standard_description: Prove that, given a system of two equations in two variables, replacing one equation by the sum of that equation and a multiple of the other produces a system with the same solutions.
problem_activity: Warm-up
The purpose of this warm-up is for students to revisit ideas they learned in the previous lesson about balanced hangers:
You can add or subtract the same thing on each side and the hanger stays in balance.
You can divide each side by the same number and the hanger stays in balance.
Launch
Give students 2 minutes of quiet work time followed by a whole-class discussion.
Student Facing
Figures A, B, C, and D show the result of simplifying the hanger in Figure A by removing equal weights from each side.
[IMG]
Here are some equations. Each equation represents one of the hanger diagrams.
\(2(x + 3y) = 4x + 2y\)
\(2y = x\)
\(2(x + 3y) + 2z = 2z + 4x + 2y\)
\(x + 3y = 2x + y\)
Write the equation that goes with each figure:
A:
B:
C:
D:
Each variable (
\(x\)
,
\(y\)
, and
\(z\)
) represents the weight of one shape. Which goes with which?
Explain what was done to each equation to create the next equation. If you get stuck, think about how the hangers changed.
Student Response
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Anticipated Misconceptions
Some students may think a variable stands for more than one object. Tell these students that a variable only stands for one object, as it also only represents one number.
Activity Synthesis
Ask students to explain how they decided on the matching equation. As students discuss the final questions, highlight responses that emphasize the same objects being removed from each side creates the next figure in line. The exception to this is the move from Hanger B to Hanger C, where the number of objects on each side is halved. Ask students to explain why this is an okay move, even though different objects are being removed from each side (1 square and 3 triangles on the left, 2 squares and 1 triangle on the right). Refer to MLR 2 (Collect and Display).


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