A kosarad üres
Solving problems with simultaneous
equations (systems of equations)
equations (systems of equations)
If we cannot find any direct relation between the missing amounts in the exercises, then it is practical to consider them as distinct unknowns. Then we set up equations with the help of these.
Example 1
In a biological experiment the data were processed with the help of two computers. One of the machines could process \latex{ 500 } samples a day, while the other one could process \latex{ 1,000 } samples. The two machines working sequentially and continually managed to analyse \latex{ 8,000 } samples in \latex{ 10 } days.
How many samples did they analyse separately?
Solution
Let \latex{ x } denote the number of samples analysed by the first, and let \latex{ y } denote the number of samples analysed by the second machine. The total number of samples: x + y = \latex{ 8,000 }.
The first machine analyses one sample in \latex{\frac{1}{500}} days, while the second machine analyses it in \latex{\frac{1}{1000}} days. Since together they have been working for 10 days, thus analysing the x and y pieces of samples in total took this amount of time:
\latex{\frac{x}{500} +\frac{y}{1,000} =10}
The simultaneous equations to solve:
\latex{\begin{rcases}x+y=8,000\\ \frac{x}{500}+\frac{y}{1000}=10\end{rcases}}.
We solve them with the help of the method of equal coefficients. By multiplying the second equation by 1000:
\latex{\begin{rcases}x+y=8,000\\ 2x+y=10,000\end{rcases}}.
By subtracting the first equation from the second one \latex{ x = 2,000 } results.
By substituting it into the first equation we get \latex{ y = 6,000 }.
Therefore the first machine did the analysis of \latex{ 2,000 }, and the second one of \latex{ 6,000 } samples.
The resulting values meet the conditions formulated in the exercise.
Example 2
If we lengthen the two pairs of parallel sides of a rectangle by \latex{ 10 } \latex{ cm } each, then its area will be \latex{ 1,000 } \latex{ cm^2 } greater, but if we shorten one pair of parallel sides by \latex{ 10 } \latex{ cm }, and we lengthen the other pair of parallel sides by \latex{ 10 } \latex{ cm }, then its area will be \latex{ 400 } \latex{ cm^2 } less. Give the length of the sides of the rectangle.
Solution
Let \latex{ x } and \latex{ y } denote the length of the sides of the original rectangle measured in cm. According to the text of the exercise the following relations exist:
Area \latex{ (cm^2) }
Change of area \latex{ (cm^2) }
Original rectangle
Rectangle with increased area
Rectangle with decreased area
\latex{ xy }
\latex{ (x + 10)(y + 10) }
\latex{ (x + 10)(y - 10) }
\latex{ 1,000 }
\latex{ -400 }
\latex{\begin{rcases}(x+10)(y+10)=xy+1000\\ (x+10)(y-10)=xy-400\end{rcases}}.
By expanding the brackets and rearranging the equations:
\latex{\begin{rcases}\underline{\begin{alignat*}{4}xy+10x+10y+100=xy+1000\\xy-10x+10y-100=xy-400\end{alignat*}}\end{rcases},\\ \begin{rcases}\underline{\begin{alignat*}{4}10x+10y=900\\ -10x+10y=-300\end{alignat*}}\\ \end{rcases},\\ \begin{rcases}\begin{alignat*}{4}x+y=90\\ -x+y=-30\end{alignat*}\end{rcases}.}
By adding the two equations \latex{ y = 30 } results, and from this we get \latex{ x = 60 }. Therefore the sides of the rectangle in question are \latex{ 60 } \latex{ cm } and \latex{ 30 } \latex{ cm } long.
Checking makes sure of the correctness of the solution: The area of the original rectangle: \latex{ 60 × 30 = 1,800 \,cm^2 }. The area of the rectangle with increased area: \latex{ 70 × 40 = 2,800 \, cm^2 }. The area of the rectangle with decreased area: \latex{ 70 × 20 = 1,400 \,cm^2 }. The changes of areas correspond to the exercise.
Example 3
One exterior angle of a triangle is \latex{ 130º }, the difference of the two interior angles not adjacent to it is \latex{ 10º }. Give the angles of the triangle.
Solution
Let the two interior angles in question be denoted by \latex{\alpha} and \latex{\beta}. We know that their sum gives the measure of the exterior angle not adjacent to them.
Based on this and on the conditions formulated in the exercise we get the following simultaneous equations:
\latex{\begin{alignat*}{4}\begin{rcases}\alpha+\beta=130^{\circ}\\ \alpha-\beta=10^{\circ}\end{rcases}\end{alignat*}.}
Figure 30
\latex{\alpha }
\latex{\beta }
\latex{\alpha+\beta }
By adding the two equations \latex{2\alpha =140°}, i.e. \latex{\alpha =70°} results. By substituting it back into one of the equations:
\latex{\beta =60°}
Thus the angles of the triangle are: \latex{ 70º }, \latex{ 60º } and \latex{ 50º }.
Example 4
A speedboat travels \latex{ 40 } \latex{ km } upstream, and then turns around and returns to its starting point. It covered this distance in \latex{ 8 } hours counted from the start. Another time it travelled \latex{ 10 } \latex{ km } upstream and \latex{ 4 } \latex{ km } downstream at the same speed, in a total of \latex{ 1.5 } hours. Give the speed of the speedboat in still water and the speed of the river.
Solution
Let the speed of the river be denoted by \latex{ x \, km/h }, and the speed of the speedboat by \latex{ y \, km/h }. The speedboat is cruising at a speed of \latex{ y - x } \latex{ km/h } upstream, while at \latex{ y } + \latex{ x } \latex{ km/h } downstream. The below table helps in setting up the simultaneous equations.
Distance \latex{ (km) }
Speed \latex{ (km/h) }
Time
\latex{ (h) }
\latex{ (h) }
Trip \latex{ 1 } upstream
\latex{ 40 }
\latex{y-x}
Total time
Trip \latex{ 1 } downstream
Trip \latex{ 2 } upstream
Trip \latex{ 2 } downstream
\latex{ 40 }
\latex{ 10 }
\latex{ 4 }
\latex{y+x}
\latex{y-x}
\latex{y+x}
\latex{\frac{40}{y-x} }
\latex{\frac{40}{y+x} }
\latex{\frac{40}{y-x} }
\latex{\frac{40}{y+x} }
\latex{ 8 } hours
\latex{ 1.5 } hours
\latex{\begin{alignat*}{4}\begin{rcases}\frac{40}{y-x}+\frac{40}{y+x}=8\\ \frac{10}{y-x}+\frac{4}{y+x}=\frac{3}{2}\end{rcases}\end{alignat*}.}
It would be quite roundabout to eliminate any of the unknowns from these simultaneous equations. Instead of this we can reach our goal by introducing new variables. It is practical to apply this method when we discover similar expressions in the equations, because with the help of this we can often simplify the solution.
Let \latex{ a } denote \latex{\frac{1}{y-x}}, and let \latex{ b } denote \latex{\frac{1}{y+x}}. So the simultaneous equations can be written in the following form:
\latex{\begin{rcases}\begin{alignat*}{4}40a+40b=8 \\ 10a+4b=\frac{3}{2}\end{alignat*}\end{rcases}.}
The steps of solving the simultaneous equations:
By applying the method of equal coefficients and multiplying the second
equation by \latex{ 4 }:
equation by \latex{ 4 }:
\latex{\begin{rcases}\begin{alignat*}{4}40a+40b=8 \\ 40a+16b=6\end{alignat*}\end{rcases}.}
By subtracting the two equations:
\latex{24b=2},
\latex{b=\frac{1}{12}}.
From this \latex{ a } can be calculated:
\latex{10a+4\times \frac{1}{12}=\frac{3}{2}},
\latex{10a=\frac{7}{6}},
\latex{a=\frac{7}{60}}.
\latex{10a+4\times \frac{1}{12}=\frac{3}{2}},
\latex{10a=\frac{7}{6}},
\latex{a=\frac{7}{60}}.
Knowing \latex{ a } and \latex{ b } we get the following simultaneous equations for x and y:
\latex{\begin{rcases}\begin{alignat*}{4}\frac{1}{y-x}=\frac{7}{60} \\ \frac{1}{y+x}=\frac{1}{12}\end{alignat*}\end{rcases}.}
Since the reciprocals of equal numbers are also equal:
\latex{\begin{rcases}\begin{alignat*}{4}y-x=\frac{60}{7} \\ y+x=12\end{alignat*}\end{rcases}.}
By adding the two equations:
\latex{2y=\frac{144}{7}},
\latex{y=\frac{72}{7}}.
\latex{y=\frac{72}{7}}.
By subtracting the first equation from the second one:
\latex{2x=\frac{24}{7}},
\latex{x=\frac{12}{7}}.
The speed of the river is \latex{\frac{12}{7}} \latex{ km/h }, and the speed of the speedboat is \latex{\frac{72}{7}} \latex{ km/h }.
By checking we can make sure that our calculation was correct.
Exercises
{{exercise_number}}. There are \latex{ 18 } litres of \latex{ 46 }% concentrated and \latex{ 12 } litres of \latex{ 54 }% concentrated alcohol. How shall we mix them so that we get \latex{ 48.5 }% concentrated alcohol?
{{exercise_number}}. We are walking beside a tram track at a speed of \latex{ 4 } \latex{ km/h }. We are overtaken by a tram every \latex{ 12 } minutes, and we meet an oncoming tram every \latex{ 4 } minutes. Give the speed of the trams and the time interval between consecutive trams.
{{exercise_number}}. If we divide a two-digit number by the number resulting when swapping the digits of the number, then the quotient will be \latex{ 4 }, and the remainder will be \latex{ 3 }. If we divide the same number by the difference of its digits, then the quotient will be \latex{ 11 }, and the remainder will be \latex{ 5 }. Give this number.
{{exercise_number}}. One of the angles of a triangle is half of the sum of the other two angles. The two larger angles together measure \latex{ 3 } times the smallest angle. Give the angles of the triangle.
