Solve the given set of three equations to get the value of the unknowns: x + 2z = 3 x + 2y + 3z = 5 3x - 5z = -13

Correct — C. Two of the three equations contain no y, so solve that pair first.

From x + 2z = 3: x = 3 − 2z
Put it in 3x − 5z = −13: 3(3 − 2z) − 5z = −13
9 − 6z − 5z = −13, so −11z = −22 and z = 2
Back-substitute: x = 3 − 2(2) = −1

Now the middle equation, x + 2y + 3z = 5:
(−1) + 2y + 6 = 5, so 2y = 0 and y = 0

Check all three: (−1) + 4 = 3, (−1) + 0 + 6 = 5, (−3) − 10 = −13 → option (c)

Why the others are wrong

• (a)
  x = 0,y = -1,z = 2 — Test it in the shortest equation, x + 2z = 3: 0 + 4 = 4, not 3. This option carries the answer's x and y values swapped.
• (b)
  x = 2,y = 0,z = -1 — Here x and z are interchanged: x + 2z gives 2 − 2 = 0, not 3, and 3x − 5z gives 6 + 5 = 11, not −13.
• (d)
  x = -1,y = 2,z = 0 — x is right but z is not: x + 2z = −1 + 0 = −1, not 3. Only z = 2 satisfies the pair of equations that exclude y.

Three unknowns, but not three equations that each need all three — and that is the shortcut.

Equations x + 2z = 3 and 3x − 5z = −13 carry no y at all, so they are a 2 × 2 system in x and z and can be solved on their own. Once x and z are fixed, y drops out of the remaining equation in a single line.

Elimination is just as quick: 3 × (x + 2z = 3) gives 3x + 6z = 9, and subtracting 3x − 5z = −13 leaves 11z = 22.

Key facts

• When a variable is missing from two of three equations, solve those two first.
• x + 2z = 3 with 3x − 5z = −13 gives z = 2 and x = −1.
• y appears only in x + 2y + 3z = 5, so it is found last, in one step.
• A solution must satisfy every equation, not the one you happened to test.

Common traps

• Launching a full three-variable elimination when two equations already exclude y.
• Sign slips in 3(3 − 2z) − 5z, where −6z and −5z combine to −11z.
• Ticking an option whose x and z look plausible without testing all three equations.

No directly related past PYQ was found.