Solve Each System By Elimination

The Elimination Method

Learning Objective(south)

· Solve a organization 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 awarding problems using the emptying method.

Introduction

The elimination method for solving systems of linear equations uses the addition holding of equality. Yous tin can add the aforementioned value to each side of an equation.

So if you have a system: x – six = −6 and ten + y = eight, you lot can add together 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 lot are adding the aforementioned value to each side of the starting time equation.

Using Addition to Eliminate a Variable

If you add the 2 equations, x – y = −6 and x + y = 8 together, as noted in a higher place, lookout what happens.

You have eliminated the y term, and this equation tin be solved using the methods for solving equations with 1 variable.

Let'due south encounter how this system is solved using the elimination method.

Case
Trouble	Use elimination to solve the organisation. x – y = − six 10 + y = 8
			Add the equations.
	2ten = 2 x = 1		Solve for x.
	x + y = 8 1 + y = 8 y = 8 – 1 y = 7		Substitute x = 1 into i of the original equations and solve for y.
	ten – y = −6 1 – 7 = −half-dozen −six = −6 TRUE	ten + y = 8 one + seven = eight eight = 8 True	Be sure to check your answer in both equations! The answers check.
Answer	The solution is (ane, seven).

Unfortunately not all systems work out this hands. How about a system like 2x + y = 12 and − 3x + y = two. If you add together these two equations together, no variables are eliminated.

But you desire to eliminate a variable. So allow's add the reverse of i of the equations to the other equation.

2x + y =12 → 2x + y = 12 → twox + y = 12

− 3x + y = 2 → − ( − 3x + y) = − (2) → three10 – y = − 2

510 + 0y = 10

You have eliminated the y variable, and the trouble can at present exist solved. Run into the example below.

Instance
Trouble	Use elimination to solve the system. 2ten + y = 12 − 3x + y = ii
	2ten + y = 12 − 3x + y = 2		You tin can eliminate the y-variable if you lot add the reverse of one of the equations to the other equation.
	twox + y = 12 iii10 – y = − 2 5x = 10		Rewrite the second equation as its opposite. Add together.
	x = 2		Solve for x.
	2(ii) + y = 12 4 + y = 12 y = viii		Substitute y = 2 into one of the original equations and solve for y.
	2x + y = 12 ii(2) + eight = 12 4 + viii = 12 12 = 12 True	− 3x + y = 2 − 3(2) + eight = 2 − 6 + 8 = two 2 = 2 True	Be sure to bank check your reply in both equations! The answers check.
Respond	The solution is (2, 8).

The post-obit are two more examples showing how to solve linear systems of equations using elimination.

Case

Problem

Utilise elimination to solve the organisation.

− 210 + 3y = − 1

2x + 5y = 25

		−2x	+	iiiy	=	−1
		iiten	+	5y	=	25

Notice the coefficients of each variable in each equation. If y'all add these ii equations, the 10 term will be eliminated since

−twox + 2x = 0.

−2x	+	3y	=	−1
twox	+	fivey	=	25
		viiiy	=	24
		y	=	three

Add and solve for y.

2x + fivey = 25

2x + 5(iii) = 25

2ten + 15 = 25

2ten = x

x = v

Substitute y = 3 into i of the original equations.

−2x + 3y = −1

−2(v) + iii(three) = −one

−10 + nine = −1

−1 = −1

TRUE

2x + 5y = 25

2(5) + 5(three) = 25

10 + 15 = 25

25 = 25

TRUE

Check solutions.

The answers check.

Answer

The solution is (5, 3).

Example

Problem

Use emptying to solve for ten and y.

4ten + 2y = 14

5x + 2y = sixteen

		ivten	+	twoy	=	14
		5x	+	2y	=	16

Notice the coefficients of each variable in each equation. You lot will need to add the opposite of one of the equations to eliminate the variable y, as 2y + 2y = foury, but

2y + (−2y) = 0.

4x	+	twoy	=	fourteen
−vx	–	iiy	=	−xvi
−ten			=	−two
		x	=	2

Change ane of the equations to its opposite, add and solve for x.

4x + 2y = fourteen

four(two) + 2y = 14

8 + 2y = fourteen

iiy = half dozen

y = iii

Substitute ten = 2 into 1 of the original equations and solve for y.

Reply

The solution is (ii, 3).

Go ahead and check this last case—substitute (2, iii) into both equations. You become two true statements: 14 = 14 and 16 = 16!

Notice that you could have used the contrary of the first equation rather than the second equation and gotten the same upshot.

Using Multiplication and Addition to Eliminate a Variables

Many times adding the equations or adding the contrary of one of the equations will non result in eliminating a variable. Await at the system beneath.

3x + 4y = 52

510 + y = xxx

If you add the equations higher up, or add the opposite of ane of the equations, yous will get an equation that still has two variables. So allow's now use the multiplication property of equality kickoff. You lot can multiply both sides of i of the equations by a number that will effect in the coefficient of ane of the variables being the reverse of the same variable in the other equation.

This is where multiplication comes in handy. Notice that the beginning equation contains the term 4y, and the second equation contains the term y. If you multiply the second equation past −4, when you add both equations the y variables volition add up to 0.

3ten + 4y = 52 → threex + ivy = 52 → iii10 + ivy =52

510 + y = 30 → − 4(5x + y) = − 4(30) → − xxx – 4y = − 120

− 17x + 0y = − 68

Run across the instance below.

Case

Trouble

Solve for x and y .

Equation A: 3x + 4y = 52

Equation B: 5x + y = 30

		3x	+	foury	=	52
		fivex	+	y	=	thirty

Look for terms that tin be eliminated. The equations exercise not take whatsoever x or y terms with the same coefficients.

	3x	+	foury	=	52
− iv(	5x	+	y )	=	− 4	(thirty)

Multiply the 2d equation past −4 so they do have the same coefficient.

		3x	+	ivy	=	52
		− 2010	–	4y	=	− 120

Rewrite the system, and add the equations.

		− 17x			=	-68
				x	=	iv

Solve for ten.

3x + 4y = 52

iii(4) + 4y = 52

12 + foury = 52

foury = xl

y = 10

Substitute ten = 4 into one of the original equations to find y.

3x + 4y = 52

iii(iv) + 4(10) = 52

12 + 40 = 52

52 = 52

TRUE

vx + y = xxx

five(four) + 10 = xxx

20 + ten = xxx

30 = 30

TRUE

Bank check your respond.

The answers check.

Respond

The solution is (4, 10).

At that place are other ways to solve this system. Instead of multiplying one equation in guild to eliminate a variable when the equations were added, you could take multiplied both equations by different numbers.

Let's remove the variable x this time. Multiply Equation A past v and Equation B by − 3.

Example

Problem

Solve for x and y.

3x + 4y = 52

fivex + y = 30

		310	+	4y	=	52
		5ten	+	y	=	thirty

Look for terms that tin be eliminated. The equations do not accept any x or y terms with the same coefficient.

5	(iii10	+	4y)	=	5	(52)
510		+	y	=	0

15x	+	20y	=	260
5ten	+	y	=	xxx

In social club to use the elimination method, y'all have to create variables that have the aforementioned coefficient—then you can eliminate them. Multiply the top equation by 5.

fifteen10	+	20y	=	260
-three (5ten	+	y )	=	−iii	(30)

fifteenx	+	xxy	=	260
−15x	–	3y	=	−ninety

Now multiply the bottom equation by −three.

fifteenten	+	twentyy	=	+	260
−fifteenx	–	3y	=	–	90
17y			=		170
y			=		x

Side by side add the equations, and solve for y.

3x + 4y = 52

3x + four(10) = 52

iiiten + xl = 52

3x = 12

x = 4

Substitute y = 10 into one of the original equations to notice x.

Respond

The solution is (4, 10).

You arrive at the same solution as before.

These equations were multiplied past v and − 3 respectively, because that gave you terms that would add together up to 0. Exist sure to multiply all of the terms of the equation.

Felix needs to find 10 and y in the following system.

Equation A: viiy − four10 = 5

Equation B: iiiy + 4x = 25

If he wants to use the elimination method to eliminate 1 of the variables, which is the nearly efficient way for him to do so?

A) Add Equation A and Equation B

B) Add 4x to both sides of Equation A

C) Multiply Equation A by 5

D) Multiply Equation B by − 1

Show/Hide Answer

A) Add Equation A and Equation B

Correct. If Felix adds the 2 equations, the terms iv10 and − ivx will cancel out, leaving xy = thirty. Felix will then easily exist able to solve for y.

B) Add together 4x to both sides of Equation A

Incorrect. Calculation 4x to both sides of Equation A will not alter the value of the equation, but information technology will non help eliminate either of the variables—you lot will end upwardly with the rewritten equation seveny = 5 + fourten. The right respond is to add Equation A and Equation B.

C) Multiply Equation A by five

Incorrect. Multiplying Equation A by 5 yields 35y − 20x = 25, which does non help you lot eliminate any of the variables in the system. Felix may notice that at present both equations have a constant of 25, but subtracting one from another is non an efficient way of solving this trouble. Instead, it would create another equation where both variables are present. The correct answer is to add together Equation A and Equation B.

D) Multiply Equation B past − one

Incorrect. Multiplying Equation B by − 1 yields − 3y – 4x = − 25, which does not help yous eliminate any of the variables in the system. Felix may discover that at present both equations have a term of − 4x, but calculation them would not eliminate them, information technology would give you a − 8x. The right reply is to add Equation A and Equation B.

Special Situations

But as with the substitution method, the elimination method will sometimes eliminate both variables, and you end up with either a true statement or a false statement. Retrieve that a faux statement ways that at that place is no solution.

Permit's look at an example.

Case
Problem	*Solve for ten* and y. - 10 – y = -four** 10 + y = 2
	-10 – y = -iv x + y = ii 0 = −2	Add the equations to eliminate the x -term.
Answer	There is no solution.

Graphing these lines shows that they are parallel lines and as such do not share any point in common, verifying that at that place is no solution.

If both variables are eliminated and y'all are left with a true statement, this indicates that in that location are an infinite number of ordered pairs that satisfy both of the equations. In fact, the equations are the aforementioned line.

Example
Problem	Solve for x and y. x + y = 2 - x − y = - 2
10 + y = 2 - x − y = -2 0 = 0		Add the equations to eliminate the x -term.
Answer	There are an infinite number of solutions.

Graphing these two equations will help to illustrate what is happening.

Solving Application Problems Using the Emptying 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.

Instance
Problem	The sum of two numbers is 10. Their difference is 6. What are the two numbers?
	ten + y = x 10 – y = 6		Write a system of equations to model the situation. x = ane number y = the other number
	ten + y = 10 + ten – y = 6 2x = 16 x = 8		Add the equations to eliminate the y-term and so solve for ten.
	x + y = 10 8 + y = x y = 2		Substitute the value for x into one of the original equations to notice y.
	ten + y = 10 8 + two = 10 10 = 10 TRUE	10 – y = 6 8 – 2 = 6 6 = six TRUE	Check your respond by substituting x = 8 and y = 2 into the original system. The answers cheque.
Reply	The numbers are 8 and 2.

Example
Problem	A theater sold 800 tickets for Fri night's performance. One kid ticket costs $iv.l and ane adult ticket costs $vi.00.The total amount nerveless was $four,500. How many of each type of ticket were sold?
	The total number of tickets sold is 800. a + c = 800 The amount of money nerveless is $4,500 6a + 4.fivec = four,500 System of equations: a + c = 800 6a + four.5c = iv,500		Write a system of equations to model the ticket sale situation. a = number of adult tickets sold c = number of child tickets sold
	half dozen(a + c) = 6(800) 6a + iv.vc = 4,500 6a + half dozenc = iv,800 6a + 4.5c = 4,500		Use multiplication to re-write the beginning equation.
	6a + 6c = 4,800 − 6a – 4.vc = −4,500 1.fivec =300 c = 200		Add the opposite of the second equation to eliminate a term and solve for c.
	a + 200 = 800 −200 −200 a = 600		Substitute 200 in for c in 1 of the original equations.
	a + c = 800 600 + 200 = 800 800 = 800 TRUE	6a + 4.5c = 4,500 6(600) + iv.5(200) = 4,500 iii,600 + 900 = iv,500 4500 = four,500 TRUE	Check your respond by substituting a = 600 and c = 200 into the original system. The answers bank check.
Answer	600 adult tickets and 200 child tickets were sold.

Summary

Combining equations is a powerful tool for solving a system of equations. Calculation or subtracting ii equations in order to eliminate a common variable is called the elimination (or improver) 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 earlier 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.