Question

Solve the system of linear equations in three variables
5x = 2y
7y = 5z
4x + 5y + 6z = 150

• A. x = 1, y = 4, z = 3
• B. x = 4, y = 10, z = 14
• C. x = 14, y = 10, z = 4
• D. x = 10, y = 14, z = 4

Answer 1

The correct option is B x = 4, y = 10, z = 14

5x - 2y + 0.z = 0 .......(1)
0.x + 7y - 5z = 0......(2)
4x + 5y + 6z = 150.....(3)
From eqn(1) we get $x=\frac{2y}{5}$------(4)
From eqn(2) we get $z=\frac{7y}{5}$-----(5)
Substituting eqn (4) and eqn(5) in eqn(3) we get
$4\times \frac{2y}{5}+5y+6\times \frac{7y}{5}=150$
$\frac{8y+25y+42y}{5}=150$
$\frac{75y}{5}=150$
$y=10$ -----------(6)
Now,
Substituting eqn(6) in eqn(4) we get
$x=\frac{2\times 10}{5}=4$
Again,
Substituting eqn(6) in eqn(5) we get
$z=\frac{7\times 10}{5}=14$

Verification:
Substituting the values of x,y and z in eqn(3) we get
$4x+5y+6z=150$
$4\times 4+5\times 10+6\times 14$
$=16+50+84$
$=150=RHS$