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.

• 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.

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

• Elimination Method
• Substitution Method

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

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

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.

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

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

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.

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.

• 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.

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

• Elimination Method
• Substitution Method

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

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

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.

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

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

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.