A kosarad üres
First-order simultaneous equations (systems of equations) in two variables
Example 1
\latex{ 3 } years ago Brian was three times as old as Andrew; in \latex{ 3 } years he will only be twice as old. How old are they now?
Solution
In this exercise it is practical to choose the ages of Andrew and Brian as unknowns and denote these by \latex{ x } and \latex{ y } respectively.
Andrew
Brian
3 years ago
now
in 3 years
\latex{x-3}
\latex{y-3}
\latex{x}
\latex{y}
\latex{x+3}
\latex{y+3}
We can set up even two equations based on the exercise:
\latex{\begin{rcases}y-3=3(x-3)\\ y+3=2(x+3)\end{rcases}}.
These can be rearranged to the following form:
\latex{\begin{rcases}\begin{align*} y-3x=-6\\ y-2x=3\end{align*}\end{rcases}}.
We are looking for such number pair x, y which satisfies both equations.
The two equations together make up a linear system of equations (or simultaneous equations) the general form of which is:
\latex{\begin{rcases}ax+by=c\\ dx+ey=f\end{rcases}},
where \latex{ a,\, b,\, c,\, d,\, e,\, f } denote real numbers.Any of the equations of the simultaneous equations are satisfied by infinitely many number pairs separately. These number pairs can be represented by one straight line each in the coordinate system. (Figure 27)
We are looking for the number pair \latex{(x; y)} which makes both equations true. The solution is given by the coordinates of the point at the intersection point of the two straight lines. By reading these values we get the graphic solution of the simultaneous equations.
The number pair in question will be \latex{x = 9} and \latex{y = 21}. Therefore now Andrew is 9 years old, and Brian is 21 years old. These satisfy the conditions of the exercise.
Figure 27
\latex{y=2x+3}
\latex{y=3x-6}
\latex{9 }
-
\latex{ 1 }
\latex{ 1 }
\latex{ -1 }
\latex{ 21 }
-
\latex{ 3 }
-
\latex{ -6 }
We can give the solution the algebraic way too. From the first equation let us express y and let us substitute it into the second equation.
\latex{\begin{alignat*}{4}\underline{\begin{rcases}y=3x-6\\ y-2x=3\end{rcases}, }\\(3x-6)-2x=3,\\ x=9. \end{alignat*}}
By substituting this value back into the first equation:
\latex{y=3\times 9-6=21}.
The method being followed is called the substitution method.
During the substitution method we proceed as below:
- From one of the equations we express one of the unknowns.
- We substitute the expression resulting for the unknown into the other equation. Thus we get an equation in one variable which we solve the usual way.
- By substituting the resulting value back into the other equation we can determine the other unknown too.
Example 2
Let us solve the simultaneous equations \latex{\begin{rcases}\begin{align*}2x+y=3\\ 4x+2y=6\end{align*}\end{rcases}}.
Solution
The second of the simultaneous equations in the exercise is exactly double the first one. Therefore one of the equations is a consequence of the other one, i.e. it does not convey any new information. Therefore these simultaneous equations have infinitely many solutions.
If we think of the graphic solution, then the straight line defined by one of the equations coincides with the straight line defined by the other equation. (Figure 28)
\latex{y=-\frac{1}{2}x+\frac{3}{2}}
Figure 28
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ 5 }
\latex{ 6 }
\latex{ x }
\latex{ y }
\latex{ 1 }
\latex{ 2 }
\latex{ -1 }
\latex{ -2 }
\latex{ -3 }
Example 3
Let us solve the simultaneous equations \latex{\begin{rcases}\begin{align*}x-3y=3\\ 3x-9y=25\end{align*}\end{rcases}}.
Solution
By dividing the second equation by \latex{ 3 }: \latex{x-3y=\frac{25}{3}} is resulting, which contradicts the first equation. Therefore these simultaneous equations have no solution.
From the first equation
From the first equation
\latex{y=\frac{x}{3}-1},
which describes the relation between the coordinates of the points on the graph of function
\latex{x\mapsto\frac{x}{3}-1},
defined on the set of real numbers.From the second equation
The graphs of the functions defined by the two equations are parallel straight lines, therefore there is no solution. (Figure 29)
\latex{y=\frac{x}{3}-\frac{25}{9}},
i.e. the graph of the function
\latex{x\mapsto\frac{x}{3}-\frac{25}{9}}
results.The graphs of the functions defined by the two equations are parallel straight lines, therefore there is no solution. (Figure 29)
Figure 29
\latex{ x-3y=3 }
\latex{ 3x-9y=25 }
\latex{ 3 }
\latex{ 4 }
\latex{ x }
\latex{ y }
\latex{ 1 }
\latex{ -1 }
\latex{ -2 }
\latex{ -1 }
Example 4
Let us solve the simultaneous equations \latex{\;\begin{rcases}2x-4y=2\\ 2x+2y=4\end{rcases}}.
Solution
Let us realise that the coefficient of \latex{ x } is the same in the two equations. In this case we can faster get to the equation containing only one unknown.
Let us subtract the two equations from each other. By composing the differences of the left and the right sides:
\latex{\begin{alignat*}{4}(2x-4y)-(2x+2y)=2-4, \\ -6y=-2,\\ y=\frac{1}{3}.\end{alignat*}}
We substitute this into one of the original simultaneous equations, in this case into the first equation:
\latex{\begin{alignat*}{4}2x-4\times\frac{1}{3}=2,\\ 2x=\frac{10}{3},\\ x=\frac{5}{3}.\end{alignat*}}
Therefore the solution of the simultaneous equations: \latex{(x;y)=\left(\frac{5}{3}; \frac{1}{3}\right)}, which is also made sure of by checking:\latex{\begin{alignat*}{4}2\times \frac{5}{3}-4\times \frac{1}{3}=\frac{10}{3}-\frac{4}{3}=2,\\ 2\times \frac{5}{3}+2\times \frac{1}{3}=\frac{10}{3}+\frac{2}{3}=4.\end{alignat*}}
The method used here is called the method of equal coefficients. This method, which means eliminating the unknowns gradually, is often called the Gaussian method or Gaussian elimination.
The method of equal coefficients can be applied too if there are no variables with equal coefficients in the equations.
- We multiply both sides of the equations by a non-zero number each so that the absolute values of the coefficients of one of the unknowns are equal.
- We eliminate this unknown by adding or subtracting the two equations.
- We solve the equation in one variable, then by substituting it into one of the equations we calculate the other unknown.
Example 5
Let us solve the following simultaneous equations with the help of the method of equal coefficients.
\latex{\begin{rcases}\begin{align*}-3x+4y=-14\\ 2x+6y=18\end{align*}\end{rcases}}.
Solution
Let us multiply the first equation by 2, and the second equation by 3, thus we get the following simultaneous equations:
\latex{\begin{rcases}\begin{align*}-6x+8y=-28 \\ 6x+18y=54\end{align*}\end{rcases}}.
By adding the two equations:\latex{\begin{align*}(-6x+8y)+(6x+18y)=-28+54, \\ 26y=26,\\ y=1.\end{align*}}
By substituting the result in the first equation:
\latex{\begin{align*}-6x+8=-28, \\ -6x=-36,\\ x=6.\end{align*}}
So the solution of the simultaneous equations is the number pair \latex{(x; y) = (6; 1)}, which is also made sure of by checking.
Exercises
{{exercise_number}}. Solve the following systems of equations graphically.
- \latex{\begin{rcases}y-2x=1\\x+2y=7\end{rcases}}
- \latex{\begin{rcases}\begin{align*}\frac{x-y}{2}=1\\ -x+2y=0\end{align*}\end{rcases}}
- \latex{\begin{rcases}\begin{align*}\frac{x}{y}+2=\frac{3}{y}\\ 3x-y=2\end{align*}\end{rcases}}
{{exercise_number}}. Solve the following systems of equations.
- \latex{\begin{rcases}\begin{align*}x+3y=-2\\3x-y=4\end{align*}\end{rcases}}
- \latex{\begin{rcases}\begin{align*}\frac{2x}{3}-\frac{y}{5}=0\\ -\frac{3x}{4}+\frac{2y}{5}=2\end{align*}\end{rcases}}
- \latex{\begin{rcases}\begin{align*}\frac{x+2}{3}\frac{y-5}{4}=3\\ \frac{x-1}{3}+\frac{y+2}{2}=1\end{align*}\end{rcases}}
{{exercise_number}}. Solve the following systems of equations with the help of the method of equal coefficients.
- \latex{\begin{rcases}\begin{align*}3x+3y=-2\\ 3x-y=4\end{align*}\end{rcases}}
- \latex{\begin{rcases}\begin{align*}3x-2y=1\\ 2x+3y=2\end{align*}\end{rcases}}
- \latex{\begin{rcases}\begin{align*}\frac{x+2}{3}-3y=3\\ \frac{x-1}{3}+2y=1\end{align*}\end{rcases}}
{{exercise_number}}. What shall we write for the parameter a if we wanted the simultaneous equations to have
- one solution;
- no solution;
- infinitely many solutions?
\latex{\begin{rcases}\begin{align*}x-2y=1\\ 2x+ay=2\end{align*}\end{rcases}}
{{exercise_number}}. Determine parameters a and b so that the simultaneous equations have
- one solution;
- infinitely many solutions.
\latex{\begin{rcases}\begin{align*}ax+y=1-b\\ bx-y=1+2a\end{align*}\end{rcases}}
Puzzle
In the below simultaneous equations we “slightly” change coefficient a. Is it true that due to this the solution of the simultaneous equations changes only “slightly” too?
\latex{\begin{rcases}\begin{align*}ax+by=c\\dx+ey=f\end{align*}\end{rcases}}
