Systems of Linear Equations in Three Variables

3.1 Systems of Linear Equations in Three Variables
 
Definition
Two or more linear equations involving the same set of variables form a system of linear equations.
 
Visual representation of characteristics of systems of linear equations in three variables
 
General Form of a Linear Equation in Three Variables

The general form of a linear equation in three variables can be written as follows:

\(ax+by+cz=d\),

where \(a\)\(b\), and \(c\) are not equal to zero.

 
Example of Systems of Linear Equations in Three Variables
System of Equations Description
\(\begin{aligned} 2x+4y-z&=10\\ x+y&=10z^2\\ 5y-z-2x&=3 \end{aligned}\) Not systems of linear equations, because there is an equation in which the highest power of the variable is \(2\).
\(\begin{aligned} p+8q-4r&=2\\ 2(p+6r)+7q&=0\\ 10r+p&=5q \end{aligned}\) Yes, because all three equations have three variables, \(p\)\(q\), and \(r\), of power \(1\).
 
Relationship of Systems of Linear Equations in Three Variables with Axis
  • A system of linear equations in three variables has three axes, namely the \(x\)-axis, \(y\)-axis and \(z\)-axis. All three linear equations form a plane on each axis.
  • Every linear equation in two variables forms a straight line on each axis.
 
Form in Cartesian Plane

Geometrically, a linear equation in three variables forms a plane in a three-dimensional space.

Triangular plane formed by three points and connecting lines.

 
Type of Solution Description
One Solution The planes intersect at only one point
Infinite Solutions The planes intersect in a straight line
No Solution The planes do not intersect at any point
 
Methods used to solve systems of linear equations in three variables
  • Elimination Method
  • Substitution Method
 
Example \(1\)
Question

Solve the following system of linear equations using the elimination method.

\(\begin{aligned} 4x-3y+z&=-10 \\ 2x+y+3z&=0\\ -x+2y-5z&=17 \end{aligned}\)

Solution

Choose any two equations.

\(\begin{aligned} 4x-3y+z&=-10 \quad \cdots\boxed{1} \\ 2x+y+3z&=0 \quad \quad \,\, \cdots \boxed{2} \end{aligned}\)


Multiply equation \(\boxed{2}\) with \(2\) so that the coefficients of \(x\) are equal.

\(\boxed{2}\times 2:\quad 4x+2y+6z=0 \quad \cdots \boxed{3}.\)


Eliminate the variable \(x\) by subtracting \(\boxed{1}\) from \(\boxed{3}\).

\(\boxed{3}-\boxed{1}:\quad 5y+5z=10 \quad \cdots \boxed{4}.\)


Choose another set of two equations.

\(\begin{aligned} 2x+y+3z&=0 \quad\,\,\, \cdots \boxed{5} \\ -x+2y-5z&=17 \quad \cdots\boxed{6} \end{aligned}\)


Multiply equation \(\boxed{6}\) with \(2\) so that the coefficients of variable \(x\) are equal.

\(\begin{aligned} \boxed{6}\times2:-2x+4y-10z&=34 \quad\cdots\boxed{7}\\ \boxed{5}+\boxed{7}: \quad\quad\quad 5y-7z&=34 \quad\cdots\boxed{8} \\ \boxed{4}-\boxed{8}: \quad\quad\quad\quad\,\,\,\ 12z&=-24 \\ z&=-2. \end{aligned}\)


Substitute \(z=-2\) into \(\boxed{8}\).

\(\begin{aligned} 5y-7(-2)&=34 \\ 5y+14&=34\\ 5y&=20 \\ y&=4. \end{aligned}\)


Substitute \(y=4\) and \(z=-2\) into \(\boxed{1}\).

\(\begin{aligned} 4x-3(4)+(-2)&=-10\\ 4x-12-2&=-10\\ 4x&=4\\ x&=1. \end{aligned}\)

Thus, \(x=1\)\(y=4\), and \(z=-2\) are the solutions to this system of linear equations.

 
Example \(2\)
Question

Solve the following system of linear equations using the substitution method.

\(\begin{aligned} 3x-y-z&=-120 \\ y-2z&=30 \\ x+y+z&=180 \end{aligned}\)

Solution

Label all the equations.

\(\begin{aligned} 3x-y-z&=-120 \quad \cdots \boxed{1} \\ y-2z&=30 \quad\quad\,\, \cdots \boxed{2} \\ x+y+z&=180 \quad\,\,\,\ \cdots \boxed{3} \end{aligned}\)


From \(\boxed{1}\)\(z=3x-y+120 \quad\cdots\boxed{4}\).

Substitute \(\boxed{4}\) into \(\boxed{2}\).

\(\begin{aligned} y-2(3x-y+120)&=30 \\ y-6x+2y-240&=30 \\ -6x+3y&=270 \\ y&=90+2x \quad \cdots \boxed{5}. \end{aligned}\)


Substitute \(\boxed{4}\) and \(\boxed{5}\) into \(\boxed{3}\).

\(\begin{aligned} x+(90+2x)+[3x-(90+2x)+120]&=180 \\ x+2x+3x-2x+90-90+120&=180 \\ 4x&=60 \\ x&=15. \end{aligned}\)


Substitute \(x=15\) into \(\boxed{5}\).

\(\begin{aligned} y&=90+2(15) \\ &=120. \end{aligned}\)


Substitute \(x=15\) and \(y=120\) into \(\boxed{3}\).

\(\begin{aligned} 15+120+z&=180 \\ z&=45. \end{aligned}\)

Thus, \(x=15\)\(y=120\), and \(z=45\) are the solutions to this system of linear equations.

 

Systems of Linear Equations in Three Variables

3.1 Systems of Linear Equations in Three Variables
 
Definition
Two or more linear equations involving the same set of variables form a system of linear equations.
 
Visual representation of characteristics of systems of linear equations in three variables
 
General Form of a Linear Equation in Three Variables

The general form of a linear equation in three variables can be written as follows:

\(ax+by+cz=d\),

where \(a\)\(b\), and \(c\) are not equal to zero.

 
Example of Systems of Linear Equations in Three Variables
System of Equations Description
\(\begin{aligned} 2x+4y-z&=10\\ x+y&=10z^2\\ 5y-z-2x&=3 \end{aligned}\) Not systems of linear equations, because there is an equation in which the highest power of the variable is \(2\).
\(\begin{aligned} p+8q-4r&=2\\ 2(p+6r)+7q&=0\\ 10r+p&=5q \end{aligned}\) Yes, because all three equations have three variables, \(p\)\(q\), and \(r\), of power \(1\).
 
Relationship of Systems of Linear Equations in Three Variables with Axis
  • A system of linear equations in three variables has three axes, namely the \(x\)-axis, \(y\)-axis and \(z\)-axis. All three linear equations form a plane on each axis.
  • Every linear equation in two variables forms a straight line on each axis.
 
Form in Cartesian Plane

Geometrically, a linear equation in three variables forms a plane in a three-dimensional space.

Triangular plane formed by three points and connecting lines.

 
Type of Solution Description
One Solution The planes intersect at only one point
Infinite Solutions The planes intersect in a straight line
No Solution The planes do not intersect at any point
 
Methods used to solve systems of linear equations in three variables
  • Elimination Method
  • Substitution Method
 
Example \(1\)
Question

Solve the following system of linear equations using the elimination method.

\(\begin{aligned} 4x-3y+z&=-10 \\ 2x+y+3z&=0\\ -x+2y-5z&=17 \end{aligned}\)

Solution

Choose any two equations.

\(\begin{aligned} 4x-3y+z&=-10 \quad \cdots\boxed{1} \\ 2x+y+3z&=0 \quad \quad \,\, \cdots \boxed{2} \end{aligned}\)


Multiply equation \(\boxed{2}\) with \(2\) so that the coefficients of \(x\) are equal.

\(\boxed{2}\times 2:\quad 4x+2y+6z=0 \quad \cdots \boxed{3}.\)


Eliminate the variable \(x\) by subtracting \(\boxed{1}\) from \(\boxed{3}\).

\(\boxed{3}-\boxed{1}:\quad 5y+5z=10 \quad \cdots \boxed{4}.\)


Choose another set of two equations.

\(\begin{aligned} 2x+y+3z&=0 \quad\,\,\, \cdots \boxed{5} \\ -x+2y-5z&=17 \quad \cdots\boxed{6} \end{aligned}\)


Multiply equation \(\boxed{6}\) with \(2\) so that the coefficients of variable \(x\) are equal.

\(\begin{aligned} \boxed{6}\times2:-2x+4y-10z&=34 \quad\cdots\boxed{7}\\ \boxed{5}+\boxed{7}: \quad\quad\quad 5y-7z&=34 \quad\cdots\boxed{8} \\ \boxed{4}-\boxed{8}: \quad\quad\quad\quad\,\,\,\ 12z&=-24 \\ z&=-2. \end{aligned}\)


Substitute \(z=-2\) into \(\boxed{8}\).

\(\begin{aligned} 5y-7(-2)&=34 \\ 5y+14&=34\\ 5y&=20 \\ y&=4. \end{aligned}\)


Substitute \(y=4\) and \(z=-2\) into \(\boxed{1}\).

\(\begin{aligned} 4x-3(4)+(-2)&=-10\\ 4x-12-2&=-10\\ 4x&=4\\ x&=1. \end{aligned}\)

Thus, \(x=1\)\(y=4\), and \(z=-2\) are the solutions to this system of linear equations.

 
Example \(2\)
Question

Solve the following system of linear equations using the substitution method.

\(\begin{aligned} 3x-y-z&=-120 \\ y-2z&=30 \\ x+y+z&=180 \end{aligned}\)

Solution

Label all the equations.

\(\begin{aligned} 3x-y-z&=-120 \quad \cdots \boxed{1} \\ y-2z&=30 \quad\quad\,\, \cdots \boxed{2} \\ x+y+z&=180 \quad\,\,\,\ \cdots \boxed{3} \end{aligned}\)


From \(\boxed{1}\)\(z=3x-y+120 \quad\cdots\boxed{4}\).

Substitute \(\boxed{4}\) into \(\boxed{2}\).

\(\begin{aligned} y-2(3x-y+120)&=30 \\ y-6x+2y-240&=30 \\ -6x+3y&=270 \\ y&=90+2x \quad \cdots \boxed{5}. \end{aligned}\)


Substitute \(\boxed{4}\) and \(\boxed{5}\) into \(\boxed{3}\).

\(\begin{aligned} x+(90+2x)+[3x-(90+2x)+120]&=180 \\ x+2x+3x-2x+90-90+120&=180 \\ 4x&=60 \\ x&=15. \end{aligned}\)


Substitute \(x=15\) into \(\boxed{5}\).

\(\begin{aligned} y&=90+2(15) \\ &=120. \end{aligned}\)


Substitute \(x=15\) and \(y=120\) into \(\boxed{3}\).

\(\begin{aligned} 15+120+z&=180 \\ z&=45. \end{aligned}\)

Thus, \(x=15\)\(y=120\), and \(z=45\) are the solutions to this system of linear equations.