Question

7 what is the solution to the system of equations? 5x - 4y = -10 3x + 2y = 16 a x = 4, y = 2 b x = 2, y = 0 c no solution d x = 2, y = 5

Explanation:

Step1: Use elimination method

Multiply the second equation \(3x + 2y = 16\) by 2 to get \(6x + 4y = 32\).

Step2: Add to first equation

Add \(5x - 4y = -10\) and \(6x + 4y = 32\): \(5x - 4y + 6x + 4y = -10 + 32\), which simplifies to \(11x = 22\), so \(x = 2\).

Step3: Substitute x=2 into second equation

Substitute \(x = 2\) into \(3x + 2y = 16\): \(3(2) + 2y = 16\) → \(6 + 2y = 16\) → \(2y = 10\) → \(y = 5\)? Wait, no, wait, let's check option D: x=2, y=5. Wait, earlier miscalculation. Wait, let's check each option:

Option A: x=4, y=2. Plug into first equation: 5(4)-4(2)=20-8=12≠-10. So A is wrong.

Option B: x=2, y=0. 5(2)-4(0)=10≠-10. Wrong.

Option D: x=2, y=5. First equation: 5(2)-4(5)=10-20=-10 (correct). Second equation: 3(2)+2(5)=6+10=16 (correct). Wait, but the original option C was marked, but that's a mistake. Wait, let's recalculate.

Wait, the second equation: 3x + 2y = 16. Multiply by 2: 6x + 4y = 32. First equation: 5x - 4y = -10. Add them: 11x = 22 → x=2. Then substitute x=2 into 3x + 2y = 16: 6 + 2y = 16 → 2y=10 → y=5. So the solution is x=2, y=5, which is option D. Wait, maybe the original marking was wrong. Let's check each option:

Option D: x=2, y=5. First equation: 52 -45=10-20=-10 (matches). Second equation: 32 +25=6+10=16 (matches). So the correct answer is D. Wait, but the user's image had C marked, but that's incorrect. So the correct solution is x=2, y=5, which is option D.

Wait, my initial step 3 had a miscalculation. Let's redo step 3:

After x=2, substitute into 3x + 2y = 16: 3*2 + 2y = 16 → 6 + 2y = 16 → 2y = 10 → y=5. So y=5, not 0 or 2. So option D is correct.

Answer:

D. \( x = 2, y = 5 \)