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Mathematics, 22.03.2021 18:00 stacy99
When solving the system of equations 3xβ3y=1 and β2x+4y=2 algebraically, a good first step is to:
A. Multiply 3xβ3y=1 by 2, and multiply β2x+4y=2 by 3.
B. Multiply 3xβ3y=1 by 3, and multiply β2x+4y=2 by β2.
C. Rewrite both equations in slope-intercept form.
D. Add the equations together.
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When solving the system of equations 3xβ3y=1 and β2x+4y=2 algebraically, a good first step is to:
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