A Equation With No Solution

v.3: Solve Systems of Equations by Elimination

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Learning Objectives

Past the stop of this section, you volition exist able to:

• Solve a system of equations past elimination
• Solve applications of systems of equations by elimination
• Choose the virtually convenient method to solve a organisation of linear equations

Note

Before you become started, take this readiness quiz.

1. Simplify −5(six−3a).
   If you missed this trouble, review Exercise 1.10.43.
2. Solve the equation \(\frac{1}{3}10+\frac{5}{8}=\frac{31}{24}\).
   If you missed this trouble, review Practice 2.5.1.

We have solved systems of linear equations by graphing and by commutation. Graphing works well when the variable coefficients are small-scale and the solution has integer values. Substitution works well when we tin can hands solve one equation for ane of the variables and not take too many fractions in the resulting expression.

The third method of solving systems of linear equations is called the Elimination Method. When we solved a system by substitution, we started with two equations and 2 variables and reduced it to one equation with one variable. This is what we'll practice with the elimination method, too, but we'll have a unlike mode to go there.

Solve a Organisation of Equations by Emptying

The Elimination Method is based on the Addition Property of Equality. The Addition Property of Equality says that when you add together the aforementioned quantity to both sides of an equation, you withal have equality. Nosotros volition extend the Addition Property of Equality to say that when you add equal quantities to both sides of an equation, the results are equal.

For any expressions a, b, c, and d,

\[\begin{array}{lc} \text{ if } & a=b \\ \text { and } & c=d \\ \text { and so } &a+c =b+d \terminate{assortment}\]

To solve a system of equations past elimination, nosotros start with both equations in standard course. Then nosotros determine which variable will be easiest to eliminate. How do nosotros decide? We want to have the coefficients of i variable be opposites, so that we tin add the equations together and eliminate that variable.

Notice how that works when we add these two equations together:

\[\begin{array}{fifty} 3x+y=5 \\ \underline{2x-y=0} \\ 5x\quad\quad=5\end{array}\]

The y'due south add to zero and we take one equation with ane variable.

Let's endeavor another ane:

\[\left\{\begin{assortment}{l}{ten+4 y=2} \\ {2 ten+5 y=-2}\end{array}\correct.\]

This time we don't see a variable that can be immediately eliminated if we add the equations.

But if we multiply the get-go equation by −ii, nosotros volition make the coefficients of x opposites. We must multiply every term on both sides of the equation past −2.

Now nosotros encounter that the coefficients of the x terms are opposites, and so ten will exist eliminated when we add these ii equations.

Add the equations yourself—the issue should be −3y = −6. And that looks easy to solve, doesn't it? Here is what it would await similar.

We'll exercise one more:

\[\left\{\begin{array}{l}{four x-three y=10} \\ {3 x+5 y=-7}\terminate{assortment}\right.\]

It doesn't announced that we can get the coefficients of one variable to be opposites past multiplying one of the equations by a constant, unless we employ fractions. So instead, nosotros'll have to multiply both equations by a abiding.

We can make the coefficients of x exist opposites if we multiply the kickoff equation by iii and the second by −4, so we become 12x and −12x.

This gives us these two new equations:

\[\left\{\begin{aligned} 12 x-9 y &=30 \\-12 x-xx y &=28 \end{aligned}\right.\]

When we add together these equations,

\[\[\left\{\begin{array}{r}{12 x-ix y=30} \\ {\underline{-12 x-20 y=28}} \\\stop{array}\correct.\\\quad\qquad {-29 y=58}\]\]

the ten'southward are eliminated and nosotros just have −29y = 58.

Once we get an equation with just one variable, we solve it. Then we substitute that value into one of the original equations to solve for the remaining variable. And, as ever, we cheque our answer to make sure it is a solution to both of the original equations.

Now nosotros'll run into how to use emptying to solve the same system of equations we solved by graphing and by substitution.

Exercise \(\PageIndex{2}\)

Solve the arrangement by elimination. \(\left\{\begin{array}{l}{three 10+y=v} \\ {ii ten-3 y=7}\end{array}\correct.\)

Answer

(ii,−1)

Exercise \(\PageIndex{3}\)

Solve the arrangement by elimination. \(\left\{\begin{array}{l}{four x+y=-5} \\ {-2 x-two y=-2}\end{array}\right.\)

Respond

(−2,3)

The steps are listed below for easy reference.

HOW TO SOLVE A System OF EQUATIONS BY Emptying.

1. Write both equations in standard class. If any coefficients are fractions, clear them.
2. Brand the coefficients of 1 variable opposites.
   • Make up one's mind which variable you will eliminate.
   • Multiply one or both equations so that the coefficients of that variable are opposites.
3. Add together the equations resulting from Step ii to eliminate ane variable.
4. Solve for the remaining variable.
5. Substitute the solution from Step 4 into one of the original equations. And so solve for the other variable.
6. Write the solution as an ordered pair.
7. Bank check that the ordered pair is a solution to both original equations.

First we'll do an instance where we can eliminate 1 variable right away.

Practice \(\PageIndex{4}\)

Solve the organization by emptying. \(\left\{\brainstorm{array}{l}{10+y=10} \\ {x-y=12}\end{array}\right.\)

Answer

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, x.
Solve for x, the remaining variable. Substitute x = 11 into one of the original equations.
Solve for the other variable, y.
Write the solution as an ordered pair.	The ordered pair is (11, −i).
Check that the ordered pair is a solution to both original equations. \(\begin{array}{rllrll} ten+y &=&10 &x-y&=&12\\ 11+(-1) &\stackrel{?}{=}&10 & 11-(-1) &\stackrel{?}{=}&12\\ 10 &=&x \checkmark & 12 &=&12 \checkmark \end{assortment}\)
	The solution is (11, −one).

Practise \(\PageIndex{5}\)

Solve the system past emptying. \(\left\{\begin{assortment}{l}{2 x+y=5} \\ {10-y=4}\end{array}\right.\)

Answer

(iii,−1)

Exercise \(\PageIndex{6}\)

Solve the system by elimination.\(\left\{\begin{array}{l}{x+y=iii} \\ {-2 x-y=-1}\cease{array}\right.\)

Answer

(−2,5)

In Do \(\PageIndex{seven}\), we will be able to brand the coefficients of one variable opposites by multiplying one equation by a constant.

Exercise \(\PageIndex{eight}\)

Solve the system past elimination.\(\left\{\begin{assortment}{l}{4 x-3 y=one} \\ {5 x-9 y=-4}\cease{array}\right.\)

Answer

(i,1)

Exercise \(\PageIndex{9}\)

Solve the arrangement by elimination.\(\left\{\brainstorm{array}{l}{3 ten+ii y=2} \\ {6 x+5 y=8}\end{array}\correct.\)

Answer

(−2,iv)

At present we'll exercise an example where we need to multiply both equations by constants in guild to make the coefficients of ane variable opposites.

Exercise \(\PageIndex{10}\)

Solve the system by elimination. \(\left\{\begin{array}{fifty}{iv x-three y=9} \\ {7 10+2 y=-vi}\stop{array}\right.\)

Answer

In this example, we cannot multiply but one equation by any abiding to get opposite coefficients. So we will strategically multiply both equations by a constant to get the opposites.

Both equations are in standard course. To get opposite coefficients of y, nosotros will multiply the offset equation by 2 and the second equation by 3.
Simplify.
Add the two equations to eliminate y.
Solve for x. Substitute x = 0 into one of the original equations.
Solve for y.
Write the solution as an ordered pair.	The ordered pair is (0, −3).
Check that the ordered pair is a solution to both original equations. \(\begin{assortment}{rllrll} 4x-3y &=&9 &7x+2y&=&-half dozen\\ 4(0)-3(-3) &\stackrel{?}{=}&9 & 7(0)+2(-iii) &\stackrel{?}{=}&-6\\ix &=&9 \checkmark & -6 &=&-6 \checkmark \end{array}\)
	The solution is (0, −3).

Exercise \(\PageIndex{11}\)

Solve the arrangement by elimination. \(\left\{\brainstorm{array}{l}{3 x-4 y=-nine} \\ {five x+3 y=xiv}\end{array}\right.\)

Reply

(one,three)

Practise \(\PageIndex{12}\)

Solve the system by emptying. \(\left\{\begin{array}{l}{seven x+8 y=iv} \\ {iii 10-five y=27}\end{array}\right.\)

Respond

(four,−three)

When the system of equations contains fractions, we volition first articulate the fractions past multiplying each equation by its LCD.

Do \(\PageIndex{13}\)

Solve the system by emptying. \(\left\{\begin{assortment}{l}{x+\frac{one}{2} y=6} \\ {\frac{3}{2} x+\frac{ii}{3} y=\frac{17}{2}}\end{assortment}\correct.\)

Answer

In this example, both equations have fractions. Our start step will be to multiply each equation by its LCD to clear the fractions.

To articulate the fractions, multiply each equation by 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 by −four.
Simplify and add. Substitute 10 = 3 into one of the original equations.
Solve for y.
Write the solution as an ordered pair.	The ordered pair is (iii, six).
Check that the ordered pair is a solution to both original equations. \(\brainstorm{array}{rllrll} x+\frac{1}{2}y &=&6 &\frac{3}{2}x+\frac{2}{3}y&=&\frac{17}{2}\\ iii+\frac{one}{2}(vi) &\stackrel{?}{=}&half-dozen &\frac{three}{2}(iii) + \frac{2}{3}(half-dozen)&\stackrel{?}{=}&\frac{17}{ii}\\ iii + 3 &\stackrel{?}{=}&6 & \frac{9}{2 }+4 &\stackrel{?}{=} & \frac{17}{2}\\ vi &=&6 \checkmark & \frac{9}{2} + \frac{viii}{2} &\stackrel{?}{=} & \frac{17}{2}\\ && & \frac{17}{two} &=&\frac{17}{2} \checkmark \end{assortment}\)
	The solution is (three, six).

Exercise \(\PageIndex{14}\)

Solve the organisation by emptying. \(\left\{\brainstorm{array}{l}{\frac{1}{3} ten-\frac{ane}{2} y=1} \\ {\frac{iii}{four} 10-y=\frac{five}{2}}\end{array}\right.\)

Respond

(half-dozen,2)

Practice \(\PageIndex{15}\)

Solve the arrangement past elimination. \(\left\{\begin{array}{l}{10+\frac{iii}{5} y=-\frac{1}{v}} \\ {-\frac{1}{2} 10-\frac{ii}{3} y=\frac{5}{6}}\terminate{assortment}\right.\)

Reply

(1,−2)

In the Solving Systems of Equations by Graphing nosotros saw that non all systems of linear equations have a unmarried ordered pair equally a solution. When the two equations were really the same line, there were infinitely many solutions. Nosotros called that a consequent system. When the 2 equations described parallel lines, there was no solution. We called that an inconsistent system.

Practise \(\PageIndex{sixteen}\)

Solve the arrangement past elimination.\(\left\{\begin{assortment}{l}{3 x+4 y=12} \\ {y=3-\frac{3}{4} x}\finish{array}\right.\)

Answer

\(\begin{array} {ll} & \left\{\begin{aligned} three x+four y &=12 \\ y &=3-\frac{3}{four} x \end{aligned}\right. \\\\\text{Write the second equation in standard form.} & \left\{\begin{assortment}{l}{3 ten+4 y=12} \\ {\frac{three}{4} x+y=3}\end{array}\correct.\\ \\ \text{Clear the fractions past multiplying thesecond equation by 4.} & \left\{\begin{aligned} 3 x+4 y &=12 \\ 4\left(\frac{three}{iv} x+y\right) &=4(three) \end{aligned}\right. \\\\ \text{Simplify.} & \left\{\brainstorm{array}{fifty}{iii x+4 y=12} \\ {3 x+4 y=12}\end{assortment}\right.\\\\ \text{To eliminate a variable, nosotros multiply thesecond equation by −1.} & \left\{\begin{array}{c}{iii x+iv y=12} \\ \underline{-3 x-iv y=-12} \end{assortment}\correct.\\ &\qquad\qquad\quad 0=0 \\ \text{Simplify and add.} \end{array}\)

This is a true statement. The equations are consistent but dependent. Their graphs would be the same line. The organization has infinitely many solutions.

After we cleared the fractions in the 2d equation, did you notice that the two equations were the aforementioned? That means nosotros accept coincident lines.

Exercise \(\PageIndex{17}\)

Solve the arrangement by elimination. \(\left\{\brainstorm{array}{l}{v x-3 y=15} \\ {y=-5+\frac{5}{iii} x}\end{array}\correct.\)

Answer

infinitely many solutions

Exercise \(\PageIndex{18}\)

Solve the system past elimination. \(\left\{\brainstorm{array}{l}{ten+two y=half dozen} \\ {y=-\frac{1}{two} x+iii}\end{assortment}\correct.\)

Answer

infinitely many solutions

Exercise \(\PageIndex{19}\)

Solve the arrangement by elimination. \(\left\{\begin{array}{l}{-half-dozen x+15 y=10} \\ {2 10-5 y=-v}\terminate{array}\right.\)

Answer

\(\begin{assortment} {ll} \text{The equations are in standard grade.}& \left\{\begin{aligned}-6 x+15 y &=10 \\ ii x-5 y &=-5 \end{aligned}\correct. \\\\ \text{Multiply the 2nd equation by 3 to eliminate a variable.} & \left\{\begin{array}{l}{-6 10+15 y=10} \\ {3(2 x-5 y)=3(-5)}\terminate{array}\correct. \\\\ \text{Simplify and add.} & \left\{\begin{aligned}{-6 10+xv y =10} \\ \underline{6 x-xv y =-15} \end{aligned}\right. \\ & \qquad \qquad \quad0\neq 5 \end{array}\)

This statement is false. The equations are inconsistent and and so their graphs would be parallel lines.

The system does not take a solution.

Practise \(\PageIndex{20}\)

Solve the arrangement by elimination. \(\left\{\begin{array}{l}{-3 x+2 y=8} \\ {9 ten-6 y=xiii}\finish{array}\correct.\)

Answer

no solution

Do \(\PageIndex{21}\)

Solve the organisation by emptying. \(\left\{\brainstorm{array}{l}{7 x-3 y=-2} \\ {-14 x+6 y=eight}\end{array}\right.\)

Answer

no solution

Solve Applications of Systems of Equations by Elimination

Some applications problems interpret straight into equations in standard grade, so we volition use the elimination method to solve them. Equally before, nosotros employ our Problem Solving Strategy to help usa stay focused and organized.

Exercise \(\PageIndex{22}\)

The sum of ii numbers is 39. Their difference is nine. Find the numbers.

Reply

\(\begin{array} {ll} \textbf{Step 1. Read}\text{ the problem}& \\ \textbf{Step ii. Identify} \text{ what nosotros are looking for.} & \text{We are looking for two numbers.} \\\textbf{Step 3. Name} \text{ what we are looking for.} & \text{Allow n = the first number.} \\ & \text{ chiliad = the 2d number} \\\textbf{Step iv. Translate} \text{ into a arrangement of equations.}& \\ & \text{The sum of 2 numbers is 39.} \\ & n+chiliad=39\\ & \text{Their difference is 9.} \\ & n−m=9 \\ \\ \text{The organisation is:} & \left\{\begin{array}{50}{n+one thousand=39} \\ {north-m=9}\end{array}\right. \\\\ \textbf{Pace 5. Solve} \text{ the system of equations. } & \\ \text{To solve the organization of equations, use} \\ \text{elimination. The equations are in standard} \\ \text{class and the coefficients of m are} & \\ \text{opposites. Add.} & \left\{\begin{assortment}{l}{northward+thou=39} \\ \underline{northward-thousand=9}\cease{assortment}\correct. \\ &\quad 2n\qquad=48 \\ \\\text{Solve for n.} & north=24 \\ \\ \text{Substitute n=24 into one of the original} &n+m=39 \\ \text{equations and solve class.} & 24+m=39 \\ & chiliad=xv \\ \textbf{Step half-dozen. Check}\text{ the answer.} & \text{Since 24+15=39 and 24−15=9, the answers bank check.}\\ \textbf{Step vii. Reply} \text{ the question.} & \text{The numbers are 24 and 15.} \terminate{array}\)

Exercise \(\PageIndex{23}\)

The sum of 2 numbers is 42. Their deviation is 8. Find the numbers.

Respond

The numbers are 25 and 17.

Exercise \(\PageIndex{24}\)

The sum of two numbers is −15. Their difference is −35. Notice the numbers.

Answer

The numbers are −25 and 10.

Exercise \(\PageIndex{26}\)

Malik stops at the grocery shop to purchase a bag of diapers and 2 cans of formula. He spends a total of $37. The side by side calendar week he stops and buys 2 bags of diapers and 5 cans of formula for a total of $87. How much does a bag of diapers price? How much is 1 can of formula?

Answer

The bag of diapers costs $11 and the can of formula costs $13.

Practice \(\PageIndex{27}\)

To get her daily intake of fruit for the day, Sasha eats a banana and viii strawberries on Wednesday for a calorie count of 145. On the following Wednesday, she eats 2 bananas and 5 strawberries for a full of 235 calories for the fruit. How many calories are there in a banana? How many calories are in a strawberry?

Answer

There are 105 calories in a assistant and 5 calories in a strawberry.

Cull the Nearly Convenient Method to Solve a System of Linear Equations

When you will have to solve a system of linear equations in a later math class, yous volition ordinarily not be told which method to apply. You volition demand to make that determination yourself. So you'll want to choose the method that is easiest to exercise and minimizes your chance of making mistakes.

Exercise \(\PageIndex{28}\)

For each organization of linear equations decide whether information technology would be more convenient to solve it by substitution or elimination. Explain your respond.

1. \(\left\{\brainstorm{assortment}{l}{3 10+8 y=40} \\ {7 x-iv y=-32}\end{array}\right.\)
2. \(\left\{\brainstorm{array}{l}{5 x+half dozen y=12} \\ {y=\frac{2}{3} x-one}\end{assortment}\right.\)

Answer

one. \(\left\{\begin{array}{l}{iii ten+viii y=40} \\ {seven x-iv y=-32}\end{assortment}\right.\)

Since both equations are in standard class, using elimination volition be most convenient.

2. \(\left\{\begin{array}{50}{v x+6 y=12} \\ {y=\frac{ii}{3} x-i}\end{assortment}\right.\)

Since one equation is already solved for y, using substitution volition be most convenient.

Exercise \(\PageIndex{29}\)

For each arrangement of linear equations, decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

1. \(\left\{\begin{array}{l}{iv x-v y=-32} \\ {three ten+2 y=-1}\cease{array}\right.\)
2. \(\left\{\begin{array}{l}{x=2 y-1} \\ {3 ten-five y=-7}\end{array}\right.\)

Reply

1. Since both equations are in standard grade, using elimination volition be most convenient.
2. Since one equation is already solved for twenty, using exchange volition exist well-nigh convenient.

Exercise \(\PageIndex{xxx}\)

For each system of linear equations, decide whether it would exist more than convenient to solve it by substitution or emptying. Explicate your answer.

1. \(\left\{\begin{array}{l}{y=two x-1} \\ {iii x-four y=-vi}\end{array}\right.\)
2. \(\left\{\brainstorm{array}{l}{6 x-2 y=12} \\ {iii x+7 y=-13}\finish{array}\right.\)

Reply

1. Since one equation is already solved for yy, using exchange will be virtually convenient;
2. Since both equations are in standard grade, using elimination will exist most convenient.

Key Concepts

• To Solve a System of Equations past Emptying
  1. Write both equations in standard form. If whatsoever coefficients are fractions, clear them.
  2. Make the coefficients of 1 variable opposites.
     • Decide which variable y'all will eliminate.
     • Multiply one or both equations so that the coefficients of that variable are opposites.
  3. Add the equations resulting from Step 2 to eliminate one variable.
  4. Solve for the remaining variable.
  5. Substitute the solution from Step 4 into ane of the original equations. Then solve for the other variable.
  6. Write the solution as an ordered pair.
  7. Check that the ordered pair is a solution to both original equations.

A Equation With No Solution,

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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