This is a systems of equations question testing the elimination method. Choice B (4) is correct — add the two equations to eliminate y: (3x + y) + (x − y) = 14 + 2 → 4x = 16 → x = 4. (If desired, verify: y = x − 2 = 2, and 3(4) + 2 = 14 ✓.) Choice A (2) is the y-value, not x — the student may have solved for y and stopped, or substituted back into the wrong equation. Choice C (8) comes from solving 4x = 16 as x = 16/2 = 8, dividing by 2 instead of 4. Choice D (12) likely comes from adding the right-hand sides to get 16, then adding 14 − 2 = 12 via some misalignment. Pro tip: When one variable has matching coefficients with opposite signs (y and −y here), adding the equations eliminates that variable immediately. Always double-check by substituting both values back into BOTH original equations.

Set the costs equal to find when taxis charge the same. We have 2x + 3 = 3x + 1. Subtract 2x from both sides: 3 = x + 1. Subtract 1 from both sides: x = 2. At 2 miles, both taxis cost $7.

Use elimination to solve this system. Add the two equations: (4x + y) + (2x - y) = 10 + 2, which gives 6x = 12, so x = 2. This matches choice B.

Use the elimination method by adding $2x + 3y = 13$ and $2x - 3y = 1$. This eliminates y, giving $4x = 14$, so $x = 14/4 = 3.5$. Substitute into $2x - 3y = 1$: $2(3.5) - 3y = 1$, $7 - 3y = 1$, $-3y = -6$, $y = 2$. Alternatively, subtract the equations to get $6y = 12$, $y = 2$ directly. Choice C of 3 might come from misadding to $4x = 12$.

Use elimination by adding the equations directly. Adding 5x - y = 14 and 2x + y = 7 gives 7x = 21, so x = 3. Substitute x = 3 into 2x + y = 7: 2(3) + y = 7, so y = 1. The solution is (3, 1).

Use elimination to solve this system with $x + y = 13$ and $x - y = 5$. Adding the equations eliminates $y$: $2x = 18$, so $x = 9$. Substitute back: $9 + y = 13$, so $y = 4$. The solution is $(9, 4)$.

Use elimination to solve this system. Adding x + y = 9 and x - y = 1 eliminates y: 2x = 10, so x = 5. Substituting into x + y = 9: 5 + y = 9, so y = 4. The solution is (5, 4).

Use elimination by adding the equations. Adding 3x + 4y = 1 and 6x - 4y = 11 gives 9x = 12, so x = 4/3. Substitute into 3x + 4y = 1: 3(4/3) + 4y = 1, which gives 4 + 4y = 1, so 4y = -3 and y = -3/4.

Use elimination method by adding the equations. Adding 2x + y = 11 and x - y = 1 gives 3x = 12, so x = 4. Substitute x = 4 into x - y = 1: 4 - y = 1, so y = 3. The solution is (4, 3). Choice B reverses the x and y values.

Use elimination: add equations $x + 2y = 10$ and $3x - 2y = 6$. The y terms cancel, giving $4x = 16$, so $x = 4$. Substitute $x = 4$ into $x + 2y = 10$: $4 + 2y = 10$, so $y = 3$. The solution is $ (4, 3) $.