Q.13
The solution for the following set of equations is,
5x + 4y + 10z = 13
x + 3y + z = 7
4x – 2y + z = 0
- (A) x = 2, y = 1, z = 1
- (B) x = 1, y = 2, z = 0
- (C) x = 1, y = 0, z = 2
- (D) x = 0, y = 1, z = 2
Solve 5x + 4y + 10z = 13 Linear Equations System
Linear equations systems like 5x + 4y + 10z = 13, x + 3y + z = 7, and 4x – 2y + z = 0 test elimination and substitution skills in exams. The correct solution is x = 1, y = 2, z = 0 (Option B), verified by plugging values into all equations. This article breaks down the solving method and checks every option for MCQ prep.
Solving Method
Label equations as:
- 5x + 4y + 10z = 13 (Eq1)
- x + 3y + z = 7 (Eq2)
- 4x – 2y + z = 0 (Eq3)
Subtract Eq2 from Eq3: (4x – 2y + z) – (x + 3y + z) = 0 – 7 → 3x – 5y = -7 [Eq4]
Eliminate z from Eq1 and Eq2: Multiply Eq2 by 10 → 10x + 30y + 10z = 70. Subtract Eq1: 5x + 26y = 57 [Eq6]
From Eq4: x = (5y – 7)/3. Substitute into Eq6: 5*(5y-7)/3 + 26y = 57. Multiply by 3: 25y – 35 + 78y = 171 → 103y = 206 → y = 2.
Then x = (5*2 – 7)/3 = 1. From Eq2: 1 + 3*2 + z = 7 → z = 0.
Option Verification
| Option | Values | Eq1 Check | Eq2 Check | Eq3 Check | Result |
|---|---|---|---|---|---|
| (A) | x=2, y=1, z=1 | 5(2)+4(1)+10(1)=24≠13 | 2+3+1=6≠7 | 8-2+1=7≠0 | Wrong |
| (B) | x=1, y=2, z=0 | 5(1)+4(2)+10(0)=13✓ | 1+6+0=7✓ | 4-4+0=0✓ | Correct |
| (C) | x=1, y=0, z=2 | 5+0+20=25≠13 | 1+0+2=3≠7 | 4-0+2=6≠0 | Wrong |
| (D) | x=0, y=1, z=2 | 0+4+20=24≠13 | 0+3+2=5≠7 | 0-2+2=0✓ | Wrong |
Why Other Options Fail
- (A) x=2, y=1, z=1: Eq1: 10+4+10=24≠13. Fails immediately.
- (C) x=1, y=0, z=2: Eq1: 5+0+20=25≠13. Incorrect.
- (D) x=0, y=1, z=2: Eq1: 0+4+20=24≠13; Eq2: 0+3+2=5≠7. Wrong.
