2
algebra

Solve by elimination (you may need to multiply one equation): {3x+2y=16x+y=6\begin{cases} 3x + 2y = 16 \\ x + y = 6 \end{cases}

A

(3,3)(3, 3)

B

(5,1)(5, 1)

C

(4,2)(4, 2)

D

(2,4)(2, 4)

Correct Answer: C

Choice C is the correct answer. To eliminate a variable, we need matching coefficients. Let's eliminate yy by multiplying the second equation by 2.

Step 1: Multiply the second equation by 2: 2(x+y)=2(6)2(x + y) = 2(6)2x+2y=122x + 2y = 12

Step 2: Subtract from the first equation: (3x+2y)(2x+2y)=1612(3x + 2y) - (2x + 2y) = 16 - 12x=4x = 4

Step 3: Substitute x=4x = 4 into the second equation: 4+y=64 + y = 6y=2y = 2

Solution: (4,2)(4, 2)

Verification: 3(4)+2(2)=12+4=163(4) + 2(2) = 12 + 4 = 16

💡 Strategic Tip: Multiply equations to create matching or opposite coefficients for elimination.

Choice A is incorrect because3(3)+2(3)=9+6=15 eq163(3) + 2(3) = 9 + 6 = 15 \ eq 16.

Choice B is incorrect because3(5)+2(1)=15+2=17 eq163(5) + 2(1) = 15 + 2 = 17 \ eq 16.

Choice D is incorrect because3(2)+2(4)=6+8=14 eq163(2) + 2(4) = 6 + 8 = 14 \ eq 16.