y&=315 \tag{4} \\ \hline 143x\;\;\;\;\;\;\;\;\;\;&=715 \tag{3)+(4}\\ x&=5 \end{align*}\)

Step 4:

Substitute \(x=5\) into Equation \((1)\).

\[\begin{align*} 2x+9y&=100\\ 2(5)+9y&=100\\ 9y&=100-10\\ 9y&=90\\ y&=10 \end{align*}\]

Step 5:

Verify the solution \((x,y)=(5,10)\) in Equation \((2)\).

\[\begin{align*} \text{LS}&=15x-4y\\ &=15(5)-4(10)\\ &=75-40\\ &=35 \end{align*}\]

\(\text{RS}=35\)

\[\text{LS}=\text{RS}\]

Therefore, \(\text{LS}=\text{RS}\) and \((5,10)\) is the solution to the system of equations.


Check Your Understanding 4


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