Exercises: systems with two variables

1
Test your knowledge!
1. Put the steps of the equation system solving process in the correct order!
  To do this, enter the numbers of the steps in the solution field one after the other without spaces (e.g. 123).
  Plug the variable into an equation.
  Set equal the equations $\mathrm{I}$ and $\mathrm{II}$ .
  Solve the resulting equation for one variable.
2. Which of the following describes an intersection point?
  The point with equal function values
  An arbitrary point within the coordinate system
  The vertex of a parabola is also called the intersection point.
3. What do you need llnear systems of equations for?
  To determine intersection points of two lines - which cannot always be done graphically.
  To determine intersection points of two circles - which cannot always be done graphically.

Solve the following systems of equations!

$\hphantom{\mathrm{I}}\mathrm{I}) \quad 3x + 4 = 2y$

$\mathrm{II}) \quad 4y = 2x + 10$

$\hphantom{\mathrm{I}} \mathrm{I}) \quad y - 1 = 2x + 3$

$\mathrm{II}) \quad 2y - 2 = 5x - 1$

$\hphantom{\mathrm{I}} \mathrm{I}) \quad 2x + 3y = 4x - 5$
$\mathrm{II}) \quad 3x - 2y = 2y + 8$

Für diese Aufgabe benötigst Du folgendes Grundwissen: Elimination by equating coefficients
Solving linear systems by equating coefficients
$\hphantom{\mathrm{I}}\mathrm{I}) \quad 2x+3y = 4x -5$
$\mathrm{II}) \quad 3x-2y = 2y+8$
1. Solve both equations for one variable (for instance y)
$\def\arraystretch{1.25} %%\begin{array}{lrl} \hphantom{\mathrm{I}}\mathrm{I}) & 2x+3y &= &4x -5 \\ \hphantom{\mathrm{I}}\mathrm{I})' & 3y &= &2x -5 \\ \hphantom{\mathrm{I}}\mathrm{I})'' & y &= &\frac{2}{3}x - \frac{5}{3} \end{array}%%$ $\def\arraystretch{1.25} \begin{array}{lrll} \hphantom{\mathrm{I}}\mathrm{I}) & 2x+3y &= &4x -5 \quad \mid -2x\\ \hphantom{\mathrm{I}}\mathrm{I}') & 3y &= &2x -5 \quad \mid :3\\ \hphantom{\mathrm{I}}\mathrm{I}'') & y &= &\frac{2}{3}x - \frac{5}{3} \end{array}$
$\def\arraystretch{1.25} \begin{array}{lrcll} \mathrm{II}) & 3x-2y &=&2y+8 &| +2y -8 \\ \mathrm{II}') & 3x-8 &=&4y &| :4\\ \mathrm{II}'') & \frac{3}{4}x-2 &=&y \end{array}$
2. Set equal
Set equal $\mathrm{I''}$ and $\mathrm{II''}$ .
$\def\arraystretch{1.25} \begin{array}{lrl}\Rightarrow &\frac{3}{4}x-2 &= &\frac{2}{3}x-\frac{5}{3} \end{array}$
3. Solve for that one variable
Solve for $x$ .
$\def\arraystretch{1.25} \begin{array}{rcll} \frac{3}{4}x-2 &= &\frac{2}{3}x-\frac{5}{3} &| -\frac{2}{3}x \\\frac{3}{4}x-\frac{2}{3}x-2 &= &-\frac{5}{3} \quad\quad &|+2\\\frac{9}{12}x-\frac{8}{12}x &= &-\frac{5}{3} +2 \\\frac{1}{12}x &= &\frac{1}{3} \quad \quad &|\cdot12\\x &= &4 \end{array}$
4. Plug into one equation to find the other variable
Plug $x$ into $\mathrm{II'}$ .
$\def\arraystretch{1.25} \begin{array}{rl} \frac{3}{4}\cdot (4)-2 &= &y \\3-2 &= &y \\y &= &1\end{array}$
Now, you can write down the solution set again
$L = \{(4{,}1)\}$
Do you have a question?
Write it here 👉

3
Solve the linear systems of equations using the substitution method.
1. $\hphantom{\mathrm{I}}\mathrm{I} \quad 3x + 4 = y$
  $\mathrm{II} \quad 4y -3x = 9$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Substitution method
  First you have to solve one of the two equations for one variable. In the present case you can see that equation $\mathrm{I}$ is already solved for $y$ .
  $\;$ Plug $y = 3 x + 4$ from equation $\mathrm{I}$ into $\mathrm{II}$
  $\mathrm{II} \quad 4\cdot(3x + 4) -3x = 9$
  Multiply out the parenthesis and solve for $x$ .
  $\def\arraystretch{1.25} \begin{array}{crcll}\mathrm{II}&\quad12x+16-3x&=&9&|-16\\&9x&=&-7&|:9\\&x&=&-\frac79\end{array}$
  $\;$ Plug $x=-\frac 79$ from $\mathrm{II}$ into $\mathrm{I}$
  $\def\arraystretch{2} \begin{array}{crcll}\mathrm{I}&3\cdot\left(-\dfrac{7}{9}\right) + 4 &=& y\\&y&=&\frac53\end{array}$
  And finally write down the solution set.
  $\mathbb{L}=\left\{\left(-\dfrac{7}{9}\left|\dfrac{5}{3}\right.\right)\right\}$ .
  Do you have a question?
  Write it here 👉
2. $\hphantom{\mathrm{I}}\mathrm{I} \quad 3s - 4t = 4$
  $\mathrm{II} \quad 4s + t = -2$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Substitution method
  First you have to solve one of the two equations for $s$ or $t$ . It is, for instance, a good idea to solve equation $\mathrm{II}$ for $t$ .
  Solve equation for $t$ .
  $\def\arraystretch{1.25} \begin{array}{crcll}\mathrm{II} & 4s + t &=& -2&|-4s\\&t&=&-2-4s\end{array}$
  $\;$ Plug $t = - 2 - 4 s$ from $\mathrm{II}$ into $\mathrm{I}$
  $\hphantom{\mathrm{I}}\mathrm{I} \quad 3s - 4\cdot(-2 - 4s) = 4$
  Multiply out the parenthesis and solve for $s$ .
  $\def\arraystretch{1.25} \begin{array}{crcll}\mathrm{I}&3s+8+16s&=&4&|-8\\&19s&=& -4&|:19\\&s&=&-\frac{4}{19}\end{array}$
  Substitute $s = -\frac{4}{19}$ into the transformed second equation to determine $t$ .
  $\def\arraystretch{1.25} \begin{array}{crcl}\mathrm{II} &t&=&-2-4\cdot(-\frac{4}{19})\\&t&=&-\frac{22}{19}\end{array}$
  And finally, write down the solution set.
  $\mathbb{L}=\left\{(s|t)=\left(-\frac{4}{19}|-\frac{22}{19}\right)\right\}$ .
  Do you have a question?
  Write it here 👉
4
Solve the following system of equations using the addition method!
$\def\arraystretch{1.25} \begin{array}{lrrll}\mathrm{I} &4x+2y &= &4 \\\mathrm{II} &6x-3y &= &-3\end{array}$

Für diese Aufgabe benötigst Du folgendes Grundwissen: Addition method
Step 0: Tidying up the equations and selecting a variable
Since the equations are already nicely ordered this point is not needed.
In this case it is the same effort to solve the problem, no matter if you select $x$ or $y$ .
The solution is calculated for the variable $x$ .
Step 1: Multiply the equations
In equation $I$ there are $4 x$ , in equation $I I$ there are $6 x$ . So a common multiple is $12 x$ . Multiply the first equation by $3$ and the second equation by $2$ .
Equation 1
$\def\arraystretch{1.25} \begin{array}{lcccccl}\mathrm{I}& 4x&+&2y&=&4& \quad &|\cdot 3\\\mathrm{I}' &12x& + &6y &= &12& \end{array}$
Equation 2
$\def\arraystretch{1.25} \begin{array}{lcccccl}\mathrm{II}& 6x& - &3y &= &-3& \quad &|\cdot 2\\\mathrm{II}' &12x& - &6y &= &-6& \end{array}$
Step 2: Elimination of a variable by addition/subtraction.
The signs of the selected variable are the same both times. Therefore you have to subtract.
$\def\arraystretch{1.25} \begin{array}{llcccll}&\mathrm{I}' & 12x&+& 6y&=&12\\-&\mathrm{II}' & 12x&-&6y&=&-6\\\hline& & \color{#009900}0&+&12y&=& 18\end{array}$
Step 3: Determine the values of both variables.
Determine the value of $y$
$\def\arraystretch{2} \begin{array}{rrll}12y &= &18 &\quad |: 12\\\color{#FF6600}y &= &\color{#FF6600}{\dfrac{3}{2}}\end{array}$
Determine the value of $x$ by plugging $y$ into $\mathrm{II}$
$\def\arraystretch{2} \begin{array}{rrll}6x - 3 \cdot \color{#FF6600}{\dfrac{3}{2}} &= &-3 \\6x - \dfrac{9}{2} &= &-3 \quad &|+ \dfrac{9}{2}\\6x &=& \dfrac{3}{2} \quad &|: 6\\\color{#009999}x &= &\color{#009999}{\dfrac{1}{4}}\end{array}$
So the solution is given by ${\color{#009999}x} = {\color{#009999}\dfrac{1}{4}}$ and ${\color{#FF6600}y} = \color{#FF6600}{\dfrac{3}{2}}$ : $\quad \mathbb{L}=\left\{(x|y)=\left({\color{#009999}\dfrac{1}{4}}\;\bigg\vert\;{\color{#FF6600}\dfrac{3}{2}}\right)\right\}$ .
5
Test your knowledge! Which method is useful to solve the following systems of equations?
1. $\hphantom{\mathrm{I}}\mathrm{I} \quad 3x + 6y = 2$
  $\mathrm{II} \quad 4x + 2 = y$
  substitution method
  equating coefficients
  addition method
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: System of linear equations
  The equation $\mathrm{II}$ is already solved for $y$ . Therefore, the substitution method is advisable.
  The other two methods also get you to the goal, but they require further transformation steps.
  Do you have a question?
  Write it here 👉
2. $\hphantom{\mathrm{I}}\mathrm{I} \quad s = 4t -7$
  $\mathrm{II} \quad s = -2 + 3t$
  substitution method
  equating coefficients
  addition method
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: System of linear equations
  All three methods can be applied directly and are useful for this reason.
  Do you have a question?
  Write it here 👉
3. $\hphantom{\mathrm{I}}\mathrm{I} \quad 2a - 2b = 3$
  $\mathrm{II} \quad 5a + 2b = 6$
  substitution method
  equating coefficients
  addition method
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: System of linear equations
  Equation $\mathrm{I}$ contains $- 2 b$ , equation $\mathrm{II}$ contains $2 b$ . So the $b$ will drop out when adding both lines, which makes the addition method suitable.
  With the other two procedures, you have to take further transformation steps, before one can apply them.
  Do you have a question?
  Write it here 👉
6
Solve with the most appropriate method.
1. $\def\arraystretch{1.25} \begin{array}{lrcc}\mathrm{I}& e+4f&=&20\\\mathrm{II}&-3e+4f&=&-12\end{array}$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: System of linear equations
  For this exercise, the addition method is most suitable, as $+ 4 f$ occurs in both equations.
  $\def\arraystretch{1.25} \begin{array}{rrrcrlr}&\mathrm{I} &e&+&4f&=&20\\-&\mathrm{II}&-3e&+&4f&=&-12\\\hline&&4e&+&0&=&32\end{array}$
  Solve for $e$
  $\def\arraystretch{1.25} \begin{array}{llccll}4e&+&0&=&32 \quad |:4\\&&e&=&\color{#009900}8\end{array}$
  Plug $e$ into one of both equations, for instance $\mathrm{I}$
  $\def\arraystretch{1.25} \begin{array}{llccll}\color{#009900}8&+&4f&=&20&|-8&\\&&4f&=&12&|:4&\\&&f&=&\color{#cc0000}3\end{array}$
  Write down the solution set
  $\mathbb{L}=\{(e|f)=(8|3)\}$
  Do you have a question?
  Write it here 👉
2. $\def\arraystretch{1.25} \begin{array}{lrcl}\mathrm{I}& 7y&=&5+2x\\\mathrm{II}&4x-14y&=&46\end{array}$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: System of linear equations
  You get a very nice solution with the substitution method.
  The equation $\mathrm{I}$ is solved to $7 y$ and in equation $\mathrm{II}$ there is $14 y$ . You can even use the substitution method without doing any transformation.
  Transform equation $\mathrm{II}$
  $\def\arraystretch{1.25} \begin{array}{lrcl}\mathrm{II}&4x-2\cdot7y&=&46\end{array}$
  Plug $7 y$ from $\mathrm{I}$ into $\mathrm{II}$
  $\def\arraystretch{1.25} \begin{array}{lrcl}\mathrm{II}&4x-2\cdot(5 + 2x)&=&46\end{array}$
  Conclude and solve for $x$
  $\def\arraystretch{1.25} \begin{array}{lrcl}\mathrm{II}&4x-10 - 4x&=&46& \vert +10\\ \mathrm{II}&0&=&56\end{array}$
  $\def\arraystretch{1.25} \begin{array}{lrcl}\end{array}$
  As you can see now, the entries in the last line are not the same, which results in a contradiction. Hence, the system has no solution, since no number can be assigned for $x$ , such that $\mathrm{II}$ is satisfied.
  Do you have a question?
  Write it here 👉
3. $\def\arraystretch{1.25} \begin{array}{lrcl}\mathrm{I}& 3.5&=&-0.5k+2.5m\\\mathrm{II}&10m&=&14+2k\end{array}$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: System of linear equations
  You best use the addition method, since you can see with a sharp eye (this is what math lessons shall train you for) that $2 k$ is $4$ times $0.5 k$ .
  Multiply the first equation by 4
  $\def\arraystretch{1.25} \begin{array}{rrl}\mathrm{I}&3.5&=&-0.5k+2.5m&|\cdot 4\\\mathrm{II}&10m&=&14+2k\\\mathrm{I'}&14&=&-2k+10m&|+2k\\\mathrm{II}&10m&=&14+2k\\\mathrm{I'}&14+2k&=&10m&\\\mathrm{II}&10m&=&14+2k\end{array}$
  As you can see now, both equations are identical. It follows that the system has an infinite number of solutions. Graphically, this corresponds to two straight lines lying on top of each other.
  Solve the equation for one more variable to obtain the solution set.
  $10 m = 14 + 2 k \quad | : 10 \\m = 1.4 + 0.2 k$
  $\mathbb{L} = \left\{ (m|k) \;\big |\; m = 1.4 + 0.2 k \right\}$
  Do you have a question?
  Write it here 👉
7
Determine the solution sets of the following systems of equations.
1. $\def\arraystretch{1.25} \begin{array}{ccccc}\mathrm{I}&5y& -& 3x& =& 1\\\mathrm{II}& x &=& y& +& 1\end{array}$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: System of linear equations
  In this case, the substitution method is useful, since the second equation is already solved for one variable.
  $\def\arraystretch{1.25} \begin{array}{ccccc}\mathrm{I}&5y& -& 3x& =& 1\\\mathrm{II}& x &=& y& +& 1\end{array}$
  Plug equation $\mathrm{II}$ into $\mathrm{I}$
  $\mathrm{I}'\quad5y-3\left(y+1\right)=1$
  Solve for y
  $\def\arraystretch{1.25} \begin{array}{rcccc}5y-3y-3&=&1&\\2y-3&=&1&|+3\\2y&=&4&|:2\\y&=&2\end{array}$
  Plug $y = 2$ into $\mathrm{II}$ and solve for x
  $\def\arraystretch{1.25} \begin{array}{rcccc}5\cdot2-3x&=&1&|-10\\-3x&=&-9&|:(-3)\\x&=&3\end{array}$
  $L=\left\{\left(x\;\left|\;y\right.\right)\right\}=\left\{\left(3\;\left|\;2\right.\right)\right\}$
  Do you have a question?
  Write it here 👉
2. $\def\arraystretch{1.25} \begin{array}{ccccc}\mathrm{I}&4x&+&5y&=&32\\\mathrm{II}&y&=&5x&-&11\end{array}$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: System of linear equations
  In this case, the substitution method is useful, since the second equation is already solved for one variable.
  $\def\arraystretch{1.25} \begin{array}{ccccc}\mathrm{I}&4x&+&5y&=&32\\\mathrm{II}&y&=&5x&-&11\end{array}$
  Plug equation $\mathrm{II}$ into $\mathrm{I}$
  $\mathrm{I}'\quad 4x+5\cdot \left(5x-11\right)=32$
  Solve for $x$
  $\def\arraystretch{1.25} \begin{array}{rccc}4x+25x-55&=&32&\\29x-55&=&32&|+55\\29x&=&87&&|:29\\x&=&3\end{array}$
  Plug $x = 3$ into $\mathrm{II}$ and solve for $y$
  $\def\arraystretch{1.25} \begin{array}{rcl}y&=&5\cdot3-11\\y&=&4\end{array}$
  $L=\left\{\left(x\;\left|\;y\right.\right)\right\}=\left\{\left(3\;\left|\;4\right.\right)\right\}$
  Do you have a question?
  Write it here 👉
3. $\def\arraystretch{1.25} \begin{array}{cccc}\mathrm{I}&15y&-&4x&=&-50\\\mathrm{II}&x&=&y&+&7\end{array}$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: System of linear equations
  In this case, the substitution method is useful, since the second equation is already solved for one variable.
  $\def\arraystretch{1.25} \begin{array}{cccc}\mathrm{I}&15y&-&4x&=&-50\\\mathrm{II}&x&=&y&+&7\end{array}$
  Plug equation $\mathrm{II}$ into $\mathrm{I}$
  $\mathrm{I}'\quad15y-4\left(y+7\right)=-50$
  Solve for $x$
  $\def\arraystretch{1.25} \begin{array}{rcll}15y-4y-28&=&-50&\\11y-28&=&-50&|+28\\11y&=&-22&|:11\\y&=&-2\end{array}$
  Plug $y = - 2$ into $\mathrm{II}$ and solve for x
  $x = - 2 + 7$
  $x = 5$
  $L=\left\{\left(x\;\left|\;y\right.\right)\right\}=\left\{\left(5\;\left|\;-2\right.\right)\right\}$
  Do you have a question?
  Write it here 👉
4. $\def\arraystretch{1.25} \begin{array}{cccc}\mathrm{I}&3x&=&y&+&15\\\mathrm{II}&2y&-&10&=&2x\end{array}$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: System of linear equations
  In this case, the substitution method is useful, since the second equation is already solved for one variable.
  $\def\arraystretch{1.25} \begin{array}{cccc}\mathrm{I}&3x&=&y&+&15\\\mathrm{II}&2y&-&10&=&2x\end{array}$
  Divide $\mathrm{II}$ by 2 and solve for $x$
  $\mathrm{II}:2\to\mathrm{II}'\quad y-5= x$
  Plug $\mathrm{II}'$ into $\mathrm{I}$
  $\mathrm{II}'$ plugged into $\mathrm{I}$ :
  $\mathrm{I}'\quad3\left(y-5\right)=y+15$
  Then, solve $\mathrm{I}'$ for $y$
  $\def\arraystretch{1.25} \begin{array}{rcll}3y-15&=&y+15&|-y; +15\\2y&=&30&|:2\\y&=&15\end{array}$
  Finally, plug $y = 15$ into $\mathrm{II}'$ and solve for $x$
  $y = 15$ plugged into $\mathrm{II}'$ :
  $\def\arraystretch{1.25} \begin{array}{rcll}15-5&=&x\\10&=&x\end{array}$
  $L=\left\{\left(x\;\left|\;y\right.\right)\right\}=\left\{\left(10\;\left|\;15\right.\right)\right\}$
  Alternative solution: equating coefficients
  Another option is to use the method of equating coefficients, because on the left side of $\mathrm{I}$ and on the right side of $\mathrm{II}$ there is almost the same term.
  $\def\arraystretch{1.25} \begin{array}{cccc}\mathrm{I}&3x&=&y&+&15\\\mathrm{II}&2y&-&10&=&2x\end{array}$
  Multiply $\mathrm{II}$ by $\frac32$ to get a $3 x$ on the right hand side.
  $\def\arraystretch{1.25} \begin{array}{cccc}\mathrm{I}&3x&=&y&+&15\\\mathrm{II}'&3y&-&15&=&3x\end{array}$
  Set the right side of $\mathrm{I}$ equal to the left side of $\mathrm{II}'$ and solve for $x$ .
  $\def\arraystretch{1.25} \begin{array}{rcll}3x-15&=&x+5&|-x;\;+15\\2x&=&20&|:2\\x&=&10\end{array}$
  Plug $x = 10$ into $\mathrm{I}$ (or $\mathrm{II}$ ) and solve for $y$ .
  $\def\arraystretch{1.25} \begin{array}{rcll}3\cdot 10&=&y+15&\\30&=&y+15&|-15\\15&=&y\end{array}$
  $L=\left\{\left(x\;\left|\;y\right.\right)\right\}=\left\{\left(10\;\left|\;15\right.\right)\right\}$
  Alternative solution: combining addition and substitution method
  The addition method can also be used here. To apply it, the equations are first transformed, so that the appropriate terms are placed one below the other:
  $\def\arraystretch{1.25} \begin{array}{cccc}\mathrm{I}&3x&=&y&+&15\\\mathrm{II}&2x&=&2y&-&10\end{array}$
  Subtract the second equation from the first equation.
  $\def\arraystretch{1.25} \begin{array}{cccc}\mathrm{I}'&x&=&-y&+&25\\\mathrm{II}&2x&=&2y&-&10\end{array}$
  Since the first equation is now solved for $x$ , we can use the substitution method again.
  Plug $\mathrm{I}'$ into $\mathrm{II}$ and solve for $y$ .
  $\def\arraystretch{1.25} \begin{array}{crcll}\mathrm{II}'&2\cdot(-y+25)&=&2y-10&\\&-2y+50&=&2y-10&|-2y \;\;|-50\\&-4y&=&-60&|:(-4)\\&y&=&15\end{array}$
  Plug $y = 15$ into $\mathrm{I}'$ and solve for $x$ .
  $x = - 15 + 25$
  $x = 10$
  $L=\{(x|y)\}=\{(10|15)\}$
  Do you have a question?
  Write it here 👉
8
Solve the following systems of equations with 2 equations and 2 variables first graphically and then by computation.
1. $\def\arraystretch{1.25} \begin{array}{ccccc}\mathrm{I}&y& -& 3x& =& 1\\\mathrm{II}& x &+& y &=& 1\end{array}$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: System of linear equations
  Linear system with two variables
  A linear system of equations is composed of several linear equations with common unknowns (variables), all of which are to be satisfied.
  Graphical solution
  To solve the system of equations graphically, you can resolve the individual lines for $y$ and draw the corresponding lines into a coordinate system. Then you only need to read off the coordinates of the intersection of the two lines.
  $\def\arraystretch{1.25} \begin{array}{ccccc}\mathrm{I}&y -3x&=&1&|+3x\\\mathrm{I}&y& =& 3x+1\\\mathrm{II}& x + y &=& 1&|-x\\\mathrm{II}& y &=&-x+ 1\end{array}$
  The intersection point is at $x = 0$ and $y = 1$ . Thus the solution set is $L=\left\{\left(0\left|\;1\right.\right)\right\}$ .
  Solution by computation
  In this case, the elimination by equating coefficients is a good choice, since you have already solved both equations for $y$ in order to solve them graphically.
  $\def\arraystretch{1.25} \begin{array}{ccccc}\mathrm{I}&y& =& 3x+1\\\mathrm{II}& y &=&-x+ 1\end{array}$
  Set $\mathrm{I}$ and $\mathrm{II}$ equal and solve for $x$ .
  $\def\arraystretch{1.25} \begin{array}{rcccc}3x+1&=&-x+1&|-1\\3x&=&-x&|+x\\4x&=&0&|:4\\x&=&0\end{array}$
  Plug the obtained value for $x$ into one of the equations, e.g. $\mathrm{II}$ .
  $y = 0 + 1 = 1$
  Now, you can write down the solution set.
  $L=\left\{\left(0\left|1\right.\right)\right\}$
  Do you have a question?
  Write it here 👉
2. $\def\arraystretch{1.25} \begin{array}{ccccc}\mathrm{I}&2y& +& 5x& =& 3\\\mathrm{II}& x &-& y &=& 1\end{array}$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: System of linear equations
  Linear system with two variables
  A linear system of equations is composed of several linear equations with common unknowns (variables), all of which are to be satisfied.
  Graphical solution
  To solve the system of equations graphically, you can resolve the individual lines for $y$ and draw the corresponding lines into a coordinate system. Then you only need to read off the coordinates of the intersection of the two lines.
  $\def\arraystretch{1.25} \begin{array}{ccccc}\mathrm{I}&2y+5x&=&3&|-5x\\\mathrm{I}&2y& =&-5x+3&|:2\\\mathrm{I}&y& =& -2{.}5x+1{.5}\\\mathrm{II}& x - y &=& 1&|-x\\\mathrm{II}& -y &=&-x+ 1&|\cdot(-1)\\\mathrm{II}& y &=&x- 1\end{array}$
  The intersection is at $x\approx0.71$ and $y\approx-0.29$ . Thus the solution set is $L=\left\{\left(0{.}71\left|\;-0{.}29\right.\right)\right\}$ .
  Solution by computation
  In this case, the elimination by equating coefficients is a good choice, since you have already solved both equations for $y$ in order to solve them graphically.
  $\def\arraystretch{1.25} \begin{array}{ccccc}\mathrm{I}&y& =& -2{.}5x+1{.5}\\\mathrm{II}& y &=&x- 1\end{array}$
  Set $\mathrm{I}$ and $\mathrm{II}$ equal and solve for $x$ .
  $\def\arraystretch{2} \begin{array}{rcccc}-2{.}5x+1{.}5&=&x-1&|-x\\-3{.}5x+1{.5}&=&-1&|-1{.}5\\-3{.}5x&=&-2{.}5&|:(-3{.}5)\\x&=&\dfrac57\approx 0{.}71\end{array}$
  Plug the obtained value for $x$ into one of the equations, e.g. $\mathrm{II}$ .
  $y=\frac{5}{7}-1=-\frac{2}{7}\approx-0.29$
  Now, you can write down the solution set.
  $L=\left\{\left(\dfrac57\left|-\dfrac27\right.\right)\right\}$
  Do you have a question?
  Write it here 👉
3. $\def\arraystretch{1.25} \begin{array}{ccccc}\mathrm{I}&5y& -& 3x& =& 10\\\mathrm{II}& 4x &+& 5y &=& 16\end{array}$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: System of linear equations
  Linear system with two variables
  A linear system of equations is composed of several linear equations with common unknowns (variables), all of which are to be satisfied.
  Graphical solution
  To solve the system of equations graphically, you can resolve the individual lines for $y$ and draw the corresponding lines into a coordinate system. Then you only need to read off the coordinates of the intersection of the two lines.
  $\def\arraystretch{1.25} \begin{array}{ccccc}\mathrm{I}&5y-3x&=&10&|+3x\\\mathrm{I}&5y& =&3x+10&|:5\\\mathrm{I}&y& =& 0{.}6x+2\\\mathrm{II}& 4x+5y &=& 16&|-4x\\\mathrm{II}& 5y &=&-4x+ 16&|:5\\\mathrm{II}& y &=&-0{.}8x+3{.}2\end{array}$
  The intersection is at $x\approx0{.}86$ and $y\approx2{.}51$ . Thus the solution set is $L=\left\{\left(0{.}86\left|\;2{.}51\right.\right)\right\}$ .
  Solution by computation
  In this case, the elimination by equating coefficients is a good choice, since you have already solved both equations for $y$ in order to solve them graphically.
  $\def\arraystretch{1.25} \begin{array}{ccccc}\mathrm{I}&y& =& 0{.}6x+2\\\mathrm{II}& y &=&-0{.}8x+3{.}2\end{array}$
  Set $\mathrm{I}$ and $\mathrm{II}$ equal and solve for $x$ .
  $\def\arraystretch{2} \begin{array}{rcccc}0{.}6x+2&=&-0{.}8x+3{.}2&|-2\\0{.}6x&=&-0{.}8x+1{.}2&|+0{.}8x\\1{.}4x&=&1{.}2&|:1{.}4\\x&=&\dfrac67\approx0{.}86\end{array}$
  Plug the obtained value for $x$ into one of the equations, e.g. $\mathrm{I}$ .
  $y=0.6\cdot\frac{6}{7}+2=\frac{88}{35}\approx2.51$
  Now, you can write down the solution set.
  $L=\left\{\left(\dfrac67\left|\dfrac{88}{35}\right.\right)\right\}$
  Do you have a question?
  Write it here 👉
9
Test your knowledge!
1. Which of the following is a system of linear equations? Check all the answers that apply.
  $\def\arraystretch{1.25} \begin{array}{rrll}\mathrm{I} &2& = &a& + &b& + &c&\\\mathrm{II} &3& = &a& - &b& + &c&\\\mathrm{III} &1& = &&&&&c&\end{array}$
  $\def\arraystretch{1.25} \begin{array}{rrll}\mathrm{I} &x& + &y& = &2&\\\mathrm{II} &5& - &y& = &4&\end{array}$
  $\def\arraystretch{1.25} \begin{array}{rrll}\mathrm{I} &x& = &1\\\mathrm{II} &y& = &2\end{array}$
  $\def\arraystretch{1.25} \begin{array}{rrll}\mathrm{I} &x& = &0\\\mathrm{II} &x& = &2&\end{array}$
  $\def\arraystretch{1.25} \begin{array}{rrll}\mathrm{I} &\frac7x&-&\frac{12}y&=&\frac56\\\mathrm{II} &x&-&y& = &2\end{array}$
  $\def\arraystretch{1.25} \begin{array}{rrll}\mathrm{I} &-3& = &x& + &2y&\\\mathrm{II} &1& = &x^2& + &y&\end{array}$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: System of linear equations
  We have a linear system of equations if and only if:
  We have several equations that have to be fulfilled at the same time.
  Those equations are linear, that means every variable occurs at most with power 1
  Now, let's check:
  $\def\arraystretch{1.25} \begin{array}{rrll}\mathrm{I} &\frac7x&-&\frac{12}y&=&\frac56\\\mathrm{II} &x&-&y& = &2\end{array}$
  Is not a linear system, since $x$ and $y$ have power $- 1$ in the first line.
  $\def\arraystretch{1.25} \begin{array}{rrll}\mathrm{I} &-3& = &x& + &2y&\\\mathrm{II} &1& = &x^2& + &y&\end{array}$
  Is not a linear system, since $x$ has power $2$ in the second line.
  All other systems are indeed linear, since $x$ and $y$ only appear with power $1$ .
  Do you have a question?
  Write it here 👉
2. How many solutions has the following linear system of equations?
  $\def\arraystretch{1.25} \begin{array}{rrll}\mathrm{I} &-3& = &x& + &2y&\\\mathrm{II} &1& = &x& + &2y&\end{array}$
  none
  exactly one
  infinitely many
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Solvability of linear systems
  You may show that the equations $\mathrm{I}$ and $\mathrm{II}$ cannot be satisfied at the same time. The best way to do this is by subtraction.
  Subtract equation $\mathrm{II}$ from $\mathrm{I}$ .
  $\def\arraystretch{1.25} \begin{array}{rlrl}\mathrm{I} - \mathrm{II} \;\rightarrow &-4 &= &0 \end{array}$
  The equation is obviously wrong. Thus, the given system of equations is also "wrong" and has no solution.
  Thus the solution set is $\mathbb{L}=\{\}$ .
  Do you have a question?
  Write it here 👉
3. How many solutions does the following system of equations have?
  $\def\arraystretch{1.25} \begin{array}{rrll}\mathrm{I} &\frac{9}{2}x&-&\frac{3}{2}y&=&3\\\mathrm{II} &3x& = &2&+&y\end{array}$
  none
  exactly one
  infinitely many
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Solvability of linear systems
  You need to show that equations $\mathrm{I}$ and $\mathrm{II}$ are equivalent. That is, any $x$ and $y$ that solves equation $\mathrm{I}$ also solves equation $\mathrm{II}$ (and vice versa).
  One way to show this is by the substitution method.
  First, solve $\mathrm{II}$ for y .
  $\def\arraystretch{1.25} \begin{array}{rrll}\mathrm{II} &3x& = &2&+&y &\vert -2\end{array}$
  $\def\arraystretch{1.25} \begin{array}{lrll}\mathrm{II}' &3x&-&2& = &y\end{array}$
  Now substitute $\mathrm{II}'$ into equation $\mathrm{I}$ . You get the new equation $\mathrm{I'}$ .
  $\def\arraystretch{1.25} \begin{array}{rrll}\mathrm{I'} &\frac{9}{2}x&-&\frac{3}{2}(3x - 2)&=&3\end{array}$
  Summarize the left side of the equation.
  $\def\arraystretch{1.25} \begin{array}{rrll}\mathrm{I'} &3&=&3\end{array}$
  The equation $\mathrm{I'}$ is true and independent of $x$ . So there are infinitely many solutions.
  $\mathbb{L}=\{(x|y)|\;y=3x-2\}$
  Do you have a question?
  Write it here 👉
4. How many solutions does the following linear system of equations have?
  $\def\arraystretch{1.25} \begin{array}{rrll}\mathrm{I} &2x&+&y&=&\frac56\\\mathrm{II} &x&-&2y& = &2\end{array}$
  none
  exactly one
  infinitely many
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Solvability of linear systems
  A linear system of equations has exactly one solution, if it can be solved unambiguously for $x$ and $y$ .
  In order to find this out, the substitution method is useful, because for example equation $\mathrm{I}$ can be solved for $y$ very easily.
  First solve equation $\mathrm{I}$ for $y$ .
  $\def\arraystretch{1.25} \begin{array}{rrll}\mathrm{I} &2x&+&y& = &\frac{5}{6} & \vert -2x\end{array}$
  $\def\arraystretch{1.25} \begin{array}{rrll}\mathrm{I'} &y& = &\frac{5}{6}&-&2x\end{array}$
  Now plug $\mathrm{I'}$ into $\mathrm{II}$ . You get the new equation $\mathrm{II'}$ .
  $\def\arraystretch{1.25} \begin{array}{rrll}\mathrm{II'} &x&-&2(\frac56 - 2x)&=&2\end{array}$
  Summarize the left side.
  $\def\arraystretch{1.25} \begin{array}{rrll}\mathrm{II'} &5x&-&\frac53&=&2\end{array}$
  Solve for $x$ .
  $x = \frac{11}{15}$
  Plug $x=\frac{11}{15}$ into $\mathrm{I}$ , in order to get $y$ .
  $\def\arraystretch{1.25} \begin{array}{rrll}\mathrm{I} &2\cdot(\frac{11}{15})&+&y& = &\frac{5}{6}\end{array}$
  Solve for $y$ .
  $y = -\frac{19}{30}$
  So there is exactly one solution.
  Do you have a question?
  Write it here 👉

Determine the solution sets of the following systems of linear equations.

$\def\arraystretch{1.25} \begin{array}{lrcll}(\text I)&2y&=&2x-40\\(\text {II})&3x&=&10-2y\end{array}$

$\def\arraystretch{1.25} \begin{array}{l}(\text I)\;\;\frac 12 x-\frac35y=3\\(\text{II})\;\frac 14x+y=8\end{array}$

Für diese Aufgabe benötigst Du folgendes Grundwissen: System of linear equations

You can solve linear systems of equations using the substitution method, the addition method, or by equating coefficients.

The substitution method is best suited here.

Substitution method

Given is: $\def\arraystretch{1.25} \begin{array}{l}(I)\;\;\frac 12x-\frac{3}5y=3\\(\text{II})\;\frac 14x+y=8\end{array}$

Rewrite $(I I)$ such that $y$ is on one side.

$\displaystyle (\text {II})\;\;\;\; \frac 14 x+y$	$=$	$\displaystyle 8\;\;\;$	$\displaystyle -\frac{1}{4}x$
$\displaystyle \left(\text{II'}\right)\ \ \ \ \ \ \ \ \ \ \ \ y$	$=$	$\displaystyle 8-\frac{1}{4}x$

Plug $y=8-\frac 14 x$ into $(\text I)$ and solve for $x$ .

$\displaystyle \left(\text{I}\right)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{2}x-\frac{3}{5}y$	$=$	$\displaystyle 3$
	↓	plug in $y=8-\frac 14 x$
$\displaystyle \frac{1}{2}x-\frac{3}{5}\cdot\left(8-\frac{1}{4}x\right)$	$=$	$\displaystyle 3$
$\displaystyle \frac{1}{2}x-\frac{24}{5}+\frac{3}{20}x$	$=$	$\displaystyle 3$	$\displaystyle +\frac{24}{5}$
$\displaystyle \frac{1}{2}x+\frac{3}{20}x$	$=$	$\displaystyle 3+\frac{24}{5}$
	↓	compute the fractions
$\displaystyle \frac{10}{20}x+\frac{3}{20}x$	$=$	$\displaystyle 3+4\ \frac{4}{5}$
$\displaystyle \frac{13}{20}x$	$=$	$\displaystyle 7\ \frac{4}{5}$	$\displaystyle \cdot\frac{20}{13}$
$\displaystyle x$	$=$	$\displaystyle 12$

plug $x = 12$ into $\left(\text{II'}\right)$ .

$\displaystyle \left(\text{II'}\right)\ \ y$	$=$	$\displaystyle 8-\frac{1}{4}x$
	↓	plug in $x = 12$
	$=$	$\displaystyle 8-\frac{1}{4}\cdot12$
	$=$	$\displaystyle 8-3$
	$=$	$\displaystyle 5$

Write down the solution set, first the solution for $x$ , then for $y$ .

$L=\left\{\left(12\;\left|\;5\right.\right)\right\}$

Solution by other methods

Do you have a question?

Write it here 👉

11
A hotel has 105 beds, which are located in 40 two- and three-bed rooms. How many two-and-three-bed rooms does the hotel have?
Solve the problem by using a system of equations!
Für diese Aufgabe benötigst Du folgendes Grundwissen: System of linear equations
Extracting equations from the problem setting
What is known? The sum of the number of two-bed rooms $x$ and the number of three-bed rooms $y$ is 40. This statement gives you equation $I$ :
$\displaystyle \mathrm{I}\quad x+y=40_{ }$
In a double room there are two beds. Thus, there are $2\cdot x$ beds in $x$ two-bed rooms and correspondingly, there are $3\cdot x$ beds in the three-bed rooms. In total, the hotel has 105 beds, so you can set up equation $I I$ :
$\displaystyle \mathrm{II} \quad 2x + 3y =105$
You have now obtained the following system of linear equations with two equations and two unknowns:
$\displaystyle \def\arraystretch{1.25} \begin{array}{cl} \mathrm{I}\quad x & + & y & = & 40 & \\ \mathrm{II}\quad 2x & + & 3y & =& 105 \end{array}$
Solve the system
To solve this system of equations, you have three solution methods at your disposal:
Substitution method
Elimination by equating coefficients
Addition method
Solving by substitution method
Solve equation $\mathrm{II}$ for one of the two variables and substitute this variable into equation $\mathrm{II}$ . It doesn't matter whether you solve $\mathrm{II}$ for $x$ or $y$ .
$\mathrm{I'}\quad y=40-x$ . plugged into $\mathrm{II}:$
$\displaystyle \def\arraystretch{1.25} \begin{array}{rcl}2\cdot x +3\cdot(40-x)&=&105&\vert\text{multiply out bracket}\\\\2x+120-3x&=&105&\vert\text{conclude}\\\\-x+120&=&105&\vert-120\\\\-x&=&-15&\vert\cdot (-1)\\\\x&=&15\end{array}$
Plug $x = 15$ into equation $\mathrm{I'}$ : $y = 40 - 15 = 25$
So the solution set of the linear system is : $\mathbb{L}=\{(15\vert 25)\}$
Answer: The hotel has 15 two-bed rooms and 25 theree-bed rooms.
Solving by equating coefficients
Solve both equations for the same variable. You have already solved the equation $\mathrm{I}$ for $y$ in the substitution method:
$\mathrm{I'}\quad y=40-x$ . Now, also $\mathrm{II}$ must be solved for $y$ :
$\displaystyle \def\arraystretch{2} \begin{array}{rcl}2\cdot x +3\cdot y&=&105&\vert-2 x\\\\3y&=&105-2x&\vert:3\\\\\mathrm{II'}\quad y&=&35-\dfrac{2}{3}x\end{array}$
Now set equal the two right sides of the equations $\mathrm{I'}$ and $\mathrm{II'}$ :
$\displaystyle \def\arraystretch{2} \begin{array}{rcl}40- x &=&35-\dfrac{2}{3}x&\vert+x\\\\40&=&35+\dfrac{1}{3}x&\vert-35\\\\5&=&\dfrac{1}{3}x&\vert\cdot3\\\\x&=&15\end{array}$
Plug $x = 15$ into $\mathrm{I'}$ : $y = 40 - 15 = 25$
So the solution set of your system of equations is : $\mathbb{L}=\{(15\vert 25)\}$
Answer: The hotel has 15 two-bed rooms and 25 theree-bed rooms.
Solvin by the addition method
$\displaystyle \def\arraystretch{1.25} \begin{array}{cccccccl} \mathrm{I}\quad x & + & y & = & 40 & \\ \mathrm{II}\quad 2x & + & 3y & =& 105 \end{array}$
The goal is that by adding the two equations, one of the unknown variables cancels out. In this system of equations you can, for instance, multiply equation $\mathrm{I}$ by $(- 2)$ , which gets you $\mathrm{I'}$ . By adding $\mathrm{I'}$ and $\mathrm{II}$ , you get the solution for $y$ :
$\displaystyle \def\arraystretch{1.25} \begin{array}{cl} \mathrm{I'}\quad &-2x& - &2y & = & -80 \\\mathrm{II}\quad &2x& + &3y & = & 105\\\mathrm{I'}+\mathrm{II}&0x&+&y&=&25\end{array}$
Plug $y = 25$ into $\mathrm{I}$ : $x+25=40\Rightarrow x=15$
So the solution set of the linear system is : $\mathbb{L}=\{(15\vert 25)\}$
Answer: The hotel has 15 two-bed rooms and 25 theree-bed rooms.
Construct a linear system of equations from the problem setting.
Tip: Choose the variable $x$ for the number of two-bed rooms and the variable $y$ for the number of three-bed rooms.
12
A farmer keeps chickens and rabbits in his barn. He counts a total of 120 legs. There are three times more chickens than rabbits. How many chickens and rabbits does the farmer have?
Solve by suing a system of equations!
Für diese Aufgabe benötigst Du folgendes Grundwissen: System of linear equations
Transforming the problem setting into equations
You may denote the number of chickens by some variable $x$ and the number of rabbits by a variable $y$ . What is known? A chicken has two legs, a rabbit has four legs. Thus $x$ chickens have $2 \cdot x$ legs and $y$ rabbits have $4 \cdot y$ legs. This statement gives you equation $\mathrm{I}$ :
$\displaystyle \mathrm{I}\quad 2\cdot x + 4 \cdot y =120$
"There are three times more chickens than rabbits" must also be cast into an equation. Since $y$ is the number of rabbits, you must multiply $y$ by $3$ to get the number $x$ of chickens. So equation $\mathrm{II}$ reads:
$\displaystyle \mathrm{II} \quad x =3\cdot y$
You have now obtained the following system of linear equations with two equations and two variables:
$\displaystyle \def\arraystretch{1.25} \begin{array}{cl} \mathrm{I}\quad 2x & + &4 y & = & 120 & \\ \mathrm{II}\quad x&=&3y& \end{array}$
To solve this system of equations, you have three methods at your disposal:
Substitution method
Elimination by equating coefficients
Addition method
Here the substitution method is suitable, since the equation $\mathrm{II}$ is already solved for $x$ .
Solution by substitution
Plug equation $\mathrm{II}\;\; x =3\cdot y$ into $\mathrm{I}$ :
$\def\arraystretch{1.25} \begin{array}{rcl}2\cdot (3y) +4y&=&120&\vert\text{ multiply out bracket}\\\\6y+4y&=&120&\vert\text{ conclude}\\\\10y&=&120&\vert:10\\\\y&=&12&\end{array}$
Plug $y = 12$ into equation $\mathrm{II}$ : $x=3 \cdot12=36$
So the solution set of your system of equations is : $\mathbb{L}=\{(36\vert 12)\}$
Answer: The farmer has $36$ chickens and $12$ rabbits.
When solving systems of equations, a verification is useful at the end of the calculation. In this case: $36$ chickens have $36\cdot 2 = 72$ legs and $12$ rabbits have $12\cdot 4 =48$ legs. The sum of the legs is $120$ , as indicated in the problem setting. The number of chickens was supposed to be three times the number of rabbits. This statement is also correct, since $36 = 3 \cdot12$ . So the given solution set is indeed a solution of the linear system.
Construct a linear system of equations from the problem setting.
Hint: Choose the variable $x$ for the number of chickens and the variable $y$ for the number of rabbits.

Determine the solution sets of the following systems of nonlinear equations.

$\def\arraystretch{1.25} \begin{array}{rcll}(\text{I})&\frac 4x+\frac 8y&=&\frac53\\(\text{II})&\frac 2x-\frac 4y&=&-\frac16\end{array}$

where $x,y \neq 0$

Für diese Aufgabe benötigst Du folgendes Grundwissen: System of linear equations

You may solve linear systems of equations using the substitution method, the addition method or by equating coefficients. Such methods can also be effective for some non-linear system of equations.

Here the addition method is suitable. If you transform the two equations so that the terms with $x$ or $y$ are identical, you can subtract the two transformed equations from each other and only have to solve one equation with one variable.

Solution by addition method

Using the addition method you get to the following solution:

Given is: $\def\arraystretch{1.25} \begin{array}{rcll}(\text{I})&\frac 4x+\frac 8y&=&\frac53\\(\text{II})&\frac 2x-\frac 4y&=&-\frac16\end{array}$

The equation $(\text{II})$ contains a term $-\frac{4}{y}$ and equation $(\text I)$ is a multiple of $-\frac{4}{y}$ , namely $\frac{8}{y}$ . So you can multiply $(\text{II})$ by $2$ and add both equations to eliminate $\frac{8}{y}$ .

$\displaystyle \left(\text{II}\right)\ \ \frac{2}{x}-\frac{4}{y}$	$=$	$\displaystyle -\frac{1}{6}$	$\displaystyle \cdot2$
$\displaystyle \left(\text{II}'\right)\ \ \frac{4}{x}-\frac{8}{y}$	$=$	$\displaystyle -\frac{2}{6}$

So the new system of equations is:

$\def\arraystretch{1.25} \begin{array}{rcll}(\text{I})&\frac 4x+\frac 8y&=&\frac53\\(\text{II}')&\frac 4x-\frac 8y&=&-\frac26\end{array}$

$(\text I) + (\text{II}')$ :

$\def\arraystretch{1.25} \begin{array}{rcll}(\text{I})&\frac 4x+\frac 8y&=&\frac53\\+\;\;(\text{II}')&\frac 4x-\frac 8y&=&-\frac26\end{array}$

$\overline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{8}{x}\;\;\;\;\;\;\;\;\;=\;\frac{5}3-\frac{2}{6}\;\;\;}$

Now solve the new equation for $x$ .

$\displaystyle \frac{8}{x}$	$=$	$\displaystyle \frac{5}{3}-\frac{2}{6}$
$\displaystyle \frac{8}{x}$	$=$	$\displaystyle \frac{5}{3}-\frac{1}{3}$
$\displaystyle \frac{8}{x}$	$=$	$\displaystyle \frac{4}{3}$
	↓	invert the fractions
$\displaystyle \frac{x}{8}$	$=$	$\displaystyle \frac{3}{4}$	$\displaystyle \cdot8$
$\displaystyle x$	$=$	$\displaystyle \frac{3\cdot8}{4}$
$\displaystyle x$	$=$	$\displaystyle 6$

Now, plug $x$ into one of the above equations, e.g. $(\text{I})$ , and solve for $y$ .

$\displaystyle \left(\text{I}\right)\ \ \frac{4}{x}+\frac{8}{y}$	$=$	$\displaystyle \frac{5}{3}$
	↓	plug in $x = 6$
$\displaystyle \frac{4}{6}+\frac{8}{y}$	$=$	$\displaystyle \frac{5}{3}$	$\displaystyle -\frac{4}{6}$
$\displaystyle \frac{8}{y}$	$=$	$\displaystyle \frac{5}{3}-\frac{4}{6}$
$\displaystyle \frac{8}{y}$	$=$	$\displaystyle \frac{5}{3}-\frac{2}{3}$
$\displaystyle \frac{8}{y}$	$=$	$\displaystyle \frac{3}{3}$
$\displaystyle \frac{8}{y}$	$=$	$\displaystyle 1$	$\displaystyle \cdot y$
$\displaystyle y$	$=$	$\displaystyle 8$

Write down the solution set. $L=\left\{\left(x\;\left|\;y\right.\right)\right\}=\left\{\left(6\;\left|\;8\right.\right)\right\}$ .

Do you have a question?

Write it here 👉

$\def\arraystretch{1.25} \begin{array}{rrcl}(\text I)&\frac{7}{x}-\frac{12}{y}&=&\frac{5}{6}\\\left(\text{II}\right)&\frac{4}{y}+\frac{5}{2}&=&\frac{9}{x}\end{array}$

where $x,y \neq 0$

Für diese Aufgabe benötigst Du folgendes Grundwissen: System of linear equations

You may solve linear systems of equations using the substitution method, the addition method or by equating coefficients. Such methods can also be effective for some non-linear system of equations.

Solution by the addition method

Using the addition method you get to the following solution:

Given is: $\def\arraystretch{1.25} \begin{array}{rrcl}(\text I)&\frac{7}{x}-\frac{12}{y}&=&\frac{5}{6}\\\left(\text{II}\right)&\frac{4}{y}+\frac{5}{2}&=&\frac{9}{x}\end{array}$

Transform $(\text{II})$ such that on one side you have only variables and on the other side only numbers.

$\displaystyle (\text{II})\;\;\;\;\;\;\;\;\;\frac4y+\frac52$	$=$	$\displaystyle \frac{9}{x}\ \ \ \ \ \ \$	$\displaystyle -\frac{9}{x}-\frac{5}{2}$
$\displaystyle (\text{II}')\;\;\;\;-\frac{9}{x}+\frac{4}{y}$	$=$	$\displaystyle -\frac{5}{2}$

The new system of equations is now:

$\def\arraystretch{1.25} \begin{array}{rrcl}(\text I)&\frac{7}{x}-\frac{12}{y}&=&\frac{5}{6}\\\left(\text{II}'\right)&-\frac{9}{x}+\frac{4}{y}&=&-\frac{5}{2}\end{array}$

Now, equation $(\text{II}')$ contains the term $\frac{4}{y}$ and equation $(\text I)$ is a multiple of $\frac{4}{y}$ , namely $-\frac{12}{y}$ . So you multiply $(\text{II}')$ by $3$ and add both equations in order to eliminate $\frac{12}{y}$ .

$\displaystyle \left(\text{II}'\right)\ \ \ \ -\frac{9}{x}+\frac{4}{y}$	$=$	$\displaystyle -\frac{5}{2}$	$\displaystyle \cdot 3$
$\displaystyle \left(\text{II}''\right)\ \ -\frac{27}{x}+\frac{12}{y}$	$=$	$\displaystyle -\frac{15}{2}$

$(\text I) +(\text{II}'')$ :

$\left(I\right)\frac7x-\frac{12}y=\frac56$

$\left(II\right)-\frac{27}{x}+\frac{12}{y}=-\frac{15}{2}$

$\def\arraystretch{1.25} \begin{array}{rrcr}(\text I)&\frac{7}{x}-\frac{12}{y}&=&\frac{5}{6}\\+\left(\text{II}''\right)&-\frac{27}{x}+\frac{12}{y}&=&-\frac{15}{2}\end{array}$

$\overline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;-\frac{20}x\;=\;-\frac{40}6\;\;\;}$

Now, you solve $-\frac{20}x\;=\;-\frac{40}6$ , by inverting the fraction.

$\displaystyle -\frac{20}{x}$	$=$	$\displaystyle -\frac{40}{6}$
$\displaystyle -\frac{x}{20}$	$=$	$\displaystyle -\frac{6}{40}$	$\displaystyle \cdot \left(-20\right)$
$\displaystyle x$	$=$	$\displaystyle \frac{6\cdot 20}{2\cdot 20}$
	↓	shorten
$\displaystyle x$	$=$	$\displaystyle \frac{6}{2}$
$\displaystyle x$	$=$	$\displaystyle 3$

Now, plug $x$ into one of the above equations, for instance $(\text{II})$ , and solve for $y$ .

$\displaystyle \left(\text{II}\right)\ \ \ \frac{4}{y}+\frac{5}{2}$	$=$	$\displaystyle \frac{9}{x}$
	↓	plug in $x = 3$
$\displaystyle \frac{4}{y}+\frac{5}{2}$	$=$	$\displaystyle \frac{9}{3}$	$\displaystyle -\frac{5}{2}$
$\displaystyle \frac{4}{y}$	$=$	$\displaystyle \frac{9}{3}-\frac{5}{2}$
$\displaystyle \frac{4}{y}$	$=$	$\displaystyle \frac{18}{6}-\frac{15}{6}$
$\displaystyle \frac{4}{y}$	$=$	$\displaystyle \frac{3}{6}$
$\displaystyle \frac{4}{y}$	$=$	$\displaystyle \frac{1}{2}$
	↓	invert the fraction
$\displaystyle \frac{y}{4}$	$=$	$\displaystyle \frac{2}{1}$	$\displaystyle \cdot 4$
$\displaystyle y$	$=$	$\displaystyle 8$

Write down the solution set. $L=\left\{\left(x\;\left|\;y\right.\right)\right\}=\left\{\left(3\;\left|\;8\right.\right)\right\}$

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$\def\arraystretch{1.25} \begin{array}{l}(\text I)\;\;\frac{4}{3x+1}=\frac{2}{3y-13}\\(\text{II})\;\frac 2{5x-10}=\frac{4}{7y-6}\end{array}$

where $x\notin\left\{-\frac{1}{3};2\right\}$ and $y\notin\left\{\frac{13}{3};\frac{6}{7}\right\}$

Für diese Aufgabe benötigst Du folgendes Grundwissen: System of linear equations

This non-linear system of equations can be transformed into a linear system of equations. And you may solve linear systems using the substitution method, the addition method or by equating coefficients.

Given is: $\def\arraystretch{1.25} \begin{array}{l}(\text{I})\;\;\frac{4}{3x+1}=\frac{2}{3y-13}\\(\text{II})\;\frac 2{5x-10}=\frac{4}{7y-6}\end{array}$

Transform $(\text{I})$ and $(\text{II})$ , such that both equations no longer contain a fraction.

$\displaystyle \left(\text{I}\right)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{4}{3x+1}$	$=$	$\displaystyle \frac{2}{3y-13}$	$\displaystyle \cdot \left(3x+1\right)\cdot \left(3y-13\right)$
	↓	multiply by the denominators
$\displaystyle 4\cdot \left(3y-13\right)$	$=$	$\displaystyle 2\cdot \left(3x+1\right)$
$\displaystyle \left(\text{I}'\right)\ \ \ \ \ \ \ \ 12y-52$	$=$	$\displaystyle 6x+2$

$\displaystyle \left(\text{II}\right)\ \ \ \ \ \ \ \ \ \ \frac{2}{5x-10}$	$=$	$\displaystyle \frac{4}{7y-6}$	$\displaystyle \cdot \left(5x-10\right)\cdot \left(7y-6\right)$
	↓	multiply by the denominators
$\displaystyle 2\cdot \left(7y-6\right)$	$=$	$\displaystyle 4\cdot \left(5x-10\right)$
$\displaystyle \left(\text{II}'\right)\ \ \ \ \ 14y-12$	$=$	$\displaystyle 20x-40$

Thus, the linear system of equations is:

$\def\arraystretch{1.25} \begin{array}{l}(\text{I}')\;\;12y-52=6x+2\\(\text{II}')\;14y-12=20x-40\end{array}$

Now you can use the known procedures for solving the system of equations.

Solution by the substitution method

With the substitution method you get this solution:

Given is: $\def\arraystretch{1.25} \begin{array}{l}(\text I')\;\;12y-52=6x+2\\(\text{II}')\;14y-12=20x-40\end{array}$

Transform $(\text{I}')$ , such that $x$ appears only one one side.

$\displaystyle \left(\text{I}'\right)\ \ \ \ \ 12y-52$	$=$	$\displaystyle 6x+2$	$\displaystyle -2$
$\displaystyle 12y-54$	$=$	$\displaystyle 6x$	$\displaystyle :6$
$\displaystyle \left(\text{I}''\right)\ \ \ \ \ \ \ \ 2y-9$	$=$	$\displaystyle x$

Plug $x = 2 y - 9$ into $(\text{II}')$ .

$\displaystyle \left(\text{II}'\right)\ \ \ 14y-12$	$=$	$\displaystyle 20x-40$
	↓	plug in $x = 2 y - 9$
$\displaystyle 14y-12$	$=$	$\displaystyle 20\cdot \left(2y-9\right)-40$
	↓	Solve for $y$
$\displaystyle 14y-12$	$=$	$\displaystyle 40y-180-40$
$\displaystyle 14y-12$	$=$	$\displaystyle 40y-220$	$\displaystyle -14y+220$
$\displaystyle 208$	$=$	$\displaystyle 26y$	$\displaystyle :26$
$\displaystyle y$	$=$	$\displaystyle 8$

Plug $y = 8$ into $(\text I')$ , in order to find $x$ .

$\displaystyle \left(\text{I}'\right)\ \ x$	$=$	$\displaystyle 2y-9$
	↓	plug in $y = 8$
	$=$	$\displaystyle 2\cdot 8-9$
	$=$	$\displaystyle 16 - 9$
	$=$	$\displaystyle 7$

Write down the solution set. $L=\left\{\left(x\left|y\right.\right)\right\}=\left\{\left(7\left|8\right.\right)\right\}$

Additionally: Check your solution

Plug $x = 7$ and $y = 8$ into equations $(\text I)$ and $(\text{II})$ and check if the equation is satisfied.

$\displaystyle (\text I)\;\;\frac{4}{3x+1}$	$=$	$\displaystyle \frac{2}{3y-13}$
	↓	plug in $x = 7$ and $y = 8$
$\displaystyle \frac{4}{3\cdot 7+1}$	$=$	$\displaystyle \frac{2}{3\cdot 8-13}$
$\displaystyle \frac{4}{21+1}$	$=$	$\displaystyle \frac{2}{24-13}$
$\displaystyle \frac{4}{22}$	$=$	$\displaystyle \frac{2}{11}$
$\displaystyle \frac{2}{11}$	$=$	$\displaystyle \frac{2}{11}\ \ \ \ \checkmark$

$\displaystyle (\text{II})\;\frac 2{5x-10}$	$=$	$\displaystyle \frac{4}{7y-6}$
	↓	plug in $x = 7$ and $y = 8$
$\displaystyle \frac{2}{5\cdot 7-10}$	$=$	$\displaystyle \frac{4}{7\cdot 8-6}$
$\displaystyle \frac{2}{35-10}$	$=$	$\displaystyle \frac{4}{56-6}$
$\displaystyle \frac{2}{25}$	$=$	$\displaystyle \frac{4}{50}$
$\displaystyle \frac{2}{25}$	$=$	$\displaystyle \frac{2}{25}\ \ \ \ \ \ \ \checkmark$

Both equations are satisfied, so our solution is correct.

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$\def\arraystretch{1.25} \begin{array}{rrcl}(\text I)&\frac{3}{2x-1}-\frac{8}{3y+2}&=&-\frac{1}{5}\\\left(\text{II}\right)&\frac5{2x-1}+\frac4{3y+2}&=&\frac8{15}\end{array}$

where $x\ne\frac{1}{2}$ and $y \neq -\frac 23$

Für diese Aufgabe benötigst Du folgendes Grundwissen: System of linear equations

You may solve linear systems of equations using the substitution method, the addition method or by equating coefficients. Such methods can also be effective for some non-linear system of equations.

Solution using the addition method

Given is: $\def\arraystretch{1.25} \begin{array}{rrcl}(\text I)&\frac3{2x-1}-\frac8{3y+2}&=&-\frac{1}{5}\\\left(\text{II}\right)&\frac5{2x-1}+\frac4{3y+2}&=&\frac{8}{15}\end{array}$

The equation $(\text{II})$ contains the term $\frac{4}{3y+2}$ and $(\text I)$ contains a multiple of $\frac{4}{3y+2}$ , namely $-\frac{8}{3y+2}$ . So you multiply $(\text{II})$ by $2$ and add both equations in order to eliminate $\frac{8}{3y+2}$ .

$\displaystyle \left(\text{II}\right)\ \ \frac{5}{2x-1}+\frac{4}{3y+2}$	$=$	$\displaystyle \frac{8}{15}\ \ \ \ \$	$\displaystyle \cdot2$
$\displaystyle \left(\text{II}'\right)\ \ \frac{10}{2x-1}+\frac{8}{3y+2}$	$=$	$\displaystyle \frac{16}{15}$

So the new system of equations is:

$\def\arraystretch{1.25} \begin{array}{rrcl}(\text I)&\frac3{2x-1}-\frac8{3y+2}&=&-\frac{1}{5}\\\left(\text{II}'\right)&\frac{10}{2x-1}+\frac8{3y+2}&=&\frac{16}{15}\end{array}$

$(\text I)+(\text{II}')$ :

$\def\arraystretch{1.25} \begin{array}{rrcl}(\text I)&\frac3{2x-1}-\frac8{3y+2}&=&-\frac{1}{5}\\+\left(\text{II}'\right)&\frac{10}{2x-1}+\frac8{3y+2}&=&\frac{16}{15}\end{array}$

$\overline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{13}{2x-1}\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\;-\frac{1}5+\frac{16}{15}\;\;\;}$

Now solve the equation resulting from the addition for $x$ .

$\displaystyle \frac{13}{2x-1}$	$=$	$\displaystyle -\frac{1}{5}+\frac{16}{15}$
$\displaystyle \frac{13}{2x-1}$	$=$	$\displaystyle -\frac{3}{15}+\frac{16}{15}$
$\displaystyle \frac{13}{2x-1}$	$=$	$\displaystyle \frac{13}{15}$
	↓	invert the fractions
$\displaystyle \frac{2x-1}{13}$	$=$	$\displaystyle \frac{15}{13}$	$\displaystyle \cdot13$
$\displaystyle 2x-1$	$=$	$\displaystyle 15$	$\displaystyle +1$
$\displaystyle 2x$	$=$	$\displaystyle 16$	$\displaystyle :2$
$\displaystyle x$	$=$	$\displaystyle 8$

Now, plug $x$ into one of the above equations, e.g., $(\text{I})$ , and solve for $y$ .

$\displaystyle \left(\text{I}\right)\ \ \frac{3}{2x-1}-\frac{8}{3y+2}$	$=$	$\displaystyle -\frac{1}{5}$
	↓	plug in $x = 8$
$\displaystyle \frac{3}{2\cdot8-1}-\frac{8}{3y+2}$	$=$	$\displaystyle -\frac{1}{5}$
$\displaystyle \frac{3}{15}-\frac{8}{3y+2}$	$=$	$\displaystyle -\frac{1}{5}$
$\displaystyle \frac{1}{5}-\frac{8}{3y+2}$	$=$	$\displaystyle -\frac{1}{5}$	$\displaystyle -\frac{1}{5}$
$\displaystyle -\frac{8}{3y+2}$	$=$	$\displaystyle -\frac{2}{5}$
	↓	invert the fractions
$\displaystyle -\frac{3y+2}{8}$	$=$	$\displaystyle -\frac{5}{2}$	$\displaystyle \cdot\left(-8\right)$
$\displaystyle 3y+2$	$=$	$\displaystyle \frac{5\cdot8}{2}$
$\displaystyle 3y+2$	$=$	$\displaystyle 20$	$\displaystyle -2$
$\displaystyle 3y$	$=$	$\displaystyle 18$	$\displaystyle :3$
$\displaystyle y$	$=$	$\displaystyle 6$

Write down the solution set, using first the solution for $x$ , then for $y$ .

$L=\left\{\left(8\;\left|\;6\right.\right)\right\}$

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$\displaystyle 1.5x+2$	$=$	$\displaystyle 0.5x+2.5\;\;$	$\displaystyle -0.5x$
$\displaystyle x+2$	$=$	$\displaystyle 2.5$	$\displaystyle -2$
$\displaystyle x$	$=$	$\displaystyle 0.5$

$\displaystyle 0.5y-2$	$=$	$\displaystyle 0.4y-0.2\;\;$	$\displaystyle +2$
$\displaystyle 0.5y$	$=$	$\displaystyle 0.4y+1.8$	$\displaystyle -0.4y$
$\displaystyle 0.1y$	$=$	$\displaystyle 1.8$	$\displaystyle :0{,}1$
$\displaystyle y$	$=$	$\displaystyle 18$

$\displaystyle \left(\text{II}\right)\ \ \ \ \ 3x$	$=$	$\displaystyle 10-\textcolor{orange}{2y}$
	↓	from eq $(\text I)$ , plug in $\textcolor{orange}{2y=2x-40}$
$\displaystyle 3x$	$=$	$\displaystyle 10-(\textcolor{orange}{2x-40})$
	↓	multiply out the bracket
$\displaystyle 3x$	$=$	$\displaystyle 10-(\textcolor{orange}{2x-40})$	$\displaystyle +2x$
	↓	solve for $x$
$\displaystyle 5x$	$=$	$\displaystyle 50$	$\displaystyle :5$
$\displaystyle x$	$=$	$\displaystyle 10$

$\displaystyle \left(\text{I}\right)\ \ \ \ 2y$	$=$	$\displaystyle 2\textcolor{green}{x}-40$
	↓	plug in $\textcolor{green}{x=10}$
$\displaystyle 2y$	$=$	$\displaystyle 2\cdot\textcolor{green}{10}-40$
$\displaystyle 2y$	$=$	$\displaystyle 20-40$
$\displaystyle 2y$	$=$	$\displaystyle -20$	$\displaystyle :2$
$\displaystyle y$	$=$	$\displaystyle -10$

Exercises: systems with two variables

Solving linear systems by equating coefficients

1. solve both equations for y

2. Set equal

3. Solve equation for x

4. Plug in x to find y

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Solving linear systems by equating coefficients

1. Solve both equations for x

2. Set equal

3. Solve for y

4. Plug in y to find x

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Solving linear systems by equating coefficients

1. Solve both equations for one variable (for instance y)

2. Set equal

3. Solve for that one variable

4. Plug into one equation to find the other variable

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Step 0: Tidying up the equations and selecting a variable

Step 1: Multiply the equations

Equation 1

Equation 2

Step 2: Elimination of a variable by addition/subtraction.

Step 3: Determine the values of both variables.

Determine the value of yyy

Determine the value of xxx by plugging yyy into II\mathrm{II}II

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Alternative solution: equating coefficients

Alternative solution: combining addition and substitution method

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Linear system with two variables

Graphical solution

Solution by computation

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Linear system with two variables

Graphical solution

Solution by computation

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Linear system with two variables

Graphical solution

Solution by computation

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Substitution method

Solution by other methods

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Substitution method

Solution by other methods

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Extracting equations from the problem setting

Solve the system

Solving by substitution method

Solving by equating coefficients

Solvin by the addition method

Transforming the problem setting into equations

Solution by substitution

Solution by addition method

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Solution by the addition method

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Solution by the substitution method

Additionally: Check your solution

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Solution using the addition method

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Determine the value of $y$

Determine the value of $x$ by plugging $y$ into $\mathrm{II}$