How Do You Know There Is One Solution in an Equation
5.3: Solve Systems of Equations by Elimination
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Past the cease of this department, you will be able to:
- Solve a organisation of equations by elimination
- Solve applications of systems of equations by elimination
- Choose the most user-friendly method to solve a arrangement of linear equations
Before you go started, have this readiness quiz.
- Simplify −v(6−3a).
If y'all missed this problem, review Practise 1.ten.43. - Solve the equation \(\frac{1}{iii}x+\frac{5}{viii}=\frac{31}{24}\).
If you missed this problem, review Exercise 2.5.1.
We have solved systems of linear equations past graphing and by substitution. Graphing works well when the variable coefficients are small and the solution has integer values. Exchange works well when nosotros tin can easily solve one equation for one of the variables and non have too many fractions in the resulting expression.
The third method of solving systems of linear equations is chosen the Emptying Method. When we solved a arrangement by exchange, we started with ii equations and 2 variables and reduced it to one equation with one variable. This is what we'll practise with the elimination method, as well, but we'll have a unlike mode to get there.
Solve a System of Equations by Emptying
The Elimination Method is based on the Addition Property of Equality. The Add-on Property of Equality says that when you add together the aforementioned quantity to both sides of an equation, you lot yet accept equality. Nosotros will extend the Addition Belongings of Equality to say that when you add together equal quantities to both sides of an equation, the results are equal.
For any expressions a, b, c, and d,
\[\begin{array}{lc} \text{ if } & a=b \\ \text { and } & c=d \\ \text { and then } &a+c =b+d \end{assortment}\]
To solve a system of equations by emptying, nosotros starting time with both equations in standard grade. Then we make up one's mind which variable will exist easiest to eliminate. How exercise we decide? We want to take the coefficients of ane variable be opposites, and then that we can add the equations together and eliminate that variable.
Notice how that works when we add these ii equations together:
\[\begin{array}{l} 3x+y=v \\ \underline{2x-y=0} \\ 5x\quad\quad=5\end{array}\]
The y's add to zero and we take one equation with 1 variable.
Permit'due south try another i:
\[\left\{\brainstorm{array}{50}{x+iv y=ii} \\ {2 x+5 y=-2}\cease{array}\right.\]
This fourth dimension we don't see a variable that tin can exist immediately eliminated if we add together the equations.
Merely if we multiply the get-go equation by −2, nosotros will make the coefficients of x opposites. We must multiply every term on both sides of the equation by −2.
Now nosotros see that the coefficients of the x terms are opposites, so 10 volition be eliminated when we add these two equations.
Add together the equations yourself—the outcome should exist −3y = −half-dozen. And that looks easy to solve, doesn't it? Here is what it would wait like.
We'll do one more:
\[\left\{\begin{array}{l}{4 ten-3 y=10} \\ {iii ten+v y=-7}\end{array}\right.\]
It doesn't appear that we can get the coefficients of one variable to exist opposites by multiplying ane of the equations past a constant, unless we use fractions. So instead, we'll have to multiply both equations by a abiding.
We tin can brand the coefficients of x be opposites if we multiply the first equation by 3 and the second by −4, so we get 12x and −12x.
This gives usa these 2 new equations:
\[\left\{\begin{aligned} 12 x-9 y &=30 \\-12 ten-20 y &=28 \stop{aligned}\correct.\]
When nosotros add these equations,
\[\[\left\{\begin{array}{r}{12 x-9 y=xxx} \\ {\underline{-12 x-20 y=28}} \\\end{array}\right.\\\quad\qquad {-29 y=58}\]\]
the x's are eliminated and we simply have −29y = 58.
Once we become an equation with just 1 variable, nosotros solve information technology. Then we substitute that value into 1 of the original equations to solve for the remaining variable. And, as e'er, we cheque our answer to brand sure information technology is a solution to both of the original equations.
Now we'll see how to utilize elimination to solve the same organisation of equations nosotros solved by graphing and by commutation.
Solve the system past elimination. \(\left\{\begin{array}{l}{iii x+y=5} \\ {two x-3 y=7}\end{assortment}\right.\)
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(2,−1)
Solve the organisation by elimination. \(\left\{\begin{array}{50}{4 x+y=-five} \\ {-2 x-2 y=-two}\stop{array}\right.\)
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(−2,3)
The steps are listed below for like shooting fish in a barrel reference.
- Write both equations in standard class. If any coefficients are fractions, articulate them.
- Brand the coefficients of ane variable opposites.
- Decide which variable you will eliminate.
- Multiply one or both equations so that the coefficients of that variable are opposites.
- Add the equations resulting from Step 2 to eliminate 1 variable.
- Solve for the remaining variable.
- Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
- Write the solution every bit an ordered pair.
- Check that the ordered pair is a solution to both original equations.
Start nosotros'll practice an example where we can eliminate 1 variable right away.
Solve the system by elimination. \(\left\{\begin{array}{50}{x+y=10} \\ {x-y=12}\finish{array}\right.\)
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Both equations are in standard form. The coefficients of y are already opposites. Add the two equations to eliminate y.
The resulting equation has only 1 variable, 10.
Solve for ten, the remaining variable. Substitute x = 11 into one of the original equations.
Solve for the other variable, y. 
Write the solution every bit an ordered pair. The ordered pair is (11, −ane). Bank check that the ordered pair is a solution
to both original equations.\(\begin{assortment}{rllrll} 10+y &=&ten &x-y&=&12\\ 11+(-one) &\stackrel{?}{=}&10 & 11-(-1) &\stackrel{?}{=}&12\\ 10 &=&10 \checkmark & 12 &=&12 \checkmark \end{array}\)
The solution is (xi, −ane).
Solve the organisation by elimination. \(\left\{\begin{array}{l}{2 x+y=5} \\ {x-y=four}\end{array}\correct.\)
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(iii,−1)
Solve the system by elimination.\(\left\{\begin{array}{l}{x+y=3} \\ {-2 10-y=-one}\cease{array}\right.\)
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(−2,5)
In Exercise \(\PageIndex{7}\), we will exist able to make the coefficients of i variable opposites by multiplying 1 equation by a abiding.
Solve the organization by emptying.\(\left\{\begin{array}{l}{4 ten-3 y=1} \\ {5 x-9 y=-4}\terminate{assortment}\right.\)
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(1,one)
Solve the organization by elimination.\(\left\{\brainstorm{array}{50}{3 x+two y=ii} \\ {6 10+5 y=8}\cease{array}\right.\)
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(−2,four)
Now we'll exercise an example where we need to multiply both equations by constants in order to make the coefficients of 1 variable opposites.
Solve the system by emptying. \(\left\{\begin{array}{l}{4 x-iii y=nine} \\ {vii x+2 y=-half-dozen}\end{assortment}\correct.\)
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In this instance, we cannot multiply but i equation by any constant to get opposite coefficients. So we volition strategically multiply both equations by a constant to get the opposites.
Both equations are in standard class. To get opposite
coefficients of y, nosotros will multiply the offset equation by 2
and the second equation by iii.
Simplify. 
Add the 2 equations to eliminate y. 
Solve for x. Substitute x = 0 into i of the original equations.
Solve for y. 
Write the solution as an ordered pair. The ordered pair is (0, −three). Check that the ordered pair is a solution to
both original equations.\(\brainstorm{assortment}{rllrll} 4x-3y &=&ix &7x+2y&=&-half dozen\\ iv(0)-3(-3) &\stackrel{?}{=}&ix & seven(0)+2(-3) &\stackrel{?}{=}&-half-dozen\\9 &=&9 \checkmark & -6 &=&-6 \checkmark \end{assortment}\)
The solution is (0, −3).
Solve the system by emptying. \(\left\{\brainstorm{array}{l}{three 10-4 y=-9} \\ {5 x+3 y=14}\end{array}\right.\)
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(1,iii)
Solve the system by elimination. \(\left\{\brainstorm{assortment}{fifty}{seven x+viii y=4} \\ {three ten-5 y=27}\terminate{array}\correct.\)
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(4,−3)
When the system of equations contains fractions, we will first clear the fractions past multiplying each equation by its LCD.
Solve the arrangement by emptying. \(\left\{\begin{array}{l}{x+\frac{1}{2} y=6} \\ {\frac{3}{2} x+\frac{2}{iii} y=\frac{17}{2}}\end{array}\right.\)
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In this instance, both equations have fractions. Our commencement pace volition be to multiply each equation by its LCD to articulate the fractions.
To articulate the fractions, multiply each equation past its LCD. 
Simplify. 
Now we are ready to eliminate one of the variables. Detect that
both equations are in standard form.We can eliminate y multiplying the top equation past −iv. 
Simplify and add together. Substitute ten = 3 into one of the original equations.
Solve for y. 
Write the solution every bit an ordered pair. The ordered pair is (3, 6). Bank check that the ordered pair is a solution
to both original equations.\(\brainstorm{array}{rllrll} ten+\frac{one}{2}y &=&6 &\frac{iii}{2}10+\frac{2}{3}y&=&\frac{17}{ii}\\ three+\frac{1}{two}(6) &\stackrel{?}{=}&6 &\frac{three}{2}(three) + \frac{ii}{3}(six)&\stackrel{?}{=}&\frac{17}{two}\\ iii + 3 &\stackrel{?}{=}&half-dozen & \frac{nine}{2 }+4 &\stackrel{?}{=} & \frac{17}{ii}\\ 6 &=&six \checkmark & \frac{ix}{ii} + \frac{eight}{2} &\stackrel{?}{=} & \frac{17}{2}\\ && & \frac{17}{2} &=&\frac{17}{2} \checkmark \end{array}\)
The solution is (iii, vi).
Solve the system by elimination. \(\left\{\brainstorm{array}{l}{\frac{ane}{3} x-\frac{one}{2} y=one} \\ {\frac{3}{four} ten-y=\frac{5}{2}}\end{assortment}\right.\)
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(vi,2)
Solve the arrangement by elimination. \(\left\{\begin{assortment}{l}{ten+\frac{3}{5} y=-\frac{i}{5}} \\ {-\frac{i}{2} x-\frac{2}{3} y=\frac{5}{half dozen}}\end{array}\right.\)
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(1,−two)
In the Solving Systems of Equations by Graphing we saw that not all systems of linear equations have a single ordered pair as a solution. When the two equations were really the same line, there were infinitely many solutions. Nosotros called that a consistent system. When the two equations described parallel lines, there was no solution. We called that an inconsistent system.
Solve the system by elimination.\(\left\{\brainstorm{array}{l}{three x+4 y=12} \\ {y=iii-\frac{three}{four} x}\end{array}\correct.\)
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\(\begin{array} {ll} & \left\{\begin{aligned} 3 x+4 y &=12 \\ y &=three-\frac{3}{four} x \end{aligned}\right. \\\\\text{Write the second equation in standard course.} & \left\{\brainstorm{array}{l}{3 x+4 y=12} \\ {\frac{three}{4} x+y=iii}\end{array}\right.\\ \\ \text{Clear the fractions past multiplying thesecond equation by iv.} & \left\{\begin{aligned} three 10+iv y &=12 \\ 4\left(\frac{three}{4} x+y\right) &=4(iii) \end{aligned}\right. \\\\ \text{Simplify.} & \left\{\begin{array}{l}{3 x+4 y=12} \\ {3 x+4 y=12}\end{array}\right.\\\\ \text{To eliminate a variable, nosotros multiply thesecond equation by −1.} & \left\{\brainstorm{array}{c}{iii x+4 y=12} \\ \underline{-3 10-4 y=-12} \end{array}\right.\\ &\qquad\qquad\quad 0=0 \\ \text{Simplify and add.} \end{assortment}\)
This is a true statement. The equations are consistent merely dependent. Their graphs would be the same line. The arrangement has infinitely many solutions.
After we cleared the fractions in the second equation, did y'all discover that the two equations were the same? That means we have ancillary lines.
Solve the system past emptying. \(\left\{\brainstorm{array}{50}{5 x-3 y=15} \\ {y=-5+\frac{5}{3} x}\end{array}\correct.\)
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infinitely many solutions
Solve the system past emptying. \(\left\{\begin{array}{l}{10+2 y=6} \\ {y=-\frac{ane}{2} x+iii}\end{array}\correct.\)
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infinitely many solutions
Solve the system past elimination. \(\left\{\brainstorm{assortment}{l}{-half dozen x+15 y=10} \\ {two x-5 y=-5}\end{array}\right.\)
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\(\begin{assortment} {ll} \text{The equations are in standard course.}& \left\{\brainstorm{aligned}-6 ten+fifteen y &=10 \\ 2 x-5 y &=-5 \end{aligned}\right. \\\\ \text{Multiply the second equation by iii to eliminate a variable.} & \left\{\begin{array}{50}{-half dozen x+fifteen y=10} \\ {3(ii x-5 y)=3(-5)}\end{array}\right. \\\\ \text{Simplify and add together.} & \left\{\brainstorm{aligned}{-6 x+15 y =10} \\ \underline{6 x-15 y =-xv} \end{aligned}\correct. \\ & \qquad \qquad \quad0\neq 5 \end{array}\)
This statement is false. The equations are inconsistent and so their graphs would be parallel lines.
The arrangement does non have a solution.
Solve the system past elimination. \(\left\{\begin{array}{l}{-three x+ii y=8} \\ {9 x-6 y=13}\stop{array}\right.\)
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no solution
Solve the organization by elimination. \(\left\{\begin{array}{l}{7 10-3 y=-2} \\ {-fourteen x+6 y=8}\cease{array}\right.\)
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no solution
Solve Applications of Systems of Equations by Elimination
Some applications problems interpret directly into equations in standard course, so nosotros will utilise the elimination method to solve them. As before, we use our Problem Solving Strategy to aid us stay focused and organized.
The sum of two numbers is 39. Their difference is nine. Notice the numbers.
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\(\begin{array} {ll} \textbf{Step i. Read}\text{ the problem}& \\ \textbf{Step two. Identify} \text{ what we are looking for.} & \text{We are looking for 2 numbers.} \\\textbf{Step 3. Name} \text{ what nosotros are looking for.} & \text{Let northward = the first number.} \\ & \text{ m = the 2nd number} \\\textbf{Step 4. Translate} \text{ into a system of equations.}& \\ & \text{The sum of ii numbers is 39.} \\ & due north+m=39\\ & \text{Their difference is ix.} \\ & n−m=9 \\ \\ \text{The arrangement is:} & \left\{\begin{array}{l}{n+yard=39} \\ {n-m=ix}\end{array}\correct. \\\\ \textbf{Stride 5. Solve} \text{ the organisation of equations. } & \\ \text{To solve the system of equations, apply} \\ \text{emptying. The equations are in standard} \\ \text{form and the coefficients of 1000 are} & \\ \text{opposites. Add.} & \left\{\brainstorm{array}{fifty}{northward+one thousand=39} \\ \underline{northward-m=9}\end{array}\right. \\ &\quad 2n\qquad=48 \\ \\\text{Solve for n.} & n=24 \\ \\ \text{Substitute northward=24 into i of the original} &n+grand=39 \\ \text{equations and solve form.} & 24+m=39 \\ & chiliad=15 \\ \textbf{Step 6. Cheque}\text{ the answer.} & \text{Since 24+xv=39 and 24−fifteen=9, the answers bank check.}\\ \textbf{Step 7. Reply} \text{ the question.} & \text{The numbers are 24 and 15.} \finish{array}\)
The sum of two numbers is 42. Their deviation is eight. Find the numbers.
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The numbers are 25 and 17.
The sum of two numbers is −15. Their divergence is −35. Observe the numbers.
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The numbers are −25 and 10.
Malik stops at the grocery store to buy a handbag of diapers and ii cans of formula. He spends a total of $37. The next calendar week he stops and buys 2 bags of diapers and 5 cans of formula for a total of $87. How much does a pocketbook of diapers cost? How much is 1 tin can of formula?
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The bag of diapers costs $eleven and the can of formula costs $13.
To get her daily intake of fruit for the mean solar day, Sasha eats a banana and viii strawberries on Wed for a calorie count of 145. On the following Wed, she eats ii bananas and 5 strawberries for a full of 235 calories for the fruit. How many calories are there in a assistant? How many calories are in a strawberry?
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There are 105 calories in a assistant and 5 calories in a strawberry.
Choose the Almost Convenient Method to Solve a Organisation of Linear Equations
When y'all will have to solve a arrangement of linear equations in a later math class, you volition usually not be told which method to use. You lot will need to make that decision yourself. So you lot'll want to choose the method that is easiest to do and minimizes your chance of making mistakes.
For each system of linear equations determine whether information technology would be more convenient to solve it by commutation or elimination. Explain your answer.
- \(\left\{\begin{array}{l}{iii x+8 y=40} \\ {7 x-4 y=-32}\end{array}\right.\)
- \(\left\{\brainstorm{array}{l}{5 x+half-dozen y=12} \\ {y=\frac{2}{iii} x-1}\end{assortment}\right.\)
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1. \(\left\{\begin{array}{l}{three x+8 y=40} \\ {7 ten-4 y=-32}\terminate{array}\right.\)
Since both equations are in standard class, using elimination will exist most convenient.
2. \(\left\{\begin{array}{l}{five 10+6 y=12} \\ {y=\frac{2}{3} x-1}\end{array}\right.\)
Since one equation is already solved for y, using substitution will be most user-friendly.
For each organisation of linear equations, determine whether it would be more convenient to solve it past substitution or elimination. Explain your reply.
- \(\left\{\begin{array}{l}{4 x-5 y=-32} \\ {3 x+2 y=-1}\end{array}\right.\)
- \(\left\{\begin{array}{fifty}{10=2 y-1} \\ {iii ten-5 y=-seven}\end{array}\correct.\)
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- Since both equations are in standard form, using elimination volition exist most user-friendly.
- Since one equation is already solved for twenty, using substitution volition be near user-friendly.
For each system of linear equations, decide whether it would be more than convenient to solve it past exchange or emptying. Explain your respond.
- \(\left\{\begin{assortment}{l}{y=2 x-1} \\ {3 x-4 y=-half-dozen}\end{array}\right.\)
- \(\left\{\begin{array}{l}{six x-2 y=12} \\ {three 10+7 y=-13}\end{array}\right.\)
- Answer
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- Since one equation is already solved for yy, using substitution volition be near user-friendly;
- Since both equations are in standard form, using elimination will be almost convenient.
Central Concepts
- To Solve a Organization of Equations by Elimination
- Write both equations in standard form. If any coefficients are fractions, clear them.
- Make the coefficients of one variable opposites.
- Decide which variable you will eliminate.
- Multiply ane or both equations so that the coefficients of that variable are opposites.
- Add the equations resulting from Footstep 2 to eliminate ane variable.
- Solve for the remaining variable.
- Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
- Write the solution equally an ordered pair.
- Bank check that the ordered pair is a solution to both original equations.
Source: https://math.libretexts.org/Courses/Monroe_Community_College/MTH_098_Elementary_Algebra/5%3A_Systems_of_Linear_Equations/5.3%3A_Solve_Systems_of_Equations_by_Elimination
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