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≈ 50 min · Grade 8 · Mathematics

Solving Systems by Substitution and Checking Solutions

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Checked by VeraTeach before publishing · September 2026

Teacher asked for
solving systems of linear equations by substitution
Teacher's note
“They solve for one variable and stop. Checking is the part they skip.”
Objective produced
Solve a system of two linear equations by substitution and verify the ordered-pair solution in both equations.

Materials & Prep

Copy the guided-practice systems, error-analysis example, independent problems, and exit ticket onto one handout or display them on the board. Prepare a two-column check template labeled “Equation 1” and “Equation 2.” No special materials are needed beyond the usual classroom setup. Display the opening graph of y = x + 2 and y = -x + 8 on a coordinate grid.

0–7min

Opening

Display two intersecting lines, y = x + 2 and y = -x + 8, on a coordinate grid without asking students to solve algebraically. Ask: “What point appears to lie on both lines? How do you know?” Students first estimate from the graph, then compare answers with a partner. Confirm that the lines meet at (3, 5), and that this point belongs to both lines.

ConnectExplain that a solution to a system is one ordered pair that makes both equations true. Today, students will find that shared point without relying on a graph. The key idea from the opening remains the same: a correct answer must work in both equations.

7–17min

Mini-lesson

ModelSay, “On the graph we could see the shared point. Most systems are not that easy to read, so we find it algebraically.” Display y = 2x + 1 and x + y = 10, then narrate each decision. The first equation is already solved for y, so substitute 2x + 1 wherever y appears in the second equation:

x + (2x + 1) = 10 3x + 1 = 10 3x = 9 x = 3

The common mistake is stopping here. x = 3 is only one coordinate, not the solution to the system. Substitute x = 3 into y = 2x + 1:

y = 2(3) + 1 2(3) + 1 = 7 y = 7

The ordered pair is (3, 7). Now check the actual pair in both original equations:

Equation 1: 7 = 2(3) + 1, so 7 = 7. Equation 2: 3 + 7 = 10, so 10 = 10.

EmphasizeThe check must use the original equations, not only the rearranged equation used during substitution. The solution is (3, 7) because both equations are true for that pair.

Check for understandingStudents hold up one finger if x = 3 is the complete solution and two fingers if (3, 7) is the complete solution. Call on one student to explain why the second response is correct.

17–29min

Guided practice

Worked exampleSolve and check the system with students. Ask them to state where each expression is substituted.

2x + y = 13 y = x + 1

Substitute x + 1 for y in the first equation

2x + (x + 1) = 13 3x + 1 = 13 3x = 12 x = 4

Substitute x = 4 into y = x + 1

4 + 1 = 5 y = 5

Have students complete the two-column check

Equation 1: 2(4) + 5 = 13, so 13 = 13. Equation 2: 5 = 4 + 1, so 5 = 5.

ScaffoldGive students this checklist and sentence frame: “I substituted for . I found = . The ordered pair is (, ). In Equation 1, = . In Equation 2, = .” Circulate and prompt students to label each coordinate before checking.

29–39min

Partner error analysis

Display

y = 3x - 2 x + y = 10

A student writes

x + (3x - 2) = 10 4x - 2 = 10 4x = 12 x = 3 “Therefore, the solution is x = 3.”

Ask partners to identify what is missing, finish the solution, and check the ordered pair in both original equations. Discuss that x = 3 is correct but incomplete. Find y by substitution:

y = 3(3) - 2 3(3) - 2 = 7 y = 7

The ordered pair is (3, 7). Check it: 3(3) - 2 = 7, so 7 = 7. 3 + 7 = 10, so 10 = 10.

ConfrontTest the incomplete pair (3, 3). It fails the first equation because 3 is not equal to 7, and it fails the second equation because 3 + 3 is not 10. Students revise the original claim in writing: “x = 3 is not the complete solution; I must find y and check both equations.”

39–47min

Independent solve and check

Students solve both systems independently. Require an ordered pair and a written check in both original equations.

  1. y = x + 4

2x + y = 13

  1. 3x + y = 14

y = 2x - 1

Expected results for monitoring

Problem 1: x = 3, y = 7, so the solution is (3, 7). The checks are 7 = 3 + 4 and 2(3) + 7 = 13. Problem 2: x = 3, y = 5, so the solution is (3, 5). The checks are 3(3) + 5 = 14 and 5 = 2(3) - 1.

Look forStudents who stop after finding x, substitute into an equation that is not one of the originals, or check only one equation. Redirect them to the checklist. Extension: Students explain why checking only the equation already solved for y is not enough, using one of the two systems as evidence.

47–50min

Exit ticket

Exit ticketSolve and check the system in both original equations. Show the ordered pair and both substitutions.

y = x + 2 3x + y = 14

Expectedx = 3 and y = 5, so the ordered pair is (3, 5). Both checks must show that 5 = 3 + 2 and 3(3) + 5 = 14. Collect the response to identify who still stops after finding one variable.

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