7
algebra

Solve by elimination: {5x+3y=295x3y=11\begin{cases} 5x + 3y = 29 \\ 5x - 3y = 11 \end{cases}

A

(3,143)(3, \frac{14}{3})

B

(5,43)(5, \frac{4}{3})

C

(4,3)(4, 3)

D

(2,193)(2, \frac{19}{3})

Correct Answer: C

Choice C is the correct answer. The 3y3y terms are opposites, so we add the equations.

Step 1: Add the equations: (5x+3y)+(5x3y)=29+11(5x + 3y) + (5x - 3y) = 29 + 1110x=4010x = 40x=4x = 4

Step 2: Substitute x=4x = 4 into the first equation: 5(4)+3y=295(4) + 3y = 2920+3y=2920 + 3y = 293y=93y = 9y=3y = 3

Solution: (4,3)(4, 3)

Verification: 5(4)3(3)=209=115(4) - 3(3) = 20 - 9 = 11

💡 Strategic Tip: Opposite coefficients signal that addition will eliminate the variable.

Choice A is incorrect because5(3)+3(143)=15+14=295(3) + 3(\frac{14}{3}) = 15 + 14 = 29 ✓, but 5(3)14=1 eq115(3) - 14 = 1 \ eq 11.

Choice B is incorrect because5(5)3(43)=254=21 eq115(5) - 3(\frac{4}{3}) = 25 - 4 = 21 \ eq 11.

Choice D is incorrect because5(2)+3(193)=10+19=295(2) + 3(\frac{19}{3}) = 10 + 19 = 29 ✓, but 5(2)19=9 eq115(2) - 19 = -9 \ eq 11.