A kosarad üres
Linear systems of equations in more than two unknowns
(higher level courseware)
In cases when there are several unknowns in the question we can still find the solutions similarly to the methods explained in the previous chapter. In this case we use systems of equations in several variables.
In the below exercises we are going to see a few examples of linear systems of equations in three unknowns.
Example 1
Let us solve the following simultaneous equations in three variables on the set of real numbers:
\latex{\begin{rcases}\begin{align*} 2x - 3y + z = -1 \\ 5x + y + 2z = 13 \\ x + 2y - 2z = -1 \end{align*}\end{rcases}}.
Solution
We give the solution by continually eliminating the unknowns. First we isolate \latex{z} from the first equation, and then we substitute it into the other two equations:
\latex{\begin{rcases}\begin{align*} z=-2x+3y-1\\5x+y+2\times\left(-2x+3y-1\right)=13\\x+2y-2\times\left(-2x+3y-1\right)=-1\end{align*}\end{rcases}}.
The last two equations lead to the following simultaneous equations:
\latex{\begin{rcases}\begin{align*} x + 7y = 15 \\ 5x -4 y = -3 \\ \end{align*}\end{rcases}}.
From the first equation: \latex{x = –7y + 15} results, and by substituting it into the second one:
\latex{5\times\left(-7y+15\right)-4y=-3, \\ -39y=-78, \\ y=2. }
By substituting it back respectively: \latex{x + 7 \times 2 = 15}, from which \latex{x = 1} and \latex{z = –2 \times 1+3 \times 2 – 1 = 3} result.
Therefore the solution of the simultaneous equations is \latex{x = 1}; \latex{y = 2}; \latex{z = 3}; \latex{(x; y; z) = (1; 2; 3)} in other form, which is also confirmed by checking.
⯁ ⯁ ⯁
In this case we can again apply the method of equal coefficients, for example as follows.
By subtracting the double of the first equation from the second equation:
\latex{x + 7y = 15}.
By adding the double of the first equation to the third equation:
\latex{5x – 4y = –3}.
It is enough to examine the simultaneous equations resulting from these:
\latex{\begin{rcases}\begin{align*} x + 7y = 15 \\ 5x -4 y = -3 \\ \end{align*}\end{rcases}}.
We subtract \latex{5} times the first equation from the second equation:
\latex{-39y=-78},
\latex{y=2}.
\latex{y=2}.
From here, similarly to the first solution, we can get the values of the other unknowns of the simultaneous equations.
⯁ ⯁ ⯁
While solving simultaneous equations in three or more unknowns we always try for eliminating the unknowns one by one to get to simultaneous equations in fewer unknowns.
Example 2
Give the volume of the cuboid for which it is true that by adding the lengths of two sides which meet the third adjacent side we obtain \latex{9} cm, \latex{16} cm and \latex{17} cm.
\latex{ c }
\latex{ a }
\latex{ b }
Figure 31
Solution
Let us denote the length of the sides of the cuboid according to the figure by \latex{a, b, c,} measured in centimetres. Then:
\latex{\begin{rcases}\begin{align*} a+b=9 \\ a+c=16 \\ b+c=17 \\ \end{align*}\end{rcases}}.
Let us use the opportunity that by adding the three equations we get an equation in which the variables have equal coefficients.
\latex{2a+2b+2c=42},
\latex{a+b+c=21}.
\latex{a+b+c=21}.
Since \latex{b + c = 17}, therefore \latex{a = 4}. Similarly \latex{a + c = 16}, therefore \latex{b = 5} and \latex{a + b = 9}, therefore \latex{c = 12}.
By giving the edges of the cuboid as ordered number triplet:
\latex{\left(a;b;c\right)=\left(4;5;12\right)}.
Thus the volume of the cuboid:
\latex{V=a\times b\times c=240\,cm^{3}}.
Example 3
Let us solve the following equation on the set of rational numbers:
\latex{\left(x+2y-2 \right)^{2}+\left(2x+2z+2\right)^{2}+\left(2y+3z+3\right)^{2}=0}.
Solution
Since
\latex{\left(x+2y-2\right)^2\geq0 };
\latex{\left(2x+2z-2\right)^2\geq0 },
\latex{\left(2y+3z-3\right)^2\geq0 },
\latex{\left(2x+2z-2\right)^2\geq0 },
\latex{\left(2y+3z-3\right)^2\geq0 },
therefore the sum of these expressions can be equal to \latex{0} if each of them separately are equal to \latex{0}. Based on this we get the following simultaneous equations:
\latex{\begin{rcases}\begin{align*} x+2y-2=0 \\ 2x+2z+2=0 \\ 2y+3z+3=0 \\ \end{align*}\end{rcases}}.
The difference of the first and the third equation lead to an equation in which variable \latex{y} does not appear.
\latex{x+2y-2-\left(2y+3z+3\right)=0},
\latex{x-3z-5=0}.
By subtracting the double of the resulting equation from the second equation of the original simultaneous equations:
\latex{2x+2z+2-2\times\left(x-3z-5\right)=0},
\latex{8z=-12},
\latex{z=-\frac{3}{2}}.
\latex{z=-\frac{3}{2}}.
By substituting the result in the previous equations respectively, the following values are resulting: \latex{x=\frac{1}{2}} and \latex{y=\frac{3}{4}}.
Thus the solution of the simultaneous equations:
\latex{\left(x;y;z\right)=\left\lgroup\frac{1}{2};\frac{3}{4};-\frac{3}{2}\right\rgroup}.
Exercises
{{exercise_number}}. Solve the following systems of equations.
- \latex{\begin{rcases}\begin{align*} x-2y=1 \\ y-z=2 \\ z-x=3 \end{align*}\end{rcases}}
- \latex{\begin{rcases}\begin{align*} x+2y+3z=1 \\ 2x+3y+z=2 \\ 3x+y+2z=3 \end{align*}\end{rcases}}
- \latex{\begin{rcases} \begin{align*} 2\left(x+y\right)=7-\left(x+2\right) \\ 5\left(x+1\right)=3\left(z+1\right) \\ \frac{y-1}{z-1}=\frac{1}{3} \end{align*} \end{rcases}}
{{exercise_number}}. Solve the following equations.
- \latex{\left|x+y-13\right|+\left|x-z-5\right|+\left|y-z-2\right|=0 }
- \latex{\left(2x+3y-11\right)^2+\left(3x+2z-13\right)^2+\left(4y+4z-29\right)^2=0}
- \latex{\left\lgroup\frac{4}{x}-\frac{3}{y}-1\right\rgroup^2+\left\lgroup\frac{2}{x}+\frac{3}{z}-4\right\rgroup^4+\left\lgroup\frac{3}{y}-\frac{1}{z}\right\rgroup^6=0}
{{exercise_number}}. A car is travelling on a level road at a speed of \latex{80} \latex{ km/h }, uphill at \latex{60} \latex{ km/h } and downhill at \latex{100} \latex{ km/h }. It covers the \latex{400} \latex{ km } long ride there in \latex{5} hours, while it covers it back in \latex{5} hours and \latex{16} minutes. Give the distance of each stage of the ride.
{{exercise_number}}. Three players, Andrew, Brian and Cissy are playing with the following condition: the loser has to double the money of the other two players. Andrew lost the first game, Brian the second and Cissy the third one; this way each of them have \latex{1} EUR. How much money did each of the players have before the game?
{{exercise_number}}. Can three line segments make up a triangle if the mutual sums of their length are \latex{42} cm, \latex{28} cm and \latex{20} cm?
