How to Do Elimination in Math – A Complete Guide

How to Do Elimination in Math?

To do elimination in math, it involves manipulating equations or expressions by adding, subtracting, or multiplying them to eliminate specific variables or terms, making the problem simpler and allowing for easier solution finding. It is commonly used in solving systems of linear equations and simplifying algebraic expressions.

Elimination is a fundamental technique in mathematics, particularly in solving systems of linear equations, simplifying expressions, and finding common factors. It’s a powerful tool that allows mathematicians to simplify complex problems and arrive at solutions efficiently. In this comprehensive guide, we’ll delve into the various aspects of elimination, from its basic principles to advanced applications in algebra and beyond.

Understanding the Basics of Elimination

At its core, elimination is a problem-solving strategy that aims to eliminate or remove certain variables or terms from an equation or system of equations. This simplifies the problem, making it easier to find a solution. Elimination is often used in scenarios where two or more equations are involved, such as systems of linear equations.

What Is the Elimination Method?

The elimination method is a process that uses elimination to reduce the simultaneous equations into one equation with a single variable. This leads to the system of linear equations being reduced to a single-variable equation, making it easier for us.

This is one of the most helpful tools when solving systems of linear equations.

\begin{aligned}\begin{matrix}&\underline{\begin{array}{cccc}&{\color{red} \cancel{-40x}} &+ 12 y&=-400\phantom{x}\\+&{\color{red} \cancel{40x}}&+ 2y&=-300\phantom{1}\end{array}}\\ &\begin{array}{cccc}\phantom{+xx} &\phantom{7xxx}&14y&=-700\\&&y&=\phantom{}-50\end{array}\end{matrix}\end{aligned}

Take a look at the equations shown above. By adding the equations, we’ve managed to eliminate $x$ and leave a simpler linear equation, $14y = -700$. From this, it will be easier for us to find the value of $y$ and eventually find the value of $x$. This example shows how easy it is for us to solve a system of equations by manipulating the equations.

The elimination method is possible all thanks to the following algebraic properties:

• Multiplication Properties
• Addition and Subtraction Properties

In the next section, we’ll show you how these properties are applied. We’ll also break down the process of solving a system of equations by using the elimination method.

Solving Systems of Linear Equations

One of the most common applications of elimination is in solving systems of linear equations. A system of linear equations consists of two or more linear equations with the same variables. The goal is to find values for the variables that satisfy all the equations simultaneously.

Step-by-Step Process for Using the Elimination Method

Write the Equations

Begin by writing down all the equations in the system. Ensure that the variables are aligned vertically so that like terms can be easily identified.

Choose a Variable to Eliminate

Select one of the variables to eliminate. The goal is to make the coefficients of that variable in both equations equal. This is typically achieved by multiplying one or both equations by suitable constants.

Add or Subtract Equations

Add or subtract the equations in a way that eliminates the chosen variable. This means that the terms with that variable will cancel each other out, leaving an equation with one less variable.

Solve for the Remaining Variable

After eliminating one variable, you will have an equation with a single variable to solve. Use algebraic techniques to find its value.

Back-Substitute

Once you’ve found the value of one variable, substitute it back into one of the original equations to solve for the remaining variable.

Check the Solution

Verify that the values you found for the variables satisfy all the original equations in the system. If they do, you have successfully solved the system.

Advanced Techniques and Tips

To become proficient in elimination, consider the following advanced techniques and tips:

Clear Fractions

When dealing with fractions, it’s often helpful to clear them by multiplying all terms by a common denominator. This simplifies the equations and makes elimination easier.

Look for Patterns

In more complex problems, patterns may emerge that allow you to eliminate variables or terms more efficiently. Recognizing these patterns can save time and effort.

Practice, Practice, Practice

Like any mathematical skill, mastering elimination requires practice. Work through a variety of problems and systems of equations to build your proficiency.

The Elimination Method for Systems of Linear Equations

The elimination method, also known as the addition-subtraction method, is a systematic approach to solving systems of linear equations. It involves manipulating the equations in such a way that when they are added or subtracted, one of the variables is eliminated, leaving a simpler equation with fewer variables to solve.

How To Solve System of Equations by Elimination?

To solve a system of equations, rewrite the equations so that when these two equations are added or subtracted, one or two variables can be eliminated. The goal is to rewrite the equation so that it will be easier for us to eliminate the terms.

These steps will help you rewrite the equations and apply the elimination method:

1. Multiply one or both of the equations by a strategic factor.
   • Focus on making one of the terms be the negative equivalent or be identical to the term found in the remaining equation.
   • Our goal is to eliminate the terms sharing the same variable.

2. Add or subtract the two equations depending on the result from the previous step.
   • If the terms we want to eliminate are negative equivalents of each other, add the two equations.
   • If the terms we want to eliminate are identical, subtract the two equations.
3. Now that we’re working with a linear equation, solve for the remaining variable’s value.
4. Use the known value and substitute it into either of the original equations.
   • This results in another equation with one unknown.
   • Use this equation to solve for the remaining unknown variable.

Why don’t we apply these steps to solve the system of linear equation $ \begin{array}{ccc}x&+\phantom{x}y&=5\\-4x&+3y&= -13 \end{array} $?

We’ll highlight the steps applied to help you understand the process:

1. Multiply both sides of the first equation by $4$ so that we end with $4x$.

\begin{aligned}\begin{array}{ccc}{\color{Teal}4}x&+{\color{Teal}4}y&={\color{Teal}4}(5)\\-4x&+3y&= -13 \\&\downarrow\phantom{x}\\4x&+ 4y&= 20\\ -4x&+3y&= -13\end{array} \end{aligned}

We want $4x$ on the first equation so that we can eliminate $x$ in this equation. We can also eliminate $y$ first by multiplying the first equation’s sides by $3$. That’s for you to work on your own, but for now, let’s continue by eliminating $x$.

2. Since we’re working with $4x$ and $-4x$, add the equations to eliminate $x$ and have one equation in terms of $y$.

\begin{aligned}\begin{matrix}&\underline{\begin{array}{cccc}\phantom{+xxx}\bcancel{\color{Teal}4x}&+4y &=\phantom{+}20\\+\phantom{xx}\bcancel{\color{Teal}-4x} &+ 3y&= -13\end{array}}\\ &\begin{array}{cccc}\phantom{+} & \phantom{xxxx}&7y&=\phantom{+}7\end{array}\end{matrix} \end{aligned}

3. Solve for $y$ from the resulting equation.

\begin{aligned}7y &= 7\\y &= 1\end{aligned}

4. Substitute $y =1$ into either of the equations from $\begin{array}{ccc}x&+\phantom{x}y&=5\\-4x&+3y&= -13 \end{array} $. Use the resulting equation to solve for $x$.

\begin{aligned}x + y&= 5\\ x+ {\color{Teal} 1} &= 5\\x& =4\end{aligned}

This means that the given system of linear equations is true when $x = 4$ and $y = 1$. We can also write its solution as $(4, 5)$. To double-check the solution, you can substitute these values into the remaining equation.

\begin{aligned}-4x + 3y&= -13\\-4(4) + 3(1)&= -13\\-13&= -13 \checkmark\end{aligned}

Since the equation holds true when $x = 4$ and $y =1$, this further confirms that the solution to the system of equation is indeed $(4, 5)$. When working on a system of linear equations, apply a similar process as we have done in this example. The level of difficulty may change but the fundamental concepts needed to use the elimination method remain constant.

In the next section, we’ll cover more examples to help you master the elimination method. We’ll also include word problems involving systems of linear equations to make you appreciate this technique more.

Applications of Elimination Method Beyond Linear Equations

Elimination is not limited to linear equations; it can be applied to a wide range of mathematical problems. In algebra, elimination is used to simplify expressions by removing common factors or terms. In calculus, it plays a crucial role in techniques like partial fraction decomposition. Elimination is even used in fields like probability and statistics to calculate probabilities and solve complex statistical problems.

Below we present a few examples involving the elimination method.

Example 1

Use the elimination method to solve the system of equations, $\begin{array}{ccc}4x- 6y&= \phantom{x}26 \,\,(1)\\12x+8y&= -12 \,\,(2)\end{array}$.

Solution

Inspect the two equations to see which equation would be easier for us to manipulate.

\begin{aligned} \begin{array}{ccc}4x- 6y&= \phantom{x}26\,\,(1)\\12x+8y&= -12\,\,(1)\end{array} \end{aligned}

Since $12x$ is a multiple of $4x$, we can multiply $3$ on both sides of Equation (1) so we’ll have $12x$ in the resulting equation. This leads to us having a $12x$ on both equations, making it possible for us to eliminate later.

\begin{aligned} \begin{array}{ccc}{\color{DarkOrange}3}(4x)& -{\color{DarkOrange}3}(6)y&={\color{DarkOrange}3}(26)\\12x&+8y&= -12\,\, \\&\downarrow\phantom{x}\\12x&- 18y&= 78\,\,\,\, \\ 12x&+8y&= -12\end{array}\end{aligned}

Since the two resulting equations have $12x$, subtract the two equations to eliminate $12x$. This leads to a single equation with one variable.

\begin{aligned}\begin{matrix}&\underline{\begin{array}{cccc}\phantom{+xxx}\bcancel{\color{DarkOrange}12x}& -18y &=\phantom{+}78\\-\phantom{xx}\bcancel{\color{DarkOrange}12x} &+ 8y&= -12\end{array}}\\ &\begin{array}{cccc}\phantom{+} & \phantom{xxxx}&-26y&=\phantom{+}90\end{array}\end{matrix}\end{aligned}

Find the value of $y$ using the resulting equation by dividing both sides by $-26$.

\begin{aligned}-26y&= 90\\y&= -\dfrac{90}{26}\\&= -\dfrac{45}{13}\end{aligned}

Now, substitute $y = -\dfrac{45}{13}$ into one of the equations from $\begin{array}{ccc}4x- 6y&= \phantom{x}26 \,\,(1)\\12x+8y&= -12 \,\,(2)\end{array}$.

\begin{aligned}4x – 6y&= 26\\4x -6\left(-\dfrac{45}{13}\right)&= 26\\4x + \dfrac{270}{13}&= 26\end{aligned}

Use the resulting equation to solve $x$ then write down the solution to our system of linear equations.

\begin{aligned}4x + \dfrac{270}{13}&= 26\\52x + 270&= 338\\52x&=68\\x&= \dfrac{17}{13}\end{aligned}

Hence, we have $x = \dfrac{17}{13}$ and $y = -\dfrac{45}{13}$. We can double-check our solution by substituting these values into the remaining equation and see if the equation still holds true.

\begin{aligned}12x+8y&= -12\\ 12\left({\color{DarkOrange}\dfrac{17}{13}}\right)+ 8\left({\color{DarkOrange}-\dfrac{45}{13}}\right)&= -12\\-12 &= -12 \checkmark\end{aligned}

This confirms that the solution to our system of equations is $\left(\dfrac{17}{13}, -\dfrac{45}{13}\right)$.

We’ve shown you examples where we only manipulate one equation to eliminate one term. Let’s now try out an example where we’re required to multiply different factors on both equations.

Example 2

Use the elimination method to solve the system of equations $ \begin{array}{ccc}3x- 4y&= \phantom{x}12\,\,(1)\\4x+3y&= \phantom{x}16\,\,(2)\end{array}$.

Solution

This example shows that we sometimes need to work on both linear equations before we can eliminate either $x$ or $y$. Since our first two examples show you how to eliminate the terms with $x$, let’s make it our goal to eliminate $y$ first this time.

Rewrite the terms with $y$ in both equations by multiplying $3$ on both sides of Equation (1) and $4$ on both sides of Equation (2).

\begin{aligned} \begin{array}{ccc}{\color{Orchid}3}(3x)& -{\color{Orchid}3}(4y)&={\color{Orchid}3}(12)\\{\color{Orchid}4}(4x)& -{\color{Orchid}4}(3y)&={\color{Orchid}4}(16)\,\, \\&\downarrow\phantom{x}\\9x&- 12y&= 36\,\, \\ 16x&+ 12y&= 64\,\,\end{array}\end{aligned}

Now that we have $-12y$ and $12y$ on both resulting equations, add the two equations to eliminate $y$.

\begin{aligned} \begin{matrix}&\underline{\begin{array}{cccc}\phantom{+xxx}9x& -\bcancel{\color{Orchid}12y} &=\phantom{+}36\\+\phantom{xx}16x &+ \bcancel{\color{Orchid}12y} &= \phantom{x}64\end{array}}\\ &\begin{array}{cccc}\phantom{+} &25x&\phantom{xxxxx}&=100\end{array}\end{matrix}\end{aligned}

The system of equations has now been reduced to a linear equation with $x$ as the only unknown. Divide both sides of the equation by $25$ to solve for $x$.

\begin{aligned}25x &= 100\\x&= \dfrac{100}{25}\\&= 4\end{aligned}

Substitute $x =4$ into either of the system of linear equations to solve for $y$. In our case, let’s use Equation (1).

\begin{aligned}3x-4y&= 12\\3(4) -4y&= 12\\-4y&= 0\\y &=0\end{aligned}

Hence, the solution to our system of linear equations is $(4, 0)$.

Feel free to substitute these values into either Equation (1) or Equation (2) to double-check the solution. For now, let’s try out a word problem involving systems of linear equations to help you appreciate this topic even more!

Example 3

Amy has a favorite pastry shop where she often buys donuts and coffee. On Tuesday, she paid $\$12$ for two boxes of donuts and one cup of coffee.   On Thursday, she purchased one box of donuts and two cups of coffee. She paid $\$9$ this time. How much does each box of donuts cost? How about one cup of coffee?

Solution

First, let’s set up the system of linear equations that represent the situation.

• Let $d$ represent the cost of one box of donuts.
• Let $c$ represent the cost of one cup of coffee.

Each equation’s right-hand side represents the total cost in terms of $d$ and $c$. Hence, we have $ \begin{array}{ccc}2d+ c&= \phantom{x}12\,\,(1)\\d+2c&= \phantom{xc}9\,\,(2)\end{array}$. Now that we have a system of linear equations, apply the elimination method to solve for $c$ and $d$.

\begin{aligned} \begin{array}{ccc}2d& + c\phantom{xxx}&= 12\phantom{xx}\\{\color{Green}2}(d)& +{\color{Green}2}(2c)&={\color{Green}2}(9)\,\, \\&\downarrow\phantom{x}\\2d&+ c\,\,&= 12\,\, \\ 2d&+ 4c&= 18\,\,\end{array}\end{aligned}

Once we’ve eliminated one of the variables (for our case, it’s $d$), solve the resulting equation to find $c$.

\begin{matrix}&\underline{\begin{array}{cccc}\phantom{+xxx}\bcancel{\color{Green}2d} & + c&=\phantom{+}12\\-\phantom{xx}\bcancel{\color{Green}2d} &+ 4c&= \phantom{x}18\end{array}}\\ &\begin{array}{cccc}\phantom{+} &\phantom{xxxx}&-3c&=-6\\&\phantom{xx}&c&= 2\end{array}\end{matrix}

Substitute $c = 2$ into either of the system of linear equations to solve for $d$.

\begin{aligned}2d + c &= 12\\2d + 2&= 12\\2d&= 10\\d&= 5\end{aligned}

This means that one box of donuts costs $\$5$ while a cup of coffee costs $\$2$ at Amy’s favorite pastry shop.