options: Analyze and solve pairs of simultaneous linear equations.

Solve linear equations in one variable.
problem_activity: Activity
The purpose of this activity is for students to recognize that multiplying an equation by a value results in the same solution. This will be useful when transforming equations for the elimination method of solving systems of equations in the supported Algebra lesson. Students will work in pairs and each partner is responsible for answering the questions in either column A or column B. Although each row has two different problems, they share the same answer. Ensure that students work their problems out independently and collaborate with one another when they do not arrive to the same answers. Students construct viable arguments and critique the reasoning of others (MP3) when they resolve errors by critiquing their partner's work or explaining their reasoning.
Launch
Arrange students in groups of two. In each group, ask students to decide who will work on column A and who will work on column B. If students are unfamiliar with the row game structure, demonstrate the protocol before they start working.
Student Facing
Work independently on your column. Partner A completes the questions in column A only and partner B completes the questions in column B only. Your answers in each row should match. Work on one row at a time and check if your answer matches your partner’s before moving on. If you don’t get the same answer, work together to find any mistakes.
Solve each equation.
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Student Response
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Activity Synthesis
The purpose of the discussion is to notice that equations that have been multiplied by the same value have the same solution.
Consider asking students to explain why their solutions might be the same for each row.
For each row, the two equations are equivalent. Sample responses for each row:
The equation in column B is 2 times the equation in column A.
In each column, one side of the equation is the distributed form of the same side of the equation in the other column.
The equation in column B is
\(\frac{1}{2}\)
of the equation in column A.
The equation in column B is
\(\frac{2}{3}\)
of the equation in column A.
The equation in column B has
\(4y\)
added to both sides of the equation from column A.
The equation in column B is
\(\frac{5}{2}\)
of the equation in column A.
While there may be other valid ways of connecting the equations (for example, in row 4, dividing the equation in column A by 3 and dividing the equation in column B by 2 result in the same equation), encourage students to recognize that there is a single multiple of one equation that results in the other equation.


correct_option_index: 1