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Unit 6: Lesson 3
- Systems of equations with elimination: King's cupcakes
- Elimination strategies
- Systems of equations with elimination: x-4y=-18 & -x+3y=11
- Systems of equations with elimination: potato chips
Systems of equations with elimination (and manipulation)
- Elimination method review (systems of linear equations)
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Video transcript
Elimination by Multiplication - Examples & Practice - Expii
Module 16: Linear Systems
The elimination method with multiplication, learning outcomes.
- Use the elimination method with multiplication
- Express the solution of a dependent system of equations containing two variables
Solve a system of equations when multiplication is necessary to eliminate a variable
Many times adding the equations or adding the opposite of one of the equations will not result in eliminating a variable. Look at the system below.
[latex]\begin{array}{r}3x+4y=52\\5x+y=30\end{array}[/latex]
If you add the equations above, or add the opposite of one of the equations, you will get an equation that still has two variables. So let’s now use the multiplication property of equality first. You can multiply both sides of one of the equations by a number that will allow you to eliminate the same variable in the other equation.
We do this with multiplication. Notice that the first equation contains the term [latex]4y[/latex], and the second equation contains the term [latex]y[/latex]. If you multiply the second equation by [latex]−4[/latex], when you add both equations the y variables will add up to [latex]0[/latex].
The following example takes you through all the steps to find a solution to this system.
Solve for [latex]x[/latex] and [latex]y[/latex].
Equation A: [latex]3x+4y=52[/latex]
Equation B: [latex]5x+y=30[/latex]
Multiply the second equation by [latex]−4[/latex] so they do have the same coefficient.
[latex]\begin{array}{l}\,\,\,\,\,\,\,\,\,3x+4y=52\\−4\left(5x+y\right)=−4\left(30\right)\end{array}[/latex]
Rewrite the system and add the equations.
[latex]\begin{array}{r}3x+4y=52\,\,\,\,\,\,\,\\−20x–4y=−120\end{array}[/latex]
Solve for [latex]x[/latex].
[latex]\begin{array}{l}−17x=-68\\\,\,\,\,\,\,\,\,\,\,x=4\end{array}[/latex]
Substitute [latex]x=4[/latex] into one of the original equations to find y.
[latex]\begin{array}{r}3x+4y=52\\3\left(4\right)+4y=52\\12+4y=52\\4y=40\\y=10\end{array}[/latex]
Check your answer.
[latex]\begin{array}{r}3x+4y=52\\3\left(4\right)+4\left(10\right)=52\\12+40=52\\52=52\\\text{TRUE}\\\\5x+y=30\\5\left(4\right)+10=30\\20+10=30\\30=30\\\text{TRUE}\end{array}[/latex]
The answers check.
The solution is [latex](4, 10)[/latex].
Solve the given system of equations by the elimination method.
[latex]\begin{array}{l}3x+5y=-11\hfill \\ x - 2y=11\hfill \end{array}[/latex]
Adding these equations as presented will not eliminate a variable. However, we see that the first equation has [latex]3x[/latex] in it and the second equation has [latex]x[/latex]. So if we multiply the second equation by [latex]-3,\text{}[/latex] the x -terms will add to zero.
Now, let us add them.
[latex]\begin{array}\ \hfill 3x+5y=−11 \\ \hfill −3x+6y=−33 \\ \text{_____________} \\ \hfill 11y=−44 \\ \hfill y=−4 \end{array}[/latex]
For the last step, we substitute [latex]y=-4[/latex] into one of the original equations and solve for [latex]x[/latex].
[latex]\begin{array}{c}3x+5y=-11\\ 3x+5\left(-4\right)=-11\\ 3x - 20=-11\\ 3x=9\\ x=3\end{array}[/latex]
Our solution is the ordered pair [latex]\left(3,-4\right)[/latex]. Check the solution in the original second equation.
Below is another video example of using the elimination method to solve a system of linear equations in which we multiply one of the equations be a constant.
It is worth demonstrating that there is more than one way to solve a system. Consider our first example. Instead of multiplying one equation in order to eliminate a variable when the equations were added, we could have multiplied both equations by different numbers.
Let’s remove the variable [latex]x[/latex] this time. Multiply Equation A by [latex]5[/latex] and Equation B by [latex]−3[/latex].
In order to use the elimination method, you have to create variables that have the same coefficient—then you can eliminate them. Multiply the top equation by [latex]5[/latex].
[latex]\begin{array}{r}5\left(3x+4y\right)=5\left(52\right)\\5x+y =30\,\,\,\,\,\,\,\,\,\,\,\,\\15x+20y=260\,\,\,\,\,\,\\5x+y=30\,\,\,\,\,\,\,\,\,\end{array}[/latex]
Now multiply the bottom equation by [latex]−3[/latex].
[latex]\begin{array}{r}15x+20y=260\,\,\,\,\,\,\,\,\\-3(5x+y)=−3(30)\\15x+20y=260\,\,\,\,\,\,\,\,\\−15x–3y=−90\,\,\,\,\,\,\,\end{array}[/latex]
Next add the equations, and solve for [latex]y[/latex].
[latex]\begin{array}{r}15x+20y=260\\−15x–3y=\,–90\\17y=170\\y=\,\,\,10\end{array}[/latex]
Substitute [latex]y=10[/latex] into one of the original equations to find [latex]x[/latex].
[latex]\begin{array}{r}3x+4y=52\\3x+4\left(10\right)=52\\3x+40=52\\3x=12\\x=4\,\,\,\end{array}[/latex]
You arrive at the same solution as before.
These equations were multiplied by [latex]5[/latex] and [latex]−3[/latex] respectively, because that gave you terms that would add up to [latex]0[/latex]. Be sure to multiply all of the terms of the equation.
In the next example, we will see that sometimes we have to multiply both numbers by different numbers in order for one variable to be eliminated.
Solve the given system of equations in two variables by elimination.
One equation has [latex]2x[/latex] and the other has [latex]5x[/latex]. The least common multiple is [latex]10x[/latex], so we will have to multiply both equations by a constant in order to eliminate one variable. Let’s eliminate [latex]x[/latex] by multiplying the first equation by [latex]-5[/latex] and the second equation by [latex]2[/latex].
Then, we add the two equations together.
[latex]\begin{array}\ −10x−15y=80 \\ \:\:10x−20y=60 \\ \text{______________} \\ \text{ }\:\:\:\:\:\:\:\:\:\:−35y=140 \\ \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:y=−4 \end{array}[/latex]
Substitute [latex]y=-4[/latex] into the original first equation.
[latex]\begin{array}{c}2x+3\left(-4\right)=-16\\ 2x - 12=-16\\ 2x=-4\\ x=-2\end{array}[/latex]
The solution is [latex]\left(-2,-4\right)[/latex]. Check it in the second original equation.
Below is a summary of the general steps for using the elimination method to solve a system of equations.
How To: Given a system of equations, solve using the elimination method
- Write both equations with x and y -variables on the left side of the equal sign and constants on the right.
- Write one equation above the other, lining up corresponding variables. If one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, add the equations together, eliminating one variable. If not, use multiplication by a nonzero number so that one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, then add the equations to eliminate the variable.
- Solve the resulting equation for the remaining variable.
- Substitute that value into one of the original equations and solve for the second variable.
- Check the solution by substituting the values into the other equation.
In the next example, we will show how to solve a system with fractions. As with single linear equations, the easiest way to solve is to clear the fractions first with the least common denominator.
[latex]\begin{array}{l}\dfrac{x}{3}+\dfrac{y}{6}=3\hfill \\ \dfrac{x}{2}-\dfrac{y}{4}=\text{ }1\hfill \end{array}[/latex]
First clear each equation of fractions by multiplying both sides of the equation by the least common denominator
[latex]\begin{array}{l}6\left(\dfrac{x}{3}+\dfrac{y}{6}\right)=6\left(3\right)\hfill \\ \text{ }2x+y=18\hfill \\ 4\left(\dfrac{x}{2}-\dfrac{y}{4}\right)=4\left(1\right)\hfill \\ \text{ }2x-y=4\hfill \end{array}[/latex]
Now multiply the second equation by [latex]-1[/latex] so that we can eliminate the x -variable.
[latex]\begin{array}{l}-1\left(2x-y\right)=-1\left(4\right)\hfill \\ \text{ }-2x+y=-4\hfill \end{array}[/latex]
Add the two equations to eliminate the x -variable and solve the resulting equation.
[latex]\begin{array}\ \hfill 2x+y=18 \\ \hfill−2x+y=−4 \\ \text{_____________} \\ \hfill 2y=14 \\ \hfill y=7 \end{array}[/latex]
Substitute [latex]y=7[/latex] into the first equation.
[latex]\begin{array}{l}2x+\left(7\right)=18\hfill \\ \text{ }2x=11\hfill \\ \text{ }x=\dfrac{11}{2}\hfill \end{array}[/latex]
The solution is [latex]\left(\dfrac{11}{2},7\right)[/latex]. Check it in the other equation.
[latex]\begin{array}{c}2x-y=4\\ 2(\dfrac{11}{2})-7=4\\ 11-7=4 \\ 4=4\end{array}[/latex]
In the following video, you will find one more example of using the elimination method to solve a system; this one has coefficients that are fractions.
It is possible to use the elimination method with multiplication and get a result that indicates no solutions or infinitely many solutions, just as with the other methods we have learned for finding solutions to systems. Recall that a dependent system of equations in two variables is a system in which the two equations represent the same line. Dependent systems have an infinite number of solutions because all of the points on one line are also on the other line. After using substitution or elimination, the resulting equation will be an identity such as [latex]0=0[/latex]. The last example includes two equations that represent the same line and are therefore dependent.
Find a solution to the system of equations using the elimination method .
[latex]\begin{array}{c}x+3y=2\\ 3x+9y=6\end{array}[/latex]
With the elimination method, we want to eliminate one of the variables by adding the equations. In this case, focus on eliminating [latex]x[/latex]. If we multiply both sides of the first equation by [latex]-3[/latex], then we will be able to eliminate the [latex]x[/latex] -variable.
[latex]\begin{array}{l}\text{ }x+3y=2\hfill \\ \left(-3\right)\left(x+3y\right)=\left(-3\right)\left(2\right)\hfill \\ \text{ }-3x - 9y=-6\hfill \end{array}[/latex]
Now add the equations.
[latex]\begin{array} \hfill−3x−9y=−6 \\ \hfill3x+9y=6 \\ \hfill \text{_____________} \\ \hfill 0=0 \end{array}[/latex]
We can see that there will be an infinite number of solutions that satisfy both equations.
If we rewrote both equations in slope-intercept form, we might know what the solution would look like before adding. Look at what happens when we convert the system to slope-intercept form.
[latex]\begin{array}{l}\text{ }x+3y=2\hfill \\ \text{ }3y=-x+2\hfill \\ \text{ }y=-\dfrac{1}{3}x+\dfrac{2}{3}\hfill \\ 3x+9y=6\hfill \\ \text{ }9y=-3x+6\hfill \\ \text{ }y=-\dfrac{3}{9}x+\dfrac{6}{9}\hfill \\ \text{ }y=-\dfrac{1}{3}x+\dfrac{2}{3}\hfill \end{array}[/latex]
See the graph below. Notice the results are the same. The general solution to the system is [latex]\left(x, -\dfrac{1}{3}x+\dfrac{2}{3}\right)[/latex].
In the following video, the elimination method is used to solve a system of equations. Notice that one of the equations needs to be multiplied by a negative one first. Additionally, this system has an infinite number of solutions.
Multiplication can be used to set up matching terms in equations before they are combined to aid in finding a solution to a system. When using the multiplication method, it is important to multiply all the terms on both sides of the equation—not just the one term you are trying to eliminate.
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- Revision and Adaptation. Provided by : Lumen Learning. License : CC BY: Attribution
- Ex: System of Equations Using Elimination (Infinite Solutions). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/NRxh9Q16Ulk . License : CC BY: Attribution
- Unit 14: Systems of Equations and Inequalities, from Developmental Math: An Open Program. Provided by : Monterey Institute of Technology and Education. Located at : . License : CC BY: Attribution
- Ex 2: Solve a System of Equations Using the Elimination Method. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/_liDhKops2w . License : CC BY: Attribution
How Do You Solve a System of Equations Using the Elimination by Multiplication Method?
There are many different ways to solve a system of linear equations. In this tutorial, you'll see how to solve such a system by combining the equations together in a way so that one of the variables is eliminated. Then, see how find the value of that variable and use it to find the value of the other variable. Take a look!
- linear equations
- system of equations
- 2 equations
- find intersection
- solve system of equations
- solve by elimination
- elimination
- variable elimination
- add equations
- multiply equation
- multiply by negative
- independent
- 2 variables
Background Tutorials
Definitions of systems.
What's a System of Linear Equations?
A system of equations is a set of equations with the same variables. If the equations are all linear, then you have a system of linear equations! To solve a system of equations, you need to figure out the variable values that solve all the equations involved. This tutorial will introduce you to these systems.
Evaluating Expressions
What is a Variable?
You can't do algebra without working with variables, but variables can be confusing. If you've ever wondered what variables are, then this tutorial is for you!
Further Exploration
Solving systems using elimination.
How Do You Solve a System of Equations Using the Elimination by Addition Method?
There are many different ways to solve a system of linear equations. In this tutorial, you'll see how to solve a system of linear equations by combining the equations together to eliminate one of the variables. Then, see how find the value of that variable and use it to find the value of the other variable. Take a look!
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Elimination Using Multiplication (Jump to: Lecture | Video )
Below, we have a system of equations:
3x + 2y = 6
Equations can be multiplied by a constant to allow for elimination by addition or subtraction. This process is called Elimination by Multiplication.
We multiply the first equation by "2" to prepare it for elimination by subtraction.
Now that we know "x", we can use this information to solve for "y".
Our solution is (2, 0).
Back to Top
The Elimination Method
Learning Objective(s)
· Solve a system of equations when no multiplication is necessary to eliminate a variable.
· Solve a system of equations when multiplication is necessary to eliminate a variable.
· Recognize systems that have no solution or an infinite number of solutions.
· Solve application problems using the elimination method.
Introduction
The elimination method for solving systems of linear equations uses the addition property of equality. You can add the same value to each side of an equation.
So if you have a system: x – 6 = −6 and x + y = 8, you can add x + y to the left side of the first equation and add 8 to the right side of the equation. And since x + y = 8, you are adding the same value to each side of the first equation.
Using Addition to Eliminate a Variable
If you add the two equations, x – y = −6 and x + y = 8 together, as noted above, watch what happens.
You have eliminated the y term, and this equation can be solved using the methods for solving equations with one variable.
Let’s see how this system is solved using the elimination method.
Unfortunately not all systems work out this easily. How about a system like 2 x + y = 12 and − 3 x + y = 2. If you add these two equations together, no variables are eliminated.
But you want to eliminate a variable. So let’s add the opposite of one of the equations to the other equation.
2 x + y =12 → 2 x + y = 12 → 2 x + y = 12
− 3 x + y = 2 → − ( − 3 x + y ) = − (2) → 3 x – y = − 2
5 x + 0 y = 10
You have eliminated the y variable, and the problem can now be solved. See the example below.
The following are two more examples showing how to solve linear systems of equations using elimination.
Notice the coefficients of each variable in each equation. If you add these two equations, the x term will be eliminated since
−2 x + 2 x = 0.
Add and solve for y .
2 x + 5 y = 25
2 x + 5(3) = 25
2 x + 15 = 25
Substitute y = 3 into one of the original equations.
−2x + 3y = −1
−2(5) + 3(3) = −1
−10 + 9 = −1
2(5) + 5(3) = 25
10 + 15 = 25
Check solutions.
The answers check.
The solution is (5, 3).
Notice the coefficients of each variable in each equation. You will need to add the opposite of one of the equations to eliminate the variable y , as 2 y + 2y = 4 y , but
2 y + (−2 y ) = 0.
Change one of the equations to its opposite, add and solve for x .
4 x + 2 y = 14
4(2) + 2 y = 14
8 + 2 y = 14
Substitute x = 2 into one of the original equations and solve for y .
The solution is (2, 3).
Go ahead and check this last example—substitute (2, 3) into both equations. You get two true statements: 14 = 14 and 16 = 16!
Notice that you could have used the opposite of the first equation rather than the second equation and gotten the same result.
Using Multiplication and Addition to Eliminate a Variables
Many times adding the equations or adding the opposite of one of the equations will not result in eliminating a variable. Look at the system below.
3 x + 4 y = 52
5 x + y = 30
If you add the equations above, or add the opposite of one of the equations, you will get an equation that still has two variables. So let’s now use the multiplication property of equality first. You can multiply both sides of one of the equations by a number that will result in the coefficient of one of the variables being the opposite of the same variable in the other equation.
This is where multiplication comes in handy. Notice that the first equation contains the term 4 y , and the second equation contains the term y . If you multiply the second equation by −4, when you add both equations the y variables will add up to 0.
3 x + 4 y = 52 → 3 x + 4 y = 52 → 3 x + 4 y = 52
5 x + y = 30 → − 4(5 x + y ) = − 4(30) → − 20 x – 4 y = − 120
− 17 x + 0 y = − 68
See the example below.
Look for terms that can be eliminated. The equations do not have any x or y terms with the same coefficients.
Multiply the second equation by − 4 so they do have the same coefficient.
Rewrite the system, and add the equations.
Solve for x .
3(4) + 4 y = 52
12 + 4 y = 52
Substitute x = 4 into one of the original equations to find y .
3(4) + 4(10) = 52
12 + 40 = 52
5(4) + 10 = 30
20 + 10 = 30
Check your answer.
The solution is (4, 10).
There are other ways to solve this system. Instead of multiplying one equation in order to eliminate a variable when the equations were added, you could have multiplied both equations by different numbers.
Let’s remove the variable x this time. Multiply Equation A by 5 and Equation B by − 3.
Look for terms that can be eliminated. The equations do not have any x or y terms with the same coefficient.
In order to use the elimination method, you have to create variables that have the same coefficient—then you can eliminate them. Multiply the top equation by 5.
Now multiply the bottom equation by −3.
Next add the equations, and solve for y .
3 x + 4(10) = 52
3 x + 40 = 52
Substitute y = 10 into one of the original equations to find x .
You arrive at the same solution as before.
These equations were multiplied by 5 and − 3 respectively, because that gave you terms that would add up to 0. Be sure to multiply all of the terms of the equation.
Special Situations
Just as with the substitution method, the elimination method will sometimes eliminate both v ariables, and you end up with either a true statement or a false statement. Recall that a false statement means that there is no solution.
Let’s look at an example.
Graphing these lines shows that they are parallel lines and as such do not share any point in common, verifying that there is no solution.
If both variables are eliminated and you are left with a true statement, this indicates that there are an infinite number of ordered pairs that satisfy both of the equations. In fact, the equations are the same line.
Graphing these two equations will help to illustrate what is happening.
Solving Application Problems Using the Elimination Method
The elimination method can be applied to solving systems of equations that model real situations. Two examples of using the elimination method in problem solving are shown below.
Combining equations is a powerful tool for solving a system of equations. Adding or subtracting two equations in order to eliminate a common variable is called the elimination (or addition) method. Once one variable is eliminated, it becomes much easier to solve for the other one. Multiplication can be used to set up matching terms in equations before they are combined. When using the multiplication method, it is important to multiply all the terms on both sides of the equation—not just the one term you are trying to eliminate.
Step 2: Subtract the second equation from the first.
Step 3: Solve this new equation for y .
Step 4: Substitute y = 2 into either Equation 1 or Equation 2 above and solve for x . We'll use Equation 1.
Solution: x = 1, y = 2 or (1,2).
Now study some more worked examples:
Click on the buttons below to see how to solve these equations.
y = (2 d.p.)
Substitute y into Equation 1 and solve for x: x + ( ) =
x = (2 d.p.)
Exercise 2.
Equation 1: x + y =
Equation 2 : x + y =
First determine what you will multiply each of the above equations by to get the same leading coefficients:
Subtract:
y = to 2 d.p.
x = to 2 d.p.
Simultaneous Linear Equations (definition) | Simultaneous Linear Equations Index | The Substitution Method >>
IMAGES
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COMMENTS
With this problem, there is no solution. If you multiply 3x + 2y = 18 by -2 (I chose -2 so when you add the equations together, variables cancel out), you
Learn how to solve a system of linear equations using the elimination method and having to multiply both equations by a constant.
Multiply for Elimination in Systems of Linear Equations. System of equations is {-7x. Image source: By Caroline Kulczycky. Report. Share. 1. Like.
The equations do not have any x or y terms with the same coefficient. ... In order to use the elimination method, you have to create variables that have the same
solve systems of equations using elimination with multiplication. Solving Systems of Equations. So far, we have solved systems using graphing, substitution
How Do You Solve a System of Equations Using the Elimination by Multiplication Method? Note: There are many different ways to solve a system of linear equations
We use the method of elimination to 'eliminate' one of the variables when solving a linear system, allowing us to solve for the other variable and in turn
Equations can be multiplied by a constant to allow for elimination by addition or subtraction. This process is called Elimination by Multiplication.
The elimination method for solving systems of linear equations uses the addition property of equality. You can add the same value to each side of an
The Elimination Method ... This method for solving a pair of simultaneous linear equations reduces one equation to one that has only a single variable. Once this