Linear Equation in Three Variables
Category : 10th Class
We can also use the system of linear equation for solving the linear equation in three variables by mere substitution. In this method we find one of the three variables in terms of other two from any one of the equation and substitute it in the second equation. From the second equation we obtained the second variable in terms of the other and substitute it in the third equation and solve it to get the third variable which on re-substituting in the previous steps we get the other variables. It can also be solved by elimination method.
Solve the following system of linear equation: \[3x-y+4z=3,x+2y-3z=0\, and \,6x+5y=-3.\]
(a) \[x=-\frac{39}{10},y=\frac{47}{10},z=\frac{5}{2}\]
(b) \[x=\frac{39}{10},y=\frac{47}{10},z=\frac{5}{2}\]
(c) \[x=\frac{39}{10},y=\frac{7}{10},z=\frac{25}{2}\]
(d) \[x=-\frac{9}{10},y=\frac{17}{10},z=\frac{25}{2}\]
(e) None of these
Answer: (e)
Explanation
We have,
\[3x-y+4z=3-----\left( 1 \right)\]
\[x-2y-3z=-2-----\left( 2 \right)\]
\[6x+5y-5z=-3-----\left( 3 \right)\]
Form equation (1), we have
\[-y=3-3x-4x-----\left( 4 \right)\]
Putting in equation (2) the above value we get,
\[x=\frac{8-13z}{5}-----\left( 5 \right)\]
Putting equation (5) in (4) we get,
\[y=\frac{-9+13z}{5}-----\left( 6 \right)\]
Solve the following system of linear equation: \[5x-7y+z=11,6x-8y-z=15\,and\,3x+2y-6z=7.\]
(a) \[x=2,y=3,z=2\]
(b) \[x=-2,y=5,z=-1\]
(c) \[x=1,y=-1,z=-1\]
(d) \[x=-3,y=1,z=-2\]
(e) None of these
Answer: (c)
Solve the system of the equation: \[6x+y-3z=5,x+3y-2z=5,2x+y+4z=8\]
(a) \[x=1,y=2,z=1\]
(b) \[x=-2,y=5,z=-1\]
(c) \[x=1,y=-1,z=-1\]
(d) \[x=-3,y=1,z=-2\]
(e) None of these
Answer: (a)
Solve the system of equation: \[2y-3z=0,x+3y=-4,3x+4y=3\]
(a) \[x=2,y=3,z=2\]
(b) \[x=5,y=-3,z=-2\]
(c) \[x=4,y=-3,z=-1\]
(d) \[x=-2,y=2,z=-2\]
(e) None of these
Answer: (b)
Find the solution of the system of the equation: \[x+y=8,y+z=10,x+z=12\]
(a) \[x=5,y=3,z=2\]
(b) \[x=5,y=5,z=-1\]
(c) \[x=8,y=-2,z=-1\]
(d) \[x=5,y=3,z=7\]
(e) None of these
Answer: (d)
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