How Do You Know When to Add or Subtract in Elimination

The Elimination Method

Learning Objective(southward)

· Solve a organization of equations when no multiplication is necessary to eliminate a variable.

· Solve a organisation of equations when multiplication is necessary to eliminate a variable.

· Recognize systems that have no solution or an space number of solutions.

· Solve awarding problems using the elimination method.

Introduction

The elimination method for solving systems of linear equations uses the add-on property of equality. You can add the same value to each side of an equation.

And so if y'all have a system: x – six = −half dozen and ten + 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 commencement equation.

Using Addition to Eliminate a Variable

If you add the ii equations, x – y = −half dozen and 10 + y = eight together, equally noted in a higher place, watch what happens.

You have eliminated the y term, and this equation can exist solved using the methods for solving equations with i variable.

Allow's see how this organisation is solved using the emptying method.

Case
Problem	Use emptying to solve the system.    x – y = − six x + y = 8
			Add the equations.
	iix = ii x = one		Solve for x.
	10 + y = 8 1 + y = 8 y = viii – 1 y = 7		Substitute x = i into one of the original equations and solve for y.
	x – y = −half-dozen 1 – 7 = −vi −6 = −6 TRUE	x + y = 8 i + 7 = eight 8 = 8 True	Exist sure to check your answer in both equations! The answers cheque.
Answer	The solution is (1, 7).

Unfortunately non all systems work out this hands. How about a organisation like 2x + y = 12 and − 3x + y = ii. If you add these two equations together, no variables are eliminated.

But yous want to eliminate a variable. So let'due south add together the opposite of one of the equations to the other equation.

 twox + y =12      →        2x + y = 12      →       210 + y = 12

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

                                                                                    5x + 0y = 10

You have eliminated the y variable, and the problem can now be solved. See the instance below.

Example
Problem	Employ elimination to solve the organization. 210 + y = 12 − 3ten + y =  2
	2x + y = 12 − 3x + y =  two		You can eliminate the y-variable if you add the opposite of one of the equations to the other equation.
	two10 + y = 12             iiix – y = − 2 5x = 10		Rewrite the 2d equation as its opposite. Add.
	x = 2		Solve for x.
	2(2) + y = 12 4 + y = 12 y = 8		Substitute y = ii into ane of the original equations and solve for y.
	2x + y = 12 2(two) + 8 = 12 4 + 8 = 12 12 = 12             True	− threex + y =  2 − 3(two) + 8 = two − 6 + 8 = 2 2 = 2 TRUE	Exist sure to check your answer in both equations! The answers check.
Answer	The solution is (2, 8).

The following are two more than examples showing how to solve linear systems of equations using elimination.

Example
Trouble	Use elimination to solve the system. − 210  + 3y = − i    two10  + fivey = 25
	−2x + 3y = −1 twox + 5y = 25		Detect the coefficients of each variable in each equation. If you add these two equations, the ten term will be eliminated since −2x + two10 = 0.
	−2x + 3y = −1 2x + 5y = 25 8y = 24 y = three		Add and solve for y.
	2ten + 5y = 25 2x + 5(iii) = 25 210 + fifteen = 25 twox = 10 x = five		Substitute y = 3 into 1 of the original equations.
	−2x + 3y = −ane −ii(5) + three(3) = −i −10 + 9 = −ane −i = −1 True	iix + fivey = 25 2(5) + 5(3) = 25 10 + xv = 25 25 = 25 TRUE	Bank check solutions. The answers check.
Answer	The solution is (5, 3).

Example
Trouble	Utilise elimination to solve for x and y. ivten  + 2y = 14 vx  + 2y = 16
	fourx + 2y = 14 5x + 2y = 16	Detect the coefficients of each variable in each equation. Y'all will need to add together the opposite of one of the equations to eliminate the variable y, as twoy + 2y = 4y, simply twoy + (−2y) = 0.
	410 + 2y = fourteen −510 – 2y = −16 −10 = −2 10 = 2	Change one of the equations to its reverse, add and solve for x.
	4x + twoy = 14 4(2) + 2y = xiv 8 + twoy = 14 2y = 6 y = three	Substitute x = 2 into one of the original equations and solve for y.
Answer	The solution is (2, 3).

Go ahead and check this last example—substitute (2, 3) into both equations. You go two truthful statements: 14 = 14 and sixteen = 16!

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

Using Multiplication and Add-on to Eliminate a Variables

Many times calculation the equations or adding the contrary of one of the equations will non event in eliminating a variable. Look at the arrangement beneath.

3ten + 4y = 52

vx + y = xxx

If you add the equations above, or add together the reverse of one of the equations, you will get an equation that nevertheless has two variables. So let's at present use the multiplication holding of equality outset. You can multiply both sides of one of the equations by a number that volition event in the coefficient of one of the variables being the contrary of the same variable in the other equation.

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

3x + ivy = 52    →        3x + ivy = 52                →             3x + 4y =52

5ten + y = 30      → − iv(5ten + y) = − 4(30)      → − twentyx – 4y = − 120

− 17ten + 0y = − 68

See the instance below.

Example
Problem	Solve for x and y . Equation A: iiiten + foury = 52 Equation B: 5x + y = 30
	threex + 4y = 52 5x + y = 30		Look for terms that can be eliminated. The equations exercise not take any 10 or y terms with the aforementioned coefficients.
	310 + 4y = 52 − four( 510 + y ) = − 4 (30)		Multiply the second equation past −four so they practice take the aforementioned coefficient.
	iiix + ivy = 52 − 20x – ivy = − 120		Rewrite the system, and add together the equations.
	− 17x = -68 10 = 4		Solve for x.
	3x + 4y = 52 3(4) + 4y = 52 12 + ivy = 52 4y = forty y = ten		Substitute ten = 4 into one of the original equations to find y.
	3x + 4y = 52 3(four) + 4(10) = 52 12 + xl = 52 52 = 52 Truthful	5x + y = 30 5(four) + 10 = thirty 20 + 10 = thirty thirty = xxx TRUE	Check your reply. The answers check.
Answer	The solution is (iv, ten).

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 past five and Equation B by − 3.

Example
Problem	Solve for ten and y. iiix + foury = 52 510 + y = 30
	3x + 4y = 52 5x + y = thirty	Expect for terms that can be eliminated. The equations do not accept whatsoever x or y terms with the same coefficient.
	five (3x + 4y) = v (52) 5x + y = 0 15x + 20y = 260 5x + y = 30	In order to use the emptying method, you have to create variables that have the same coefficient—then yous can eliminate them. Multiply the top equation by 5.
	fifteenx + 20y = 260     -3 (v10 + y ) = −3 (30) 15x + twentyy = 260 −15x – 3y = −90	At present multiply the bottom equation by −3.
	xvx + 20y = + 260 −15x – iiiy = – 90 17y = 170 y = 10	Next add the equations, and solve for y.
	3x + 4y = 52 threex + 4(ten) = 52 3ten + 40 = 52 iiix = 12 x = 4	Substitute y = 10 into one of the original equations to detect x.
Answer	The solution is (4, x).	You arrive at the same solution as before.

These equations were multiplied by 5 and − three respectively, considering that gave you terms that would add upwardly to 0. Be sure to multiply all of the terms of the equation.

Felix needs to observe x and y in the post-obit organization.

Equation A: 7y − 4x = 5

Equation B: 3y + fourx = 25

If he wants to utilise the elimination method to eliminate one of the variables, which is the most efficient way for him to do so?

A) Add Equation A and Equation B

B) Add 410 to both sides of Equation A

C) Multiply Equation A past 5

D) Multiply Equation B by − 1

Prove/Hide Answer

A) Add together Equation A and Equation B

Correct. If Felix adds the 2 equations, the terms 4x and − 4x volition cancel out, leaving 10y = 30. Felix will then easily be able to solve for y.

B) Add fourx to both sides of Equation A

Incorrect. Adding 4ten to both sides of Equation A will not modify the value of the equation, but information technology volition not help eliminate either of the variables—you volition cease up with the rewritten equation seveny = 5 + 4x. The correct respond is to add Equation A and Equation B.

C) Multiply Equation A by 5

Incorrect. Multiplying Equation A by 5 yields 35y − 20x = 25, which does not assistance you eliminate any of the variables in the system. Felix may notice that at present both equations take a constant of 25, but subtracting i from another is non an efficient mode of solving this problem. Instead, it would create another equation where both variables are present. The correct answer is to add Equation A and Equation B.

D) Multiply Equation B past − i

Incorrect. Multiplying Equation B by − 1 yields − threey – ivx = − 25, which does not aid y'all eliminate any of the variables in the organisation. Felix may notice that now both equations have a term of − ivten, but adding them would not eliminate them, it would give y'all a − 8x. The correct respond is to add Equation A and Equation B.

Special Situations

Just equally with the substitution method, the elimination method will sometimes eliminate both variables, and you cease upwardly with either a truthful statement or a simulated argument. Call up that a false statement means that there is no solution.

Let'due south look at an case.

Example
Trouble	Solve for x and y. - ten – y = -4 x + y = 2
	-x – y  = -4 x + y   = 2 0 = −2	Add together the equations to eliminate the x -term.
Answer	There is no solution.

Graphing these lines shows that they are parallel lines and every bit 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 in that location are an infinite number of ordered pairs that satisfy both of the equations. In fact, the equations are the same line.

Example
Problem	Solve for x and y. 10 + y = 2 - 10 − y = - 2
x + y  =  2 - x − y = -2  0 =   0		Add the equations to eliminate the ten -term.
Reply	At that place are an infinite number of solutions.

Graphing these two equations volition aid to illustrate what is happening.

Solving Application Bug Using the Emptying Method

The elimination method can be applied to solving systems of equations that model existent 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 six. What are the two numbers?
	x + y  = 10   x – y  = 6		Write a system of equations to model the situation. x = one number y = the other number
	10 + y = x   + x – y = six 210 = 16 x  = 8		Add the equations to eliminate the y-term and then solve for x.
	x + y = x viii + y = 10  y  = 2		Substitute the value for ten into i of the original equations to find y.
	10 + y  = 10 8 + two = 10 10 = 10 TRUE	10 – y = 6 8 – ii = 6 6 = 6 TRUE	Cheque your answer by substituting 10 = 8 and y = two into the original organization. The answers check.
Answer	The numbers are 8 and 2.

Instance
Problem	A theater sold 800 tickets for Fri night'southward functioning. One child ticket costs $4.50 and one adult ticket costs $six.00.The full amount nerveless was $iv,500. How many of each blazon of ticket were sold?
	The total number of tickets sold is 800. a + c  =  800 The amount of coin collected is $4,500    6a  + 4.5c  = 4,500 Organization of equations: a + c  =  800 6a + four.5c  = 4,500		Write a system of equations to model the ticket auction state of affairs. a = number of developed tickets sold c = number of child tickets sold
	6(a + c) = six(800) 6a  + 4.fivec  = 4,500 6a +   6c  = 4,800 6a  + 4.5c = 4,500		Apply multiplication to re-write the outset equation.
	6a +   sixc  = 4,800     − half dozena – 4.fivec  =  −four,500 ane.5c  =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 one of the original equations.
	a + c  =  800 600 + 200 = 800 800 = 800 TRUE	6a  + iv.5c  = 4,500 6(600) + 4.5(200) = four,500 3,600 + 900 = 4,500 4500 = 4,500 Truthful	Check your answer by substituting a = 600 and c = 200 into the original system. The answers check.
Respond	600 adult tickets and 200 child tickets were sold.

Summary

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 1. Multiplication tin can exist used to ready matching terms in equations before they are combined. When using the multiplication method, it is of import to multiply all the terms on both sides of the equation—not just the i term you are trying to eliminate.

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