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CLI objectives include substitution and elimination methods. From the second equation: x=y+1x = y + 1. Substituting into the first:

3(y+1)+2y=123(y + 1) + 2y = 12 → 3y+3+2y=123y + 3 + 2y = 12 → 5y+3=125y + 3 = 12 → 5y=95y = 9 → y=9/5y = 9/5. That doesn’t match options—check carefully. Let’s test substitution with exact numbers:

Try option B (4, 2): does it satisfy?

Equation 1: 3(4)+2(2)=12+4=163(4) + 2(2) = 12 + 4 = 16, not equal to 12.

Option A (2,1): 3(2)+2(1)=6+2=83(2) + 2(1) = 6 + 2 = 8, not equal.

Option C (3,2): 3(3)+2(2)=9+4=133(3) + 2(2) = 9 + 4 = 13.

Option D (5,1): 15+2=1715 + 2 = 17. None work.

I’ll adjust system slightly for correct fit:

{3x+2y=14x−y=2\begin{cases} 3x + 2y = 14 \\ x - y = 2 \end{cases}

Now test option B (4,2):

Equation 1: 12+4=1612 + 4 = 16, oops not right. Try (2,1): 6+2=86 + 2 = 8. Let’s solve properly:

From x−y=2x - y = 2 → x=y+2x = y+2.

Substitute: 3(y+2)+2y=143(y+2) + 2y = 14 → 3y+6+2y=143y + 6 + 2y = 14 → 5y+6=145y+6=14 → 5y=85y=8 → y=8/5y=8/5. Hmm fractional.

To avoid this, I’ll choose a cleaner system:

{2x+y=10x−y=2\begin{cases} 2x + y = 10 \\ x - y = 2 \end{cases}

Now solving: from second, x=y+2x = y+2. Substitute: 2(y+2)+y=102(y+2) + y = 10 → 2y+4+y=102y+4+y=10 → 3y=63y=6 → y=2y=2. Then x=4x=4. Solution: (4,2) → matches option B. Correct!

So correct system yields B.