Why can we subtract equations without losing the answer?

Start with shared solutions if systems are new to you. Here, you need substitution and collecting like terms.

Suppose two measurements give these constraints:

\begin{aligned} x+y&=5,\\ 2x-y&=1. \end{aligned}

The pair $(2,3)$ satisfies both. Our goal is to make that answer easy to find without guessing it.

Imagine subtracting twice the first equation from the second. Before doing the arithmetic, predict which variable will disappear. Will $(2,3)$ still satisfy the resulting equation?

Keep a record of your investigation

Record the original equations and their shared solution $(2,3)$ as your reference. Before each operation, write the predicted new row and whether the shared solution will change. Replace only the indicated row, keeping the other row fixed. Record the resulting equation and test $(2,3)$ in both equations before opening the explanation. Use the original pair to check that your comparison concerns shared solutions, not just the appearance of one line.

Change a line, keep its intersection

Use Next operation once. Compare the equations, the matrix entries, and the two lines. Then predict what dividing the second equation by $-3$ will do.

Change the equations. Keep the solution.

Step 0 of 3 · Starting system

Equation 1 (solid)x + y = 5

Equation 2 (dashed)2x − y = 1

Coefficients and right-hand sides
x	y	equals
1	1	5
2	-1	1

The green point (2, 3) satisfies both equations. Predict what happens to it when you eliminate x from the second equation.

Use Undo step to compare adjacent stages. Which operation changes the equation’s appearance without moving its line at all?

Compare your observations

Subtracting twice the first equation removes $x$ from the second. Its line becomes horizontal but still passes through $(2,3)$ . Dividing the second equation by $-3$ changes the written equation from $-3y=-9$ to $y=3$ , which describes the same horizontal line.

Where the new equation comes from

We must perform the operation on both sides:

(2x-y)-2(x+y)=1-2(5).

Collecting terms gives $-3y=-9$ , so $y=3$ . The first equation remains $x+y=5$ , giving $x=2$ .

We can organize the arithmetic in an augmented matrix:

\left[\begin{array}{cc|c} 1&1&5\\ 2&-1&1 \end{array}\right].

Each row represents an equation. The first two columns hold the coefficients of $x$ and $y$ ; the last holds the right-hand side. The vertical separator keeps those roles visible.

Writing $R_2\leftarrow R_2-2R_1$ means replacing row 2 with its old value minus twice row 1, including the last entry:

[2,-1\mid 1]-2[1,1\mid 5]=[0,-3\mid -9].

Why no solutions disappear—or sneak in

Any solution of the original pair satisfies the new second equation, because we subtract equal quantities from equal quantities.

That establishes one direction. To establish the other, notice that we kept the first equation. Adding twice it back to the new second equation recovers the original second equation. Every solution of the new pair therefore solves the old pair too.

This reversibility is why three row operations preserve solutions: swapping rows, multiplying a row by a nonzero number, and adding a multiple of one row to another.

Multiplying a row by zero is different. It turns a constraint into $0=0$ , and there is no inverse operation that recovers the lost information.

Check the answer against the original problem

Always return to the starting equations:

2+3=5,\qquad 2(2)-3=1.

Both hold. Elimination changed how we expressed the constraints while keeping their shared answer.

Design your own comparison

On paper, choose a nonzero multiplier $c$ different from the worked subtraction and replace the second equation by $R_2+cR_1$ . Keep the first equation unchanged. Predict the new coefficients, calculate them, and test the original solution.

Write an inverse operation. Then design a row change that would lose a restriction, and find a point that satisfies your altered system but not the original. Explain why merely checking that $(2,3)$ still works would miss the problem.

Write a short conclusion: what you predicted, what you observed, and how the evidence changed or supported your explanation.

Check the reasoning in your investigation

The new second row is $[2+c,-1+c\mid1+5c]$ , and subtracting $cR_1$ restores the old row. That inverse establishes equality of the solution sets. Replacing the second row by zero loses its restriction: for example, $(0,5)$ satisfies the remaining first equation but fails the original second equation.

Question 1 of 30 checked · 0 correct

Start with x + y = 7 and 3x + y = 11. What replaces the second equation after R₂ ← R₂ − 3R₁?−2y = 11−2y = −102x = 4

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