Choice B is the correct answer. We can use elimination by multiplying the second equation by 3.

Step 1: Multiply the second equation by 3: $3(2x - y) = 3(5)$$6x - 3y = 15$

Step 2: Add to the first equation: $(4x + 3y) + (6x - 3y) = 21 + 15$$10x = 36$$x = 3.6$

That gives non-integer. - First: $4(3) + 3(3) = 12 + 9 = 21$ ✓

• Second: $2(3) - 3 = 6 - 3 = 3 \ eq 5$

Step 1: Multiply the second equation by 3: $6x - 3y = 9$

Step 2: Add to the first equation: $10x = 30$$x = 3$

Step 3: Substitute into second equation: $2(3) - y = 3$$6 - y = 3$$y = 3$

Solution: $(3, 3)$

Verification: $4(3) + 3(3) = 21$ ✓ and $2(3) - 3 = 3$ ✓

💡 Strategic Tip: Choose the method that minimizes fractional arithmetic.

Choice A is incorrect because$2(2) - 4.33 = -0.33 \ eq 3$.

Choice C is incorrect because$4(4) + 3(\frac{5}{3}) = 16 + 5 = 21$ ✓, but $2(4) - \frac{5}{3} = \frac{19}{3} \ eq 3$.

Choice D is incorrect because$4(1) + 3(\frac{17}{3}) = 4 + 17 = 21$ ✓, but $2(1) - \frac{17}{3} = -\frac{11}{3} \ eq 3$.