3. Solving Systems of Equations by Elimination
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Chapter 5
3.
Solving Systems of Equations by Elimination
This lesson explores the elimination method as a technique for solving systems of equations. It emphasizes the importance of understanding the terms "consistency" and "independence" in the context of these systems. These concepts help in determining whether a system has one solution, no solutions, or infinitely many solutions. Real-world applications include budgeting for school gym equipment, calculating the number of toys in a box, and determining the profit from selling tools at a garage sale. The elimination method is presented as a reliable and straightforward approach for solving these types of problems, making it a valuable skill for students and professionals alike.
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Lesson Settings & Tools
| | 11 Theory slides |
| | 8 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Solving Systems of Equations by Elimination
Slide of 11
The aim of this lesson is to show a method to solve a system of linear equations by adding or subtracting the equations.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Challenge
How Many Toys Are There in Each Box?
Maya is researching a toy factory that currently produces robot action figures and car toys.
54x+101y=20
Also, each robot and each car are sold to the stores for $15 and $5, respectively. The factory makes $500 for each box sold to a store. This information can also be written as an equation.
15x+5y=500
These equations can be combined to form a system of linear equations.
{54x+101y=2015x+5y=500
Solve this system of equations without graphing. How many robots and cars are there in each box? {"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"Robots:","formTextAfter":null,"answer":{"text":["20"]}}
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"Cars:","formTextAfter":null,"answer":{"text":["40"]}}
Discussion
Rewriting Equations
Sometimes the equations of a system can be rewritten in a way that is easier to work with. As an example, consider the following system of linear equations.
{x+2y=321x−y=-21(I)(II)
In Equation (II), the coefficient of the variable x is a fraction. It would be easier to solve the system if all the coefficients were integers. It is possible to rewrite the system using equivalent systems. Rule
Equivalent Systems
In a system of equations, an equivalent system can be created by replacing one equation with the sum of two or more equations in the system or by replacing an equation with a multiple of itself. An equation can be also replaced by the sum of that equation and a multiple of another equation in the system.
Proof
The statement will be proved for linear equations. For non-linear equations, the proof is similar. Consider a system of linear equations.
{ax+by=cdx+ey=f(I)(II)
Let (x1,y1) be a solution to the system. Therefore, this ordered pair satisfies both equations simultaneously.
{ax1+by1=c ✓dx1+ey1=f ✓
Consider now the system formed by the following two equations. - Equation (I)
- Equation (II) plus a multiple of Equation (I)
{ax+by=cdx+ey+α(ax+by)=f+αc
It is already known that (x1,y1) is a solution to the first equation of this new system. It needs to be verified that the ordered pair is also a solution to the second equation.
{ax1+by1=c ✓dx1+ey1+α(ax1+by1)=?f+αc
Since (x1,y1) is a solution to the original system, it is known that ax1+by1=c and that dx1+ey1=f. These two expressions can be substituted into the second equation above.
Therefore, (x1,y1) is also a solution to the second equation of the new system. This means that this ordered pair is a solution to the system formed by the first equation of the original system and the sum of the first equation and a multiple of the second equation.
{ax1+by1=c ✓dx1+ey1+α(ax1+by1)=f+αc ✓
Discussion
Elimination Method
Given a system of two equations in two variables, replacing one equation with the sum of that equation and a multiple of the other equation produces an equivalent system. This fact is used to solve systems of equations by the Elimination Method. Consider an example system of linear equations.
If at any step of the method a true statement is found, then the lines represented by the equations of the system are coincidental. The system would be a dependent system and have infinitely many solutions. Conversely, if at any step a false statement is found, then the lines are parallel. In this case, the system would be inconsistent and have no solution.
{3x+2y=6y=2x−11(I)(II)
To solve the system by using the Elimination Method, there are five steps to follow.
1
Write the Equations in the Same Form
expand_more First, all like terms must be gathered on the same sides of the equations. In Equation (I), the variable terms are on the same side of the equation. However, the variable terms are on both sides of the equations in Equation (II). Like terms can be gathered on the same sides of the equations by applying the Properties of Equality.
2
Multiply an Equation
expand_more Multiply one of the equations by a constant so that one of the variable terms of the resulting equation is equal to or is the opposite of the corresponding variable term in the other equation. In this case, multiplying Equation (II) by -2 will produce opposite coefficients for the y-variable.
-2(-2x+y)=-2(-11)⇕4x−2y=22
Both the original and the resulting equations have the same solutions because they are equivalent equations. 3
Add the New Equation and the Other Equation
expand_more After the rewrites, the system of equations looks the following way.
Note that this step results in an equation in only one variable. This equation can be solved by dividing both sides by 7.
{3x+2y=64x−2y=22
Add these two equations by adding the right-hand sides together and the left-hand sides together. This way one variable will be eliminated.
3x+2y+(4x−2y)=6+22
▼
Simplify
RemovePar
Remove parentheses
3x+2y+4x−2y=6+22
CommutativePropAdd
Commutative Property of Addition
3x+4x+2y−2y=6+22
AddSubTerms
Add and subtract terms
7x=28
4
Write an Equivalent System
expand_more Substitute the value for the solved equation in one variable for any of the equations of the system. This produces an equivalent system of equations. In this case, Equation (I) will be replaced.
{3x+2y=6-2x+y=-11(I)(II)⇕{x=4-2x+y=-11
Note that the first equation is the solution value of x. 5
Solve the Equivalent System
expand_more To solve the new system, substitute the found value into the other equation. In this case, 4 will be substituted into Equation (II) for x to find the value of y.
In this system, the value of y is -3. Therefore, the solution to the system of equations, which is the point of intersection of the lines, is (4,-3), or x=4, y=-3.
Pop Quiz
Checking if the Coordinate Pair is a Solution
For each system of linear equations, verify whether the coordinate pair is a solution.
Example
The Price of Gold
A vlogger that Diego likes to watch bought silver and gold to make Olympic-style medals. The vlogger will show the process of making the medals as a multi-video series.
25s+2000g=7075
Diego notes that if he subtracts 50 times the number of ounces of gold from 5 times the number of ounces of silver, there would still be 65 ounces left. This can also be written as an equation.
5s−50g=65
A system of equations is obtained by combining these two equations.
{25s+2000g=70755s−50g=65(I)(II)
a Solve the system of equations graphically.
b Solve the system of equations using the Substitution Method.
c Solve the system of equations using the Elimination Method.
d Interpret the system in terms of consistency and independence.
Answer
a See solution.
b See solution.
c See solution.
d The system is consistent and independent.
Hint
a First write the equations in slope-intercept form.
b Isolate one variable in an equation.
c Recall that equivalent systems have the same solution.
d A system of equations can be inconsistent, consistent dependent, or consistent independent.
Solution
a To solve the system graphically, first each equation should be written in slope-intercept form. The variable s will arbitrarily be considered as the dependent variable, and g will be the independent variable.
{25s+2000g=70755s−50g=65(I)(II)
▼
(I), (II): Write in slope-intercept form
DivEqn
(I): LHS/25=RHS/25
{2525s+2000g=2835s−50g=65
WriteSumFrac
(I): Write as a sum of fractions
{2525s+252000g=2835s−50g=65
DivEqn
(II): LHS/5=RHS/5
{2525s+252000g=28355s−50g=13
WriteDiffFrac
(II): Write as a difference of fractions
{2525s+252000g=28355s−550g=13
(I), (II): ca⋅b=ca⋅b
{2525s+252000g=28355s−550g=13
(I), (II): Calculate quotient
{1s+80g=2831s−10g=13
(I), (II): Identity Property of Multiplication
{s+80g=283s−10g=13
SubEqn
(I): LHS−80g=RHS−80g
{s=-80g+283s−10g=13
AddEqn
(II): LHS+10g=RHS+10g
{s=-80g+283s=10g+13
Looking at the graph, it can be seen that the solution is about 3 ounces of gold and a bit more than 40 ounces of silver. In this case, since the point of intersection is not a lattice point, finding the exact solution is not possible.
b To solve the system using the Substitution Method, first a variable should be isolated from one of the equations. It seems easier to isolate s on Equation (II).
{25s+2000g=70755s−50g=65(I)(II)
▼
(II): Solve for s
DivEqn
(II): LHS/5=RHS/5
{25s+2000g=707555s−50g=13
WriteDiffFrac
(II): Write as a difference of fractions
{25s+2000g=707555s−550g=13
MovePartNumRight
(II): ca⋅b=ca⋅b
{25s+2000g=707555s−550g=13
CalcQuot
(II): Calculate quotient
{25s+2000g=70751s−10g=13
IdPropMult
(II): Identity Property of Multiplication
{25s+2000g=7075s−10g=13
AddEqn
(II): LHS+10g=RHS+10g
{25s+2000g=7075s=10g+13
Substitute
(I): s=10g+13
{25(10g+13)+2000g=7075s=10g+13
{g=3s=10g+13
c To solve this system using the Elimination Method, it is important to remember that equivalent systems share the same solution. Equation (I) can be divided by 5 so that the variable s has the same coefficient in both equations.
{25s+2000g=70755s−50g=65(I)(II)
▼
(I): Simplify
DivEqn
(I): LHS/5=RHS/5
{525s+2000g=14155s−50g=65
WriteSumFrac
(I): Write as a sum of fractions
{525s+52000g=14155s−50g=65
MovePartNumRight
(I): ca⋅b=ca⋅b
{525s+52000g=14155s−50g=65
CalcQuot
(I): Calculate quotient
{5s+400g=14155s−50g=65
{5s+400g=14155s−50g=65(I)(II)
SysEqnSub
(I): Subtract (II)
{5s+400g−(5s−50g)=1415−655s−50g=65
▼
(I): Solve for g
Distr
(I): Distribute -1
{5s+400g−5s+50g=1415−655s−50g=65
AddSubTerms
(I): Add and subtract terms
{450g=13505s−50g=65
DivEqn
(I): LHS/450=RHS/450
{g=35s−50g=65
d To interpret the system in terms of the independence and consistency, these concepts should be reviewed.
| Concept | Definition |
|---|---|
| Consistent System | A system of equations that has at least one solution. |
| Inconsistent System | A system of equations that has no solution. |
| Dependent System | A system of equations with infinitely many solutions. |
| Independent System | A system of equations that has exactly one solution. |
Since the system has exactly one solution, it is both consistent and independent.
Example
Buying Equipment for the School Gym
As part of his student council duties, Tadeo is in charge of getting equipment for the school gym. At the moment, the gym needs new basketballs and volleyballs. Each basketball costs $75 and each volleyball costs $50. The gym's budget is $1400.
{75b+50v=140022b+10v=364(I)(II)
a Solve the system using the Elimination Method.
b Verify the answer from Part A.
c Interpret the system in terms of the consistency and independence.
Answer
a See solution.
b See solution.
c The system is consistent and independent.
Hint
a What is the easiest way for a variable to have the same coefficient in both equations?
b Substitute the values from Part A into the original system of equations.
c A system of equations can be inconsistent, consistent independent, or consistent dependent.
Solution
a To solve the system using the Elimination Method, the coefficients of one variable have to be same to eliminate the term by adding or subtracting. Looking at the system, it can be noted that dividing Equation (I) by -5 will achieve this purpose.
{75b+50v=140022b+10v=364(I)(II)
▼
(I): Simplify
DivEqn
(I): LHS/(-5)=RHS/(-5)
{-575b+50v=-28022b+10v=364
WriteSumFrac
(I): Write as a sum of fractions
{-575b+-550v=-28022b+10v=364
MovePartNumRight
(I): ca⋅b=ca⋅b
{-575b+-550v=-28022b+10v=364
MoveNegDenomToFrac
(I): Put minus sign in front of fraction
{-575b+(-550v)=-28022b+10v=364
AddNeg
a+(-b)=a−b
{-575b−550v=-28022b+10v=364
CalcQuot
(I): Calculate quotient
{-15b−10v=-28022b+10v=364
{-15b−10v=-28022b+10v=364(I)(II)
SysEqnAdd
(I): Add (II)
{-15b−10v+(22b+10v)=-280+36422b+10v=364
▼
(I): Solve for b
RemovePar
(I): Remove parentheses
{-15b−10v+22b+10v=-280+36422b+10v=364
AddSubTerms
(I): Add and subtract terms
{7b=8422b+10v=364
DivEqn
(I): LHS/7=RHS/7
{b=1222b+10v=364
b To verify the solution, the values found in Part A need to be substituted into the system of equations.
| Equation (I) | Equation (II) | |
|---|---|---|
| Equation | 75b+50v=1400 | 22b+10v=364 |
| Substitute | 75(12)+50(10)=1400 | 22(12)+10(10)=364 |
| Simplify | 1400=1400 ✓ | 364=364 ✓ |
The values verify both equations of the system. Therefore, the solution is correct.
c To interpret the system in terms of independence and consistency, these concepts will be reviewed.
| Concept | Definition |
|---|---|
| Consistent System | A system of equations that has at least one solution. |
| Inconsistent System | A system of equations that has no solution. |
| Dependent System | A system of equations with infinitely many solutions. |
| Independent System | A system of equations that has exactly one solution. |
Since the system has exactly one solution, the system is both consistent and independent.
Example
Tools in a Toolbox
Ignacio is looking through his dad's toolbox to see if there are any duplicates that can be sold at a garage sale. He discovers that there are some extra hammers and wrenches.
{w=13−2h7h+3.50w=45.50(I)(II)
a Solve the system of equations using the Elimination Method.
b Interpret the system in terms of consistency and independence.
Answer
a See solution.
b The system is consistent and dependent.
Hint
a In Equation (I), group the variable terms on the same side.
b A system of equations can be inconsistent, consistent independent, or consistent dependent.
Solution
a To solve the system using the Elimination Method, all the variables should be on the same side of the equation in both equations. This can be achieved by adding 2h to both sides of Equation (I).
{w=13−2h7h+3.50w=45.50(I)(II)⇕{2h+w=137h+3.50w=45.50(I)(II)
Now, a variable needs to have the same or opposite coefficients in both equations. This can be achieved by multiplying Equation (I) by 3.5.
{2h+w=137h+3.5w=45.5(I)(II)
MultEqn
(I): LHS⋅3.5=RHS⋅3.5
{(2h+w)3.5=45.57h+3.5w=45.5
Distr
(I): Distribute 3.5
{7h+3.5w=45.57h+3.5w=45.5
SysEqnSub
(II): Subtract (I)
{7h+3.5w=45.57h+3.5w−(7h+3.5w)=45.5−45.5
Distr
(II): Distribute -1
{7h+3.5w=45.57h+3.5w−7h−3.5w=45.5−45.5
SubTerms
(II): Subtract terms
{7h+3.5w=45.50=0 ✓
b To interpret the system in terms of consistency and independence, these concepts should be reviewed.
| Concept | Definition |
|---|---|
| Consistent System | A system of equations that has at least one solution. |
| Inconsistent System | A system of equations that has no solution. |
| Dependent System | A system of equations with infinitely many solutions. |
| Independent System | A system of equations that has exactly one solution. |
Since the system has infinitely many solutions, the system is consistent and dependent. In a system of linear equations, if the system is dependent, then the equations that make the system are equations of coincidental lines.
Example
The Lengths of Two Tracks
LaShay is on the track and field team at school. She has two tracks nearby her home where she trains for races. She is trying to remember the lengths of the tracks.
If LaShay completes 2 laps on the first track and 1 lap on the second track, then she runs 5 miles. On the other hand, if she completes 6 laps on the first track and 3 laps on the second track, then she runs 18 miles. Letting x be the length of the first track and y the length of the second track, this information can be written as a system of linear equations.
{2x+y=56x+3y=18(I)(II)
a Solve the system of equations using the Elimination Method.
b Interpret the system in terms of consistency and independence.
Answer
a See solution.
b The system is inconsistent.
Hint
a Make the coefficients of a variable term the same in both equations.
b A system of equations can be inconsistent, consistent independent, or consistent dependent.
Solution
a To solve the system using the Elimination Method, a variable term first needs to have the same coefficient in both equations. To achieve this, the first equation will be multiplied by 3.
{2x+y=56x+3y=18(I)(II)
{6x+3y=156x+3y=18
b To interpret the system in terms of consistency and independence, these concepts should be reviewed.
| Concept | Definition |
|---|---|
| Consistent System | A system of equations that has at least one solution. |
| Inconsistent System | A system of equations that has no solution. |
| Dependent System | A system of equations with infinitely many solutions. |
| Independent System | A system of equations that has exactly one solution. |
Since the system has no solution, it is an inconsistent system.
Pop Quiz
Practice Solving Systems of Linear Equations
Solve the system of linear equations to find the values of x and y.
Closure
Finding Out How Many Toys Are in Each Box
At the beginning of this lesson, a system of equations was introduced for the number of toys in a box. In both equations, x is the number of robots and y the number of cars.
{54x+101y=2015x+5y=500(I)(II)
Solve this system without graphing. How many robots and cars are there in each box? {"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"Robots:","formTextAfter":null,"answer":{"text":["20"]}}
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"Cars:","formTextAfter":null,"answer":{"text":["40"]}}
Hint
Multiply or divide an equation by a number so that a variable has the same coefficient in both equations.
Solution
This system of linear equations can be solved without graphing using the Elimination Method. To use this method, Equation (I) will be multiplied by 10 to get rid of the fractions, while Equation (II) will be divided by 5 to make the coefficients of the y-variable equal.
Now that the variable y has the same coefficient in both equations, Equation (II) will be subtracted from Equation (I). Then, Equation (I) will be solved for the x-variable.
Therefore, there are 20 robots in each box. Now, the value of x will be substituted into Equation (II) to find the value of y.
There are 40 cars in each box.
{54x+101y=2015x+5y=500(I)(II)
▼
(I), (II): Simplify
MultEqn
(I): LHS⋅10=RHS⋅10
{(54x+101y)10=20015x+5y=500
Distr
(I): Distribute 10
{54x(10)+101y(10)=20015x+5y=500
CommutativePropMult
(I): Commutative Property of Multiplication
{54(10)x+101(10)y=20015x+5y=500
MoveRightFacToNum
(I): ca⋅b=ca⋅b
{540x+1010y=20015x+5y=500
CalcQuot
(I): Calculate quotient
{8x+1y=20015x+5y=500
DivEqn
(II): LHS/5=RHS/5
{8x+1y=200515x+5y=100
WriteSumFrac
(II): Write as a sum of fractions
{8x+1y=200515x+55y=100
MovePartNumRight
(II): ca⋅b=ca⋅b
{8x+1y=200515x+55y=100
CalcQuot
(II): Calculate quotient
{8x+1y=2003x+1y=100
(I), (II): Identity Property of Multiplication
{8x+y=2003x+y=100
Solve the system of linear equations by using the Elimination Method.
⎩⎪⎨⎪⎧-3x+4y=11-3x+2y=1
{"type":"text","form":{"type":"point2d","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false,"decimal":true,"function":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text1":"1","text2":"2"}}
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To solve a system of linear equations using the Elimination Method, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other. In this exercise, this means that either the x-terms or the y-terms must cancel each other out. 3 x+4 y=11 & (I) -3 x+2 y=1 & (II) The x-terms eliminate each other if we add the equations. Let's do it!
Now we can solve for x by substituting the value of y into Equation (I) and simplifying.
The solution to the system, which is the point of intersection of the lines, is (1,2).
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Solve the following system of equations by using the Elimination Method.
⎩⎪⎨⎪⎧2x+y=35x−9y=19
{"type":"text","form":{"type":"point2d","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false,"decimal":true,"function":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text1":"2","text2":"-1"}}
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To solve a system of linear equations by using the Elimination Method, one of the variable terms must be eliminated when the equations are added or subtracted. 2 x+ y=3 & (I) 5 x-9 y=19 & (II) Currently, none of the terms in this system will cancel out. However, if we multiply the first equation by 9, then the y-variables will have opposite coefficients in both equations. 9(2 x+ y)=9(3) 5 x-9 y=19 ⇕ 18 x+ 9y=27 5 x- 9y=19 Now the y-terms will eliminate each other if we add the equations.
We can now solve for y by substituting the value of x into the first equation and simplifying.
The solution to the system, which is the point of intersection of the lines, is (2,-1).
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Solve the system of equations by using the Elimination Method.
⎩⎪⎨⎪⎧5x−3y=562x+7y=6
{"type":"text","form":{"type":"point2d","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false,"decimal":true,"function":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text1":"10","text2":"-2"}}
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To solve a system of linear equations by using the Elimination Method, one of the variable terms must be eliminated when the equations are added or subtracted. 5 x-3 y=56 & (I) 2 x+7 y=6 & (II) Currently, none of the terms in this system will cancel out. However, if we multiply the first equation by 2 and the second equation by - 5, then the x-variable will have opposite coefficients in both equations. 2(5 x-3 y)=2(56) - 5(2 x+7 y)=- 5(6) ⇕ 10x-6 y=112 - 10x-35 y=- 30 We can see that the x-terms will eliminate each other if we add the equations. Let's do it!
We can now solve for x by substituting the value of y into Equation (I) and simplifying.
The solution to the system, which is the point of intersection of the lines, is (10,-2).
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How many solutions does the system have?
⎩⎪⎨⎪⎧5x+7y=1010x+14y=10
{"type":"choice","form":{"alts":["One solution","No solutions","Infinitely many solutions"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":1}
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To determine how many solutions this system has, we will solve it by using the Elimination Method. Doing so will result in one of three cases.
| Result of Solving by Substitution | Number of Solutions |
|---|---|
| A value for x and a value for y are determined. | One solution |
| An identity is found, for example 2=2. | Infinitely many solutions |
| A contradiction is found, for example 2≠ 3. | No solution |
This means we should solve the given system of equations, then make our conclusion based on the result.
Solving by Elimination
To solve a system of linear equations by using the Elimination Method, one of the variable terms must be eliminated when the equations are added or subtracted. 5x+ 7y=10 & (I) 10x+ 14y=10 & (II) Currently, none of the terms in this system will cancel out. However, if we multiply the first equation by - 2, then the x-variables will have opposite coefficients in both equations. - 2(5 x+7 y)=- 2(10) 10 x+14 y=10 ⇕ - 10x- 14y=- 20 10x+ 14y=10 Now, both the x- and y-terms will eliminate each other if we add Equation (I) to Equation (II).
Solving this system resulted in a contradiction. This means that the system has no solutions.
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Solving Systems of Equations by Elimination
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