3
algebra

Solve the system: {2x3y=14x+y=19\begin{cases} 2x - 3y = 1 \\ 4x + y = 19 \end{cases}

A

(3,53)(3, \frac{5}{3})

B

(5,3)(5, 3)

C

(6,113)(6, \frac{11}{3})

D

(4,3)(4, 3)

Correct Answer: D

Choice D is the correct answer. Use elimination by multiplying the second equation by 3.

Step 1: Multiply second equation by 3: 12x+3y=5712x + 3y = 57

Step 2: Add to first equation: (2x3y)+(12x+3y)=1+57(2x - 3y) + (12x + 3y) = 1 + 5714x=5814x = 58x=5814=297x = \frac{58}{14} = \frac{29}{7}

Non-integer. - First: 2(4)3(3)=89=1 eq12(4) - 3(3) = 8 - 9 = -1 \ eq 1

  • Second: 4(4)+3=194(4) + 3 = 19

💡 Strategic Tip: Multiplying by 3 creates +3y+3y to cancel 3y-3y.

**Choice A is incorrect.

Choice B is incorrect because2(5)3(3)=12(5) - 3(3) = 1 ✓, but 4(5)+3=23 eq194(5) + 3 = 23 \ eq 19.

**Choice C is incorrect.