# What is the Elimination Method?

To solve a system of equations using elimination or addition, begin by rewriting both equations in standard form Ax+By=C. Check to see if the coefficients of one pair of like variables add to zero. If not, multiply one or both of the equations by a non-zero number to make one set of like variables add to zero. Add the two equations to solve for one of the variables. Substitute this value into one of the original equations to solve for the other variable. Check your work by substituting into the other equation.

Hi, and welcome to this video on solving systems of equations using the elimination method.

When we solve systems of equations, we are looking for the point where two or more functions intersect. There are four different ways to approach these kinds of problems: graphing, substitution, elimination, and using a matrix. Today we are going to focus on how to use the elimination method to find the point where two lines cross.

The elimination method involves exactly what it sounds like, eliminating one of our terms to solve for the other. If we are using standard linear equations, in the form \(ax+by=c\), we want to find a number or numbers to multiply one or both of our equations by to get rid of either the x-term or the y-term by adding or subtracting the two equations.

Let’s look at an example so I can show you what I mean.

Find the point of intersection for these two lines: \(2x+3y=14\) and \(x–4y=18\).

First, we want to see if there is any way to eliminate one of our terms by simply adding or subtracting our equations. That is not possible for this example, so we are going to need to multiply one of our equations by some number so that we can add or subtract them and get rid of either the x-term or the y-term.

For this problem, we can multiply the second equation by 2 and then subtract to eliminate our x-terms.

So we’re gonna multiply this equation by 2.

\(2(x-4y=18)\)

Which will give us: \(2x-8y=36\). And we’re gonna subtract \(2x+3y=14\).

\(2x-2x=0\), \(-8y-3y=-11y\), and \(36-14=22\).

\(-11y=22\)

If we divide by -11 on both sides, we get that \(y=-2\).

Solving this new equation gives us that \(y=-2\). And to find our x-term, we choose either equation, and plug in this y-value to solve for *x*.

I’m going to use the equation \(x–4y=18\).

\(x-4(-2)=18\)

\(x+8=18\)

Subtract 8 from both sides.

\(x=10\)

Our x-value for this system is 10. Since we are solving for the point of intersection between these two functions, we want to write our final answer as a coordinate point. Our point of intersection for this example is \((10, -2)\).

Let’s look at another example.

\(3x+4y=6\)

\(x-y=2\)

For this example, we can multiply our bottom equation by either 3 or 4 to eliminate one of our terms. I am going to multiply by 4.

\(4(x-y=2)\)

\(4x-4y=8\)

Now, since I want my y-terms to cancel out, I am going to add my two equations and solve for x.

(So I’m just going to rewrite our equation right here, \(3x+4y=6\))

And then we’ll add the two equations.

\(3x+4x=7x\), \(4y-4y=0\), and \(6+8=14\).

\(7x=14\)

If we divide by 7 on both sides, we get that \(x=2\).

Now that we know \(x=2\), we can choose either equation to plug this value into and solve for *y*. I am going to use \(x–y=2\).

\(2-y=2\)

Subtract 2 from both sides.

\(-y=0\)

And divide by -1 on both sides.

\(y=0\)

Now all we have to do is write our answer in the form of a coordinate point, \((2, 0)\).

I want to try one more example before we go.

\(2x-7y=19\)

\(3x-11y=4\)

This example is a little bit different from our other ones because there isn’t one number we can easily multiply an equation by to eliminate a term, not without using fractions. So for this example, we are going to multiply both equations by a number in order to cancel something out.

If we multiply our first equation by 3 and our second equation by 2, we will get a 6*x* term in both equations that we can then cancel out. Let’s try it.

\(3(2x-7y=19)\;\;\;\;\;\;\;\;\;\;2(3x-11y=4)\)

\(6x-21y=27\;\;\;\;\;\;\;\;\;\;6x-22y=8\)

Now we subtract our two equations so that we have \(6x-6x=0\), and then \(-21y-(-22y)=y\) (it’s like adding that 22*y*). Now we do \(57-8=49\).

\(y=49\)

This gives us that \(y=49\). Now we’re going to use this information to plug *y* into either equation and solve for *x*. I’m going to use the equation \(2x–7y=19\).

\(2x-7(49)=19\)

\(2x-343=19\)

\(2x=362\)

\(x=181\)

So \(x=181\). Our final answer then is the point, \((181, 49)\).

Remember, when using the elimination method to solve systems of equations, the whole point is to try and get rid of one of your terms by multiplying one of the equations by a constant and then adding or subtracting your equations.

I hope this video was helpful. Thanks for watching and happy studying!

## Practice Questions

**Question #1:**

Use the elimination method to solve the system of equations.

\(3x-y=23\)

\(2x+y=13\)

**Answer:**

If the equations are added together, the y-terms will be eliminated. Adding the two equations results in \(5x=25\). Now, solve for \(x\) by dividing both sides by 5: \(x=5\). Now plug in 5 for \(x\) into one of the original equations. If the first equation is used, \(3x-y=12\) becomes \(3(5)-y=12\), or \(15-y=12\). Subtract 15 from both sides, and then divide both sides by -1: y=3. We know that \(x=5\) and \(y=3\). This means that the lines will intersect at the point \((5,3)\).

**Question #2:**

Determine the solution to the system of equations using the elimination method.

\(3x+y=10\)

\(-4x-2y=2\)

**Answer:**

The simplest way to use the elimination method to solve this system of equations is to multiply the first equation by 2 so that the \(y\)-values cancel out.

\(2(3x+y=10)\)

\(6x+2y=20\)

Notice how \(+2y\) and \(-2y\) will cancel out if the two equations are added.

\(6x\) | \(+2y\) | \(=\) | \(20\) | |

\(+\) | \(-4x\) | \(-2y\) | \(=\) | \(2\) |

\(2x\) | \(=\) | \(22\) |

Solve for \(x\) by dividing both sides by 2.

\(\frac{2x}{2}=\frac{22}{2}\)

\(x=11\)

Now plug 11 in for \(x\) into either of the original equations. If we use the first original equation, \(3x+y=10\) becomes \(3(11)+y=10\).

\(33+y=10\)

Subtract 33 from both sides.

\(33-33+y=10-33\)

\(y=-23\)

We know that \(x=11\) and \(y=-23\). The lines will intersect at the point \((11,-23)\).

**Question #3:**

Solve the system of equations using the elimination method.

\(3x+2y=7\)

\(5x-3y=37\)

**Answer:**

One approach is to multiply the first equation by 3 and the second equation by 2. This will create a situation where the \(y\)-variables cancel out when the equations are added. (Note that another approach would be to multiply the first equation by 5 and the second equation by 3. This would result in the \(x\)-variables canceling. Either approach will lead to the correct answer!) Let’s try the first approach:

\(3x+2y=7\) becomes \(9x+6y=21\).

\(5x-3y=37\) becomes \(10x-6y=74\).

Now, add the two new equations.

\(9x\) | \(+6y\) | \(=\) | \(21\) | |

\(+\) | \(10x\) | \(-6y\) | \(=\) | \(74\) |

\(19x\) | \(=\) | \(95\) |

Solve for \(x\) by dividing both sides by 19.

\(\frac{19x}{19}=\frac{95}{19}\)

\(x=5\)

Now plug in 5 for \(x\) in either equation. If we use the original first equation \(3x+2y=7\) becomes \(3(5)+2y=7\).

\(15+2y=7\)

Subtract 15 from both sides, and then divide both sides by 2.

\(15-15+2y=7-15\)

\(2y=-8\)

\(\frac{2y}{2}=\frac{-8}{2}\)

\(y=-4\)

The value of \(x\) is 5, and the value of \(y\) is -4. This means the lines will intersect at the point \((5,-4)\).

**Question #4:**

Willow spends $188 at the pet store. She purchases 7 small dog toys and 11 large dog toys. Sybil spends $236 at the same pet store. She purchases 13 small toys and 11 large toys. Determine the price of a small toy and a large toy.

Small toys cost $10 and large toys cost $12.

Small toys cost $8 and large toys cost $10.

Small toys cost $6 and large toys cost $14.

Small toys cost $8 and large toys cost $12.

**Answer:**

If we let \(x=\) the cost of small dog toys and \(y=\) the cost of large dog toys, then we can set up the following equation to describe Willow’s spending: \(7x+11y=188\).

Similarly, we can set up an equation to describe Sybil’s spending: \(13x+11y=236\).

The two equations can be set up as a system of equations, and then the elimination method can be used to determine the value of \(x\) and \(y\). If the second equation is subtracted from the first equation, the \(y\)-values will cancel out.

\(7x\) | \(+11y\) | \(=\) | \(188\) | |

\(–\) | \((13x\) | \(+11y\) | \(=\) | \(236)\) |

\(-6x\) | \(=\) | \(-48\) |

Divide both sides by -6.

\(\frac{-6x}{-6}=\frac{-48}{-6}\)

\(x=8\)

Now plug \(x=8\) into one of the original equations. If the first equation is used, \(7x+11y=188\) becomes \(7(8)+11y=188\).

\(56+11y=188\)

Subtract 56 from both sides, and then divide both sides by 11.

\(56-56+11y=188-56\)

\(11y=132\)

\(\frac{11y}{11}=\frac{132}{11}\)

\(y=12\)

Since \(x=\) the cost of small dog toys and \(y=\) the cost of large dog toys, this means that the small dog toys cost $8 and the large dog toys cost $12.

**Question #5:**

A track and field athlete is practicing for an upcoming track meet. On Monday the athlete practices the long jump 6 times and the triple jump 9 times. This takes the athlete 39 minutes. On Tuesday the athlete practices the long jump 4 times and the triple jump 4 times. This takes the athlete 20 minutes. How long does it take the athlete to practice each event one time?

It takes the athlete 1 minute to practice the long jump and 3 minutes to practice the triple jump.

It takes the athlete 2 minutes to practice the long jump and 3 minutes to practice the triple jump.

It takes the athlete 4 minutes to practice the long jump and 4 minutes to practice the triple jump.

It takes the athlete 2 minutes to practice the long jump and 3 minutes to practice the triple jump.

**Answer:**

We will let \(x=\) the number of minutes it takes to practice the long jump and \(y=\) the number of minutes it takes to practice the triple jump.

Equation #1: *On Monday the athlete practices the long jump 6 times and the triple jump 9 times. This takes the athlete 39 minutes*: \(6x+9y=39\).

Equation #2: *On Tuesday the athlete practices the long jump 4 times and the triple jump 4 times. This takes the athlete 20 minutes*: \(4x+4y=20\).

If we want the \(x\)-terms to cancel, then we can multiply the first equation by 2, and multiply the second equation by 3.

Equation #1: \(2(6x+9y=39)\) becomes \(12x+18y=78\)

Equation #2: \(3(4x+4y=20)\) becomes \(12x+12y=60\)

\(12x\) | \(+18y\) | \(=\) | \(78\) | |

\(–\) | \((12x\) | \(+12y\) | \(=\) | \(60)\) |

\(6y\) | \(=\) | \(18\) |

Divide both sides by 6.

\(\frac{6y}{6}=\frac{18}{6}\)

\(y=3\)

Now plug \(y=3\) into one of the original equations. If we use the first equation, \(6x+9y=39\) becomes \(6x+9(3)=39\).

\(6x+27=39\)

Subtract 27 from both sides and then divide both sides by 6.

\(6x+27-27=39-27\)

\(6x=12\)

\(\frac{6x}{6}=\frac{12}{6}\)

\(x=2\)

\(x=2\) which means the athlete spends 2 minutes on the long jump. \(y=3\) which means the athlete spends 3 minutes on the triple jump.