Set 7: Systems of Equations (Intermediate)

Choice C is correct. Choice C is the correct answer. The $3y$ terms are opposites, so we add the equations. Step 1: Add the equations: (4 x + 3 y) + (4 x - 3 y) = 27 + 13$8x = 40$x = 5 Step 2: Substitute $x = 5$ into the first equation: 4(5) + 3 y = 27$20+ 3 y = 273y = 7y = \frac{7}{3}$$ Solution: (5, \frac{7}{3})$Verification:$4(5) - 3(\frac{7}{3}) = 20 - 7 = 13$✓ Strategic Tip: Opposite coefficients make addition the ideal elimination strategy. Choice A is incorrect because$4(6) + 3(1) = 24 + 3 = 27$✓, but$4(6) - 3(1) = 24 - 3 = 21 \neq 13$. Choice B is incorrect because$4(7) - 3(-\frac{1}{3}) = 28 + 1 = 29 \neq 13$. Choice D is incorrect because$4(8) + 3(-\frac{5}{3}) = 32 - 5 = 27$✓, but$4(8) - 3(-\frac{5}{3}) = 32 + 5 = 37 \neq 13$.