3
algebra

Solve by elimination: {3x+y=133xy=5\begin{cases} 3x + y = 13 \\ 3x - y = 5 \end{cases}

A

(2,7)(2, 7)

B

(4,1)(4, 1)

C

(3,4)(3, 4)

D

(5,2)(5, -2)

Correct Answer: C

Choice C is the correct answer. The yy terms are opposites (+y+y and y-y), so we can add the equations to eliminate yy.

Step 1: Add the two equations: (3x+y)+(3xy)=13+5(3x + y) + (3x - y) = 13 + 56x=186x = 18x=3x = 3

Step 2: Substitute x=3x = 3 into the first equation: 3(3)+y=133(3) + y = 139+y=139 + y = 13y=4y = 4

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

Verification: 3(3)4=94=53(3) - 4 = 9 - 4 = 5

💡 Strategic Tip: When coefficients are opposites, adding eliminates that variable immediately.

Choice A is incorrect because3(2)+7=6+7=133(2) + 7 = 6 + 7 = 13 ✓, but 3(2)7=67=1 eq53(2) - 7 = 6 - 7 = -1 \ eq 5.

Choice B is incorrect because3(4)1=121=11 eq53(4) - 1 = 12 - 1 = 11 \ eq 5.

Choice D is incorrect because3(5)(2)=15+2=17 eq53(5) - (-2) = 15 + 2 = 17 \ eq 5.