Introduction / Context:
Here we are given a system of three linear equations in three variables x, y and z and asked to find the symmetric expression xy + yz + zx. Instead of trying to guess values, the fastest method is to solve the system systematically, then calculate the required combination. Such problems check both basic linear algebra skills and the ability to work with symmetric expressions once the variables are known.
Given Data / Assumptions:
- Equation 1: 2x + 3y − 5z = 18.
- Equation 2: 3x + 2y + z = 29.
- Equation 3: x + y + 3z = 17.
- x, y, z are real numbers.
- We must compute S = xy + yz + zx.
Concept / Approach:
We solve the linear system by elimination or substitution. After obtaining numerical values for x, y and z, we plug them into xy, yz and zx and add them. Because the equations are well balanced, the solution turns out to have small integer values, making the final computation straightforward. This approach is direct and avoids the complexity of trying to express xy + yz + zx without solving for x, y, z explicitly.
Step-by-Step Solution:
1. Consider the system:
2x + 3y − 5z = 18,
3x + 2y + z = 29,
x + y + 3z = 17.
2. From the third equation, express x in terms of y and z: x = 17 − y − 3z.
3. Substitute x into the second equation: 3(17 − y − 3z) + 2y + z = 29.
4. Simplify: 51 − 3y − 9z + 2y + z = 29, giving 51 − y − 8z = 29.
5. Rearranging: −y − 8z = 29 − 51 = −22, so y + 8z = 22. Call this equation (A).
6. Substitute x into the first equation: 2(17 − y − 3z) + 3y − 5z = 18.
7. Simplify: 34 − 2y − 6z + 3y − 5z = 18, giving 34 + y − 11z = 18.
8. Rearranging: y − 11z = 18 − 34 = −16, so y = 11z − 16. Call this equation (B).
9. Substitute (B) into (A): (11z − 16) + 8z = 22.
10. This gives 19z − 16 = 22, so 19z = 38, hence z = 2.
11. Now find y from (B): y = 11 * 2 − 16 = 22 − 16 = 6.
12. Use the third equation x + y + 3z = 17 to find x: x + 6 + 3 * 2 = 17, so x + 12 = 17, giving x = 5.
13. We now have x = 5, y = 6, z = 2.
14. Compute S = xy + yz + zx = (5)(6) + (6)(2) + (5)(2).
15. Evaluate: 5 * 6 = 30, 6 * 2 = 12, 5 * 2 = 10, so S = 30 + 12 + 10 = 52.
Verification / Alternative check:
We can verify the solution by substituting x = 5, y = 6, z = 2 back into all three original equations. Equation 1: 2 * 5 + 3 * 6 − 5 * 2 = 10 + 18 − 10 = 18, correct. Equation 2: 3 * 5 + 2 * 6 + 2 = 15 + 12 + 2 = 29, correct. Equation 3: 5 + 6 + 3 * 2 = 5 + 6 + 6 = 17, also correct. Thus the triple (5, 6, 2) satisfies the system, and our computed S = 52 is consistent.
Why Other Options Are Wrong:
Option 32, 64, and 46 are obtained only by incorrect arithmetic or by using wrong values of x, y and z. For example, if any of the quantities were miscalculated, the re substitution check would fail for at least one of the original equations. Since (5, 6, 2) is the unique solution to the linear system, the corresponding value S = 52 is the only valid answer.
Common Pitfalls:
Typical errors include algebra mistakes when substituting x into other equations, sign errors when rearranging, or arithmetic slips while simplifying. Some learners also forget to check their solution in all three equations, which can allow hidden mistakes to slip through. A systematic elimination procedure with a quick verification step is the best defence against such issues.
Final Answer:
The required value of xy + yz + zx is
52.
Discussion & Comments