Find the values of x, y and z, so as to satisfy the equations given below: x + y + z = 12; x + y – z = 6; x – y + z = 4
- (a)x = 5, y = 4, z = – 3
- (b)x = 5, y = – 4, z = 3
- (c)x = 5, y = – 4, z = – 3
- (d)x = 5, y = 4, z = 3
Answer
Why
Correct — D. The three equations differ only in signs, so subtract in pairs and one variable dies each time.
(1) x + y + z = 12 (2) x + y − z = 6 (3) x − y + z = 4
(1) − (2): 2z = 12 − 6 = 6, so z = 3
(1) − (3): 2y = 12 − 4 = 8, so y = 4
Back into (1): x = 12 − 4 − 3 = 5
x = 5, y = 4, z = 3 → option (d)
Why the others are wrong
- (a)x = 5, y = 4, z = – 3 — With z = −3, equation (1) reads 5 + 4 − 3 = 6, not 12. Flipping the sign on z turns equation (1) into equation (2).
- (b)x = 5, y = – 4, z = 3 — With y = −4, equation (1) reads 5 − 4 + 3 = 4, not 12 — eight short, because y has been subtracted instead of added.
- (c)x = 5, y = – 4, z = – 3 — Both signs flipped gives 5 − 4 − 3 = −2 in equation (1); this pair of values fails all three equations, not just one.
Concept
Three linear equations in three unknowns are solved by elimination: combine two equations so one variable cancels, repeat, then back-substitute.
Here the system is unusually kind. Equations (1) and (2) agree on x and y and differ only in the sign of z, so subtracting them leaves 2z alone. The same trick on (1) and (3) leaves 2y.
In a multiple-choice setting there is a faster route: substitute each option into the simplest equation. Every wrong option here already fails x + y + z = 12, so that one check settles the item.
Key facts
- Subtracting two equations that differ only in one variable's sign isolates that variable.
- Three unknowns need three independent equations, and here two subtractions plus one substitution finish the system.
- All three wrong options fail x + y + z = 12, so checking equation (1) alone identifies the answer.
Study next
Common traps
- Subtracting in the wrong order and carrying the minus sign into z.
- Adding (1) and (2) instead of subtracting, which kills z but leaves x + y = 9 with two unknowns.
- Solving correctly and then picking an option that lists the same numbers with a sign changed.
Every coefficient here is 1 and the numbers are small, which is the signal that option-testing will beat full elimination on the clock.
Related PYQs
No directly related past PYQ was found.