Math I · A-REI.5

Why Elimination Works in a System of Linear Equations

Elimination matters because it shows why a powerful shortcut for systems is logically valid, not just a mysterious cancellation trick.

Concept Algebra
Domain Reasoning with Equations and Inequalities
Read time 11 minutes

What this learning objective is really asking you to learn

This objective is not merely asking students to “do elimination.” It is asking students to understand why elimination is allowed. That difference matters. A student can memorize the steps of elimination and still feel that algebra is a list of mysterious tricks. This learning objective pushes deeper: when you replace one equation in a system with that equation plus a multiple of the other equation, why do you get a new system with the same solutions?

A system of equations is a set of conditions that must all be true at the same time. For example, the system

\[2x + 3y = 12\]
\[x - y = 1\]

does not ask for a value of \(x\) or a value of \(y\) in isolation. It asks for an ordered pair \((x, y)\) that makes both equations true. Graphically, each equation is a line. The solution to the system is the point where the two lines meet. Algebraically, it is the pair of numbers that passes both truth tests.

Elimination is based on the idea that you can combine true statements to create another true statement. If a point satisfies both equations, then it also satisfies the sum of the equations, or one equation plus any multiple of the other. For example, if \(2x + 3y = 12\) is true and \(x - y = 1\) is true, then adding the left sides and adding the right sides gives \(3x + 2y = 13\), which must also be true for the same solution. This new equation may not contain the exact same information by itself, but when paired with one of the original equations, it can describe the same intersection point.

The key phrase is “preserves solutions.” The goal is not to change the problem. The goal is to rewrite the system so that one variable becomes easier to eliminate while the solution set stays the same. This is like translating a sentence into another language without changing its meaning. The wording changes; the truth remains.

A linear combination means multiplying equations by numbers and adding them. In two-variable linear systems, this is the machinery behind elimination. If one equation has \(+3y\) and another has -3y, adding the equations removes \(y\). If the coefficients do not already match, you can multiply one or both equations first. That multiplication is not arbitrary magic. Multiplying both sides of an equation by the same nonzero number preserves equality. Adding equal quantities to equal quantities preserves equality. Elimination is built out of those logical moves.

This objective also prepares students for a much larger idea: algebraic operations are not valid because they seem convenient; they are valid because they preserve truth. The student is learning to ask, “Did this step keep the same solution set?” That question is one of the foundations of mature mathematical reasoning.

Why students should learn this math

Students should learn this because real problems rarely come with only one condition. Real life is full of simultaneous constraints. A business may need price and quantity to satisfy both a revenue goal and a supply limit. A builder may need dimensions to satisfy both area and material constraints. A phone plan may include both a fixed monthly cost and a per-gigabyte charge, and a customer may compare it to another plan. A recipe may need total volume and a certain ratio of ingredients. A trip may involve distance, time, speed, fuel, and cost. Systems of equations are the mathematics of “more than one thing must be true at once.”

But the reason A-REI.5 focuses on justification is even more important. Many students experience algebra as a place where teachers perform strange legal moves: “multiply this equation by 3,” “add the equations,” “cancel the y’s,” “replace this line,” and so on. Without justification, elimination feels like a ritual. With justification, it becomes common sense: if two statements are true, then carefully combining them creates another true statement. You are not cheating. You are preserving the same solution while changing the form of the information.

This matters in technical fields because systems of equations scale up. Engineers solve systems with many variables to analyze circuits, forces, heat flow, chemical mixtures, and structures. Economists solve systems to study equilibrium among markets. Computer graphics uses systems and matrices to transform points in space. Navigation systems combine multiple signals to estimate a location. Data science uses large systems to fit models. Machine learning relies heavily on linear algebra, where elimination becomes matrix operations and solution spaces.

Even outside advanced careers, the habit of preserving truth is valuable. Suppose a student is comparing two job offers: one has higher hourly pay but fewer hours; another has lower hourly pay but a bonus. The student may create equations for weekly income and ask when the offers are equal. Suppose a family compares two internet plans: one has a lower monthly fee but higher overage charges. A system can reveal the usage level where the plans cost the same. The logic is not abstract decoration. It helps people find decision points.

Elimination also teaches a clean kind of reasoning: reduce complexity without losing meaning. That is a powerful life skill. In a messy situation, you may have too many moving pieces. A good thinker looks for a way to combine facts so that one unknown disappears and a clearer question remains. In algebra, eliminating a variable is a controlled version of that thinking.

The “why” is therefore not simply “because you will need it in Algebra II.” The deeper reason is that elimination teaches students how to manipulate information responsibly. It shows how complex constraints can be transformed into simpler ones while keeping the same answer. That is the core of mathematical modeling, technical problem solving, and logical decision-making.

The historical machinery: from simultaneous problems to linear algebra

The problem of solving simultaneous equations is ancient. Long before modern symbolic algebra, people needed to solve problems involving several unknown quantities. Systems appear naturally in trade, land measurement, taxation, distribution, and engineering. If several goods have different values and several transactions are known, the unknown prices can often be found only by solving conditions together.

One of the most famous early traditions involving systems of linear equations appears in ancient Chinese mathematics, especially in methods that resemble what modern students call elimination. Problems were arranged in counting-board layouts that, in modern language, look much like augmented matrices. The solver would manipulate rows of numbers to simplify the system. This shows something important: elimination is not a recent classroom invention. It is a natural technique people discovered because it works.

In later European mathematics, elimination became more formal as algebraic notation improved. The development of symbolic algebra allowed mathematicians to write general equations, not just solve one numerical puzzle at a time. Eventually, elimination became connected with matrices and determinants. The name “Gaussian elimination” is often used today because of Carl Friedrich Gauss’s influential use of systematic elimination methods, especially in astronomical and geodetic computations. But the basic idea has roots across cultures and long predates the modern name.

The historical arc is useful for students because it shows that elimination is part of a much larger machine. At first, a student may see a system of two equations and two variables. Later, the same idea becomes a system of ten equations and ten unknowns, or thousands of equations handled by a computer. The equations may represent forces in a bridge, currents in an electrical network, pixels in an image, or coefficients in a statistical model. The classroom procedure is the small version of a very large tool.

The modern form of this tool lives in linear algebra. In linear algebra, a system of equations can be represented by a matrix. The legal moves of elimination are called elementary row operations. Replacing one row by that row plus a multiple of another row is one of the most important row operations. It preserves the solution set for the same reason it does in Integrated Math I: the new row contains information logically derived from the old rows, and the original information can be recovered by reversing the operation.

That reversibility is crucial. If a transformation can be reversed, it has not thrown away information. This idea appears throughout mathematics. In algebra, valid equation steps preserve solutions or are checked when they might introduce extras. In geometry, rigid motions preserve distance and angle. In functions, inverse functions undo each other. In computer science, reversible operations preserve information. A-REI.5 is one early place where students learn that mathematics cares not only about getting an answer but about preserving meaning through each transformation.

Where this fits in the big map of mathematics

This objective sits at a major junction between algebra, geometry, logic, and linear algebra. In the algebra branch, it connects to solving systems, manipulating equations, and understanding equivalence. In the geometry branch, each equation represents a line, and the system represents the intersection of lines. In the logic branch, each step must be justified as preserving truth. In the linear algebra branch, the same idea becomes row reduction and matrix methods.

It also connects backward to earlier equation-solving. When students solve a one-variable equation, they perform operations that keep the equation balanced. Add 5 to both sides. Divide both sides by 3. Distribute or factor carefully. The goal is to isolate the variable while keeping the same solution. A-REI.5 extends that idea from one equation to a whole system. Now the object being preserved is not just one equation’s solution, but the common solution of two equations.

It connects forward to graphing systems. If two lines intersect at one point, elimination should find the coordinates of that point. If two lines are parallel, elimination may produce a false statement such as \(0 = 7\), showing no solution. If two equations describe the same line, elimination may produce a true statement such as \(0 = 0\), showing infinitely many solutions. The algebra and graph are two versions of the same story.

It also connects to modeling. A system is a model with multiple conditions. Elimination is one way to extract the hidden quantities from those conditions. Later, students will see systems involving quadratics, exponentials, inequalities, and matrices. They will also see optimization problems, where constraints define a feasible region and the goal is to find the best point inside it. Linear programming, economics, engineering design, and data fitting all build on the same idea: preserve the constraints while transforming the problem into a form that reveals the answer.

The big-picture map looks like this: arithmetic teaches operations, algebra teaches unknowns, systems teach simultaneous conditions, elimination teaches truth-preserving transformation, and linear algebra turns those transformations into a general language for high-dimensional problems. A-REI.5 is a hinge point where students move from “solve this equation” to “reason about an entire system.”

How to execute the skill technically

The central proof can be understood with a general system. Suppose the original system is

\[Equation 1: ax + by = c\]
\[Equation 2: dx + ey = f\]

Now create a new second equation by replacing Equation 2 with Equation 2 plus \(k\) times Equation 1:

\[(dx + ey) + k(ax + by) = f + kc\]

This simplifies to

\[(d + ka)x + (e + kb)y = f + kc\]

The new system is

\[ax + by = c\]
\[(d + ka)x + (e + kb)y = f + kc\]

Why does this preserve solutions? First, if \((x, y)\) satisfies the original system, then \(ax + by = c\) and \(dx + ey = f\) are both true. Multiplying the first true equation by \(k\) gives \(k(ax + by) = kc\). Adding that true statement to the second true equation gives the new second equation. So every original solution is also a solution of the new system.

Second, if \((x, y)\) satisfies the new system, then it satisfies Equation 1 and the new Equation 2. Because the new Equation 2 was made by adding \(k\) times Equation 1 to the old Equation 2, we can recover the old Equation 2 by subtracting \(k\) times Equation 1 from the new Equation 2. So every solution of the new system is also a solution of the original system. The solution sets match in both directions. That is the justification.

A numerical example makes this concrete:

\[2x + 3y = 12\]
\[x - y = 1\]

Multiply the second equation by 3:

\[3x - 3y = 3\]

Now add it to the first equation:

\[2x + 3y = 12\]
\[3x - 3y = 3\]
\[5x = 15\]

So \(x = 3\). Substitute into \(x - y = 1\):

\(3 - y = 1\), so \(y = 2\).

The solution is \((3, 2)\). Check it: \(2(3) + 3(2) = 12\), and \(3 - 2 = 1\). The checking step confirms the solution, but A-REI.5 is about why the elimination step was legal in the first place.

Another way to see the logic is through equations as information. The equation \(2x + 3y = 12\) contains one relationship. The equation \(x - y = 1\) contains another. Adding a multiple of one equation to the other creates a new relationship that must be true wherever both original relationships are true. Pairing that new relationship with one original relationship gives a different description of the same intersection.

Common mistakes and how to avoid them

One common mistake is adding equations without a purpose. Elimination is not random combining. The goal is usually to create opposite coefficients so that one variable disappears. Students should ask, “Which variable am I eliminating, and what multiple will create opposites?”

Another mistake is failing to apply operations to the entire equation. If you multiply an equation by 3, every term on both sides must be multiplied by 3. Multiplying only the variable terms or only one side breaks equality. The legal move is powerful precisely because it is balanced.

A deeper mistake is replacing too much at once without keeping an equivalent system. In basic elimination, you usually keep one equation and replace the other with a combination. That preserves enough information to recover the original system. If you combine equations carelessly and discard both originals, you may lose information. For example, turning two equations into only one equation usually creates infinitely many possible points, not a unique solution.

Students also sometimes think that a newly created equation has the same graph as one of the original equations. Usually it does not. The new equation may be a different line. The claim is not that each individual line stays the same. The claim is that the system’s intersection point stays the same when the new equation is paired with the retained equation.

Finally, students may trust the procedure but not understand special cases. If elimination gives \(0 = 0\), that means the equations were dependent; they represent the same line, so there are infinitely many solutions. If elimination gives a false statement such as \(0 = 5\), the system is inconsistent; the lines are parallel and never meet. These results are not errors. They are information about the solution set.

What students should be able to say

A student who has mastered this objective should be able to say: “Elimination works because I am replacing one equation with a new equation that is logically created from the original system. Any solution of the original system must satisfy the new equation, and because the operation can be reversed, any solution of the new system also satisfies the original system. The graph of the individual equation may change, but the solution set of the system stays the same.”

That is a major shift. The student is no longer performing elimination as a trick. The student understands it as truth-preserving transformation.

Problem Library

Problems in the App From This Objective

147 problems across 12 archetypes in the app.

substitute a solution into original and transformed systems.
12 problems Warmup Practice Mixed Review Assessment
Problem 1

Verify that replacing one equation in x + y = 5, x - y = 1 with the sum of both equations preserves the solution (3,2).

Problem 2

Verify that replacing one equation in 2x + y = 9, x - y = 0 with the sum of both equations preserves the solution (3,3).

Problem 3

Verify that replacing one equation in 3x + y = 11, x - y = 1 with the sum of both equations preserves the solution (3,2).

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Problem 4

Verify that replacing one equation in x + y = 2, x - y = 0 with the sum of both equations preserves the solution (1,1).

Problem 5

Verify that replacing one equation in x + y = 3, x - y = 1 with the sum of both equations preserves the solution (2,1).

Problem 6

Verify that replacing one equation in x + y = 3, -x + y = 1 with the sum of both equations preserves the solution (1,2).

Problem 7

Verify that replacing one equation in x + 2y = 6, x - y = 3 with the sum of both equations preserves the solution (4,1).

Problem 8

Verify that replacing one equation in 2x + y = 6, x - y = -3 with the sum of both equations preserves the solution (1,4).

Problem 9

Verify that replacing one equation in 3x + y = 9, x - y = -1 with the sum of both equations preserves the solution (2,3).

Problem 10

Verify that replacing one equation in x + y = 5, x - y = 5 with the sum of both equations preserves the solution (5,0).

Problem 11

Verify that replacing one equation in x + y = 3, -x + y = 3 with the sum of both equations preserves the solution (0,3).

Problem 12

Verify that replacing one equation in 2x + y = 6, x + 2y = 6 with the sum of both equations preserves the solution (2,2).

reason about equivalent equations before elimination.
12 problems Warmup Practice Mixed Review Assessment
Problem 13

Explain why multiplying x + y = 6 by 3 preserves its solution set.

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Problem 14

Explain why multiplying 2x - y = 5 by -2 preserves its solution set.

Problem 15

Explain why multiplying 4x + y = 8 by 1/2 preserves its solution set.

Problem 16

Explain why multiplying x - 3 = 7 by 5 preserves its solution set.

Problem 17

Explain why multiplying 3a + 2b = 10 by -1 preserves its solution set.

Problem 18

Explain why multiplying x/2 + y/3 = 1 by 6 preserves its solution set.

Problem 19

Explain why multiplying 0.5x - 1.2 = 3 by 10 preserves its solution set.

Problem 20

Explain why multiplying 5x = 15 by 1/5 preserves its solution set.

Problem 21

Explain why multiplying 2x + 4 = 8 by -1/2 preserves its solution set.

Problem 22

Explain why multiplying 3x - 5 = x + 1 by 2 preserves its solution set.

Problem 23

Explain why multiplying 2(x + 1) = 8 by 3 preserves its solution set.

Problem 24

Explain why multiplying x - 7 = 0 by -4 preserves its solution set.

connect linear combination to equality-preserving operations.
12 problems Warmup Practice Mixed Review Assessment
Problem 25

Justify adding a multiple of one equation to another in 2x + y = 7, x - y = 2.

Problem 26

Justify adding a multiple of one equation to another in 3x + 2y = 12, x + y = 5.

Problem 27

Justify adding a multiple of one equation to another in 4x - y = 10, 2x + y = 8.

Problem 28

Justify adding a multiple of one equation to another in x + 2y = 5, x - 2y = 1.

Problem 29

Justify adding a multiple of one equation to another in 5x + y = 10, 3x + y = 6.

Problem 30

Justify adding a multiple of one equation to another in x + y = 7, 2x - y = 8.

Problem 31

Justify adding a multiple of one equation to another in 2x + 3y = 10, x + y = 4.

Problem 32

Justify adding a multiple of one equation to another in 3x - 2y = 1, x - y = 0.

Problem 33

Justify adding a multiple of one equation to another in 3a + 4b = 18, -3a + b = 2.

Problem 34

Justify adding a multiple of one equation to another in 7m - 2n = 15, 3m - 2n = 7.

Problem 35

Justify adding a multiple of one equation to another in x + 5y = 11, 2x - y = 4.

Problem 36

Justify adding a multiple of one equation to another in 4p + 3q = 22, p + q = 6.

Open in simulator
distinguish valid linear combinations from altered systems.
12 problems Warmup Practice Mixed Review Assessment
Problem 37

Decide whether the proposed elimination step add equations to get 3x = 9 for 2x + y = 7, x - y = 2 is valid. Explain.

Problem 38

Decide whether the proposed elimination step subtract second equation to get 2x + 2y = 6 for 3x + 2y = 10, x + y = 4 is valid. Explain.

Problem 39

Decide whether the proposed elimination step change x + y = 5 to x + y = 6 for x + y = 5, x - y = 1 is valid. Explain.

Problem 40

Decide whether the proposed elimination step multiply the second equation by 2 and subtract it from the first to get y = 2 for 2x + 3y = 10, x + y = 4 is valid. Explain.

Problem 41

Decide whether the proposed elimination step multiply the first equation by 3 to get 3x + 2y = 15 for x + 2y = 5, 3x + y = 7 is valid. Explain.

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Problem 42

Decide whether the proposed elimination step subtract the second equation from the first to get -5y = 10 for 5x - 2y = 12, 5x + 3y = 2 is valid. Explain.

Problem 43

Decide whether the proposed elimination step add equations to get 3y = 16 for 4x + y = 10, -4x + 2y = 5 is valid. Explain.

Problem 44

Decide whether the proposed elimination step multiply the first equation by 2 and the second by 3, then add them to get 13x = 26 for 2x + 3y = 1, 3x - 2y = 8 is valid. Explain.

Problem 45

Decide whether the proposed elimination step subtract the second equation from the first to get 2y = 2 for 3x - y = 7, 3x + y = 5 is valid. Explain.

Problem 46

Decide whether the proposed elimination step multiply the first equation by 2 and add it to the second to get 5y = 13 for -x + 4y = 6, 2x - 3y = 1 is valid. Explain.

Problem 47

Decide whether the proposed elimination step add 2x to the first equation to get 3x + 2y = 8 for x + 2y = 8, 3x - y = 5 is valid. Explain.

Problem 48

Decide whether the proposed elimination step add equations to eliminate y and get 3x + 4y = 12 for 2x + y = 5, x + 3y = 7 is valid. Explain.

choose a multiplier to create opposite coefficients.
12 problems Warmup Practice Mixed Review Assessment
Problem 49

Find the missing multiplier to eliminate y in 2x + 3y = 12, 5x - y = 7.

Problem 50

Find the missing multiplier to eliminate y in 4x + y = 9, 3x + 2y = 8.

Problem 51

Find the missing multiplier to eliminate x in 3x + 2y = 11, x - 5y = 4.

Problem 52

Find the missing multiplier to eliminate x in x + 2y = 5, 3x - y = 8.

Open in simulator
Problem 53

Find the missing multiplier to eliminate y in -2x + 5y = 1, 4x + y = 9.

Problem 54

Find the missing multiplier to eliminate x in 5x - 2y = 10, x + 3y = 7.

Problem 55

Find the missing multiplier to eliminate y in 3x - 4y = 1, 2x + 2y = 8.

Problem 56

Find the missing multiplier to eliminate x in -x + 3y = 6, 2x + 5y = 1.

Problem 57

Find the missing multiplier to eliminate y in 6x + y = 15, x - 3y = 2.

Problem 58

Find the missing multiplier to eliminate x in -4x + y = 10, x + 2y = 5.

Problem 59

Find the missing multiplier to eliminate y in x + 6y = 12, 3x - 2y = 4.

Problem 60

Find the missing multiplier to eliminate x in 2x - 7y = 3, x + 4y = 1.

track both sides of equations through transformations.
12 problems Warmup Practice Mixed Review Assessment
Problem 61

Compare transformations replace first equation with 2x = 6 and keep second equation and replace both equations with 2x = 6 for x + y = 5, x - y = 1. Which preserves solutions?

Problem 62

Compare transformations multiply the second equation by 2 on both sides and multiply only the x-term in the second equation by 2 for 2x + y = 8, x - y = 1. Which preserves solutions?

Problem 63

Compare transformations subtract the second equation from the first to get 2x = 4 and keep x + y = 6 and subtract only the y terms and keep all other terms unchanged for 3x + y = 10, x + y = 6. Which preserves solutions?

Problem 64

Compare transformations add the two equations to get 4x = 8 and keep x + 2y = 7 and add the two equations to get 4x = 8 and replace both original equations with 4x = 8 for x + 2y = 7, 3x - 2y = 1. Which preserves solutions?

Problem 65

Compare transformations replace the first equation with (5x - y) + (x + y) = 3 + 9 and keep the second equation and replace the first equation with 5x = 3 and keep the second equation for 5x - y = 3, x + y = 9. Which preserves solutions?

Problem 66

Compare transformations multiply the first equation by 3 and the second equation by 2 and multiply only the left side of the first equation by 3 for 2x + 3y = 12, x - y = 1. Which preserves solutions?

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Problem 67

Compare transformations replace the first equation with the sum of the first and second equations, and keep the second and third equations and replace all three equations with x + y + z = 6 for x + y + z = 6, x - y + z = 2, 2x + y - z = 3. Which preserves solutions?

Problem 68

Compare transformations divide the first equation by 2 on both sides and divide only the 4x term in the first equation by 2 for 4x - 2y = 10, x + y = 3. Which preserves solutions?

Problem 69

Compare transformations subtract the second equation from the first to get 10y = 10 and keep the second equation and subtract the second equation from the first to get 10y = 10 and replace both equations with 10y = 10 for x + 5y = 11, x - 5y = 1. Which preserves solutions?

Problem 70

Compare transformations substitute 2y + 3 for x in the second equation to get 3(2y + 3) - y = 5 and keep the first equation and substitute 2y + 3 for x in the second equation and then discard the first equation for x = 2y + 3, 3x - y = 5. Which preserves solutions?

Problem 71

Compare transformations subtract the first equation from the second to get 0 = -2 and keep the first equation and replace the first equation with 2x = 7 and the second equation with y = 5 for 2x + y = 7, 2x + y = 5. Which preserves solutions?

Problem 72

Compare transformations replace the second equation with 0 = 0 by subtracting the first from the second, and keep the first equation and replace the first equation with x = 10 and the second equation with y = 10 for x + y = 10, x + y = 10. Which preserves solutions?

communicate why combining equal quantities keeps equality true.
12 problems Warmup Practice Mixed Review Assessment
Problem 73

Explain elimination using the concrete context tickets where x is adult price and y is child price and equations 2x + y = 25, x + y = 15.

Problem 74

Explain elimination using the concrete context bags with apples x and oranges y and equations 3x + 2y = 18, x + 2y = 10.

Problem 75

Explain elimination using the concrete context rental bundles where x is the chair cost and y is the table cost and equations 4x + y = 38, 2x + y = 24.

Problem 76

Explain elimination using the concrete context orders of books (x) and notebooks (y) and equations 5x + 3y = 70, 2x + 3y = 40.

Problem 77

Explain elimination using the concrete context packages of pens (x) and pencils (y) and equations 4x + 5y = 35, 4x + 2y = 20.

Problem 78

Explain elimination using the concrete context fruit baskets with apples (x) and bananas (y) and equations 7x + 2y = 29, 5x + 2y = 21.

Problem 79

Explain elimination using the concrete context restaurant bills for pizzas (x) and drinks (y) and equations 3x + 4y = 52, 3x + y = 31.

Problem 80

Explain elimination using the concrete context batches of cookies (x) and brownies (y) and equations 6x + 8y = 100, 6x + 5y = 70.

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Problem 81

Explain elimination using the concrete context clothing bundles with shirts (x) and hats (y) and equations 2x + 3y = 65, x + 3y = 45.

Problem 82

Explain elimination using the concrete context deliveries of small boxes (x) and large boxes (y) and equations 9x + 4y = 120, 9x + y = 75.

Problem 83

Explain elimination using the concrete context orders of coffee (x) and pastries (y) and equations 5x + 6y = 80, 5x + 2y = 40.

Problem 84

Explain elimination using the concrete context garden plots with rose bushes (x) and lavender plants (y) and equations 3x + 7y = 95, x + 7y = 65.

audit multipliers, signs, and equation replacement.
12 problems Warmup Practice Mixed Review Assessment
Problem 85

Find the first invalid step in this elimination solution: 2x + y = 7 and x - y = 2, add equations, 3x = 5, x = 5/3.

Problem 86

Find the first invalid step in this elimination solution: 3x + 2y = 12 and x + y = 5, multiply second equation by -2: -2x + y = -10, add to first equation.

Problem 87

Find the first invalid step in this elimination solution: x + y = 6 and x - y = 2, add equations to get 2x = 8, replace both original equations with 2x = 8.

Problem 88

Find the first invalid step in this elimination solution: 2x + 3y = 10 and x - y = 1, multiply second equation by 3: 3x - 3y = 1, add to first equation.

Problem 89

Find the first invalid step in this elimination solution: 4x + y = 7 and 2x - 3y = 1, multiply second equation by -2: -2x + 6y = -2, add to first equation.

Problem 90

Find the first invalid step in this elimination solution: 5x + 2y = 15 and 5x - y = 3, subtract second equation from first: y = 12, y = 4.

Problem 91

Find the first invalid step in this elimination solution: x + 4y = 9 and 2x - 4y = 3, add equations: x = 12, x = 12.

Problem 92

Find the first invalid step in this elimination solution: 3x + y = 8 and x - y = 2, add 3x - y = 8 to x - y = 2, 4x - 2y = 10.

Problem 93

Find the first invalid step in this elimination solution: x + y = 5 and x - y = 1, add equations: 2x = 6, x = 3, substitute x = 3 into x - y = 1 to get 3 + y = 1, y = -2.

Problem 94

Find the first invalid step in this elimination solution: x + 2y = 7 and 3x = 9, multiply first equation by 3: 3x + 2y = 21, subtract second equation from modified first: 2y = 12.

Problem 95

Find the first invalid step in this elimination solution: 2x + 2y = 10 and x - y = 1, divide first equation by 2: x + y = 5, divide second equation by x: 1 - y/x = 1/x.

Problem 96

Find the first invalid step in this elimination solution: x + 2y = 8 and 2x + 4y = 16, multiply first equation by -2: -2x + 4y = -16, add to second equation.

Open in simulator
solve or verify both systems and compare solution sets.
12 problems Warmup Practice Mixed Review Assessment
Problem 97

Show that the original system x + y = 5, x - y = 1 and transformed system 2x = 6, x - y = 1 have the same ordered-pair solution.

Problem 98

Show that the original system 2x + y = 8, x + y = 5 and transformed system x = 3, x + y = 5 have the same ordered-pair solution.

Problem 99

Show that the original system 3x + y = 11, x + y = 5 and transformed system 2x = 6, x + y = 5 have the same ordered-pair solution.

Problem 100

Show that the original system x + y = 3, 2x + y = 4 and transformed system x = 1, x + y = 3 have the same ordered-pair solution.

Problem 101

Show that the original system x + y = 3, x - y = 1 and transformed system 2x = 4, x - y = 1 have the same ordered-pair solution.

Problem 102

Show that the original system x + y = 5, 2x - y = 7 and transformed system 3x = 12, x + y = 5 have the same ordered-pair solution.

Problem 103

Show that the original system x + y = 4, 3x - y = 0 and transformed system 4x = 4, x + y = 4 have the same ordered-pair solution.

Problem 104

Show that the original system x + y = 5, x - y = 5 and transformed system 2x = 10, x + y = 5 have the same ordered-pair solution.

Problem 105

Show that the original system x + y = 5, 2x + y = 5 and transformed system x = 0, x + y = 5 have the same ordered-pair solution.

Problem 106

Show that the original system x + y = 2, 2x + y = 1 and transformed system x = -1, x + y = 2 have the same ordered-pair solution.

Problem 107

Show that the original system x + y = 2, x - y = 4 and transformed system 2x = 6, x + y = 2 have the same ordered-pair solution.

Problem 108

Show that the original system 2x + y = 7, x - y = -1 and transformed system 3x = 6, x - y = -1 have the same ordered-pair solution.

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preserve signs while forming linear combinations.
12 problems Warmup Practice Mixed Review Assessment
Problem 109

Justify the elimination step for the standard-form system -2x + y = 5, 2x + 3y = 7, preserving negative signs.

Problem 110

Justify the elimination step for the standard-form system 3x - 2y = 4, -3x + y = -1, preserving negative signs.

Problem 111

Justify the elimination step for the standard-form system -4x + 2y = 6, 4x - y = 2, preserving negative signs.

Problem 112

Justify the elimination step for the standard-form system -5x + 2y = 10, 5x + y = 5, preserving negative signs.

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Problem 113

Justify the elimination step for the standard-form system 6x - 3y = 9, -6x - y = 3, preserving negative signs.

Problem 114

Justify the elimination step for the standard-form system -x + 4y = -2, x + 2y = -4, preserving negative signs.

Problem 115

Justify the elimination step for the standard-form system 3x - 2y = 7, x + 2y = 1, preserving negative signs.

Problem 116

Justify the elimination step for the standard-form system 2x - 5y = 1, 3x + 5y = 9, preserving negative signs.

Problem 117

Justify the elimination step for the standard-form system 4x - y = -3, 2x + y = 9, preserving negative signs.

Problem 118

Justify the elimination step for the standard-form system -7x + 3y = -10, 7x - 2y = 5, preserving negative signs.

Problem 119

Justify the elimination step for the standard-form system -2x - 3y = -5, 5x + 3y = 11, preserving negative signs.

Problem 120

Justify the elimination step for the standard-form system -8x - 2y = -12, 8x + 5y = 2, preserving negative signs.

state each transformation and why it preserves solutions.
15 problems Warmup Practice Mixed Review Assessment
Problem 121

Write a proof-style explanation for solving the system x + y = 5, x - y = 1 by elimination.

Problem 122

Write a proof-style explanation for solving the system 2x + y = 8, x - y = 1 by elimination.

Problem 123

Write a proof-style explanation for solving the system 3x + y = 11, x - y = 1 by elimination.

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Problem 124

Write a proof-style explanation for solving the system x + y = 7, x - y = 3 by elimination.

Problem 125

Write a proof-style explanation for solving the system x + 2y = 10, x - 2y = 2 by elimination.

Problem 126

Write a proof-style explanation for solving the system 2x + y = 7, x - y = 2 by elimination.

Problem 127

Write a proof-style explanation for solving the system x + 3y = 11, x - 3y = -1 by elimination.

Problem 128

Write a proof-style explanation for solving the system 3x + 2y = 13, x - 2y = -1 by elimination.

Problem 129

Write a proof-style explanation for solving the system x + y = 8, x - y = 2 by elimination.

Problem 130

Write a proof-style explanation for solving the system 4x + y = 14, x - y = 1 by elimination.

Problem 131

Write a proof-style explanation for solving the system x + y = 6, 2x - y = 3 by elimination.

Problem 132

Write a proof-style explanation for solving the system x + 2y = 7, 3x - 2y = 5 by elimination.

Problem 133

Write a proof-style explanation for solving the system x + y = 9, 2x - y = 6 by elimination.

Problem 134

Write a proof-style explanation for solving the system x + y = 10, 3x - y = 14 by elimination.

Problem 135

Write a proof-style explanation for solving the system 4x + y = 17, x - y = 3 by elimination.

recognize when multiplying by zero or replacing both equations improperly changes the system.
12 problems Warmup Practice Mixed Review Assessment
Problem 136

Decide whether the elimination action multiply one equation by 0 before adding preserves the system's information or loses/changes it.

Problem 137

Decide whether the elimination action replace one equation with the sum of both equations and keep the other equation preserves the system's information or loses/changes it.

Problem 138

Decide whether the elimination action replace both equations with the same sum equation preserves the system's information or loses/changes it.

Problem 139

Decide whether the elimination action multiply one equation by a non-zero constant preserves the system's information or loses/changes it.

Problem 140

Decide whether the elimination action swap the positions of the two equations preserves the system's information or loses/changes it.

Problem 141

Decide whether the elimination action replace one equation with 3 times itself preserves the system's information or loses/changes it.

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Problem 142

Decide whether the elimination action replace one equation with the difference of the two equations and keep the other equation preserves the system's information or loses/changes it.

Problem 143

Decide whether the elimination action replace one equation with 2 times the sum of both equations and keep the other equation preserves the system's information or loses/changes it.

Problem 144

Decide whether the elimination action remove one of the equations from the system preserves the system's information or loses/changes it.

Problem 145

Decide whether the elimination action replace one equation with a constant value, e.g., '5 = 5' preserves the system's information or loses/changes it.

Problem 146

Decide whether the elimination action replace both equations with two different linear combinations, where one is a multiple of the other preserves the system's information or loses/changes it.

Problem 147

Decide whether the elimination action replace the first equation with the sum of both equations and the second equation with the first original equation preserves the system's information or loses/changes it.