Question 1 • Algebra III A
Substitute y = 5x − 1 into the second equation, −15x − 3y = 3. This gives −15x − 3(5x − 1) = 3, so −30x + 3 = 3 and x = 0. Then y = 5(0) − 1 = −1, so the system has one solution: (0, −1).
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