Exercises: systems with two variables

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.

$\begin{array}{ccccc}\mathrm{I}& y-3x& =& 1& |+3x\\ \mathrm{I}& y& =& 3x+1& \\ \mathrm{II}& x+y& =& 1& |-x\\ \mathrm{II}& y& =& -x+1& \end{array}$

The intersection point is at $x=0$ and $y=1$. Thus the solution set is $L=\left\{\left(0|1\right)\right\}$.

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.

$\begin{array}{cccc}\mathrm{I}& y& =& 3x+1\\ \mathrm{II}& y& =& -x+1\end{array}$

Set $\mathrm{I}$ and $\mathrm{II}$ equal and solve for $x$.

$\begin{array}{rccc}3x+1& =& -x+1& |-1\\ 3x& =& -x& |+x\\ 4x& =& 0& |:4\\ x& =& 0& \end{array}$

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.

$L=\left\{\left(0|1\right)\right\}$

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.

$\begin{array}{ccccc}\mathrm{I}& 2y+5x& =& 3& |-5x\\ \mathrm{I}& 2y& =& -5x+3& |:2\\ \mathrm{I}& y& =& -2.5x+1.5& \\ \mathrm{II}& x-y& =& 1& |-x\\ \mathrm{II}& -y& =& -x+1& |\cdot (-1)\\ \mathrm{II}& y& =& x-1& \end{array}$

The intersection is at $x\approx 0.71$ and $y\approx -0.29$. Thus the solution set is $L=\left\{\left(0.71|-0.29\right)\right\}$.

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.

$\begin{array}{cccc}\mathrm{I}& y& =& -2.5x+1.5\\ \mathrm{II}& y& =& x-1\end{array}$

Set $\mathrm{I}$ and $\mathrm{II}$ equal and solve for $x$.

$\begin{array}{rccc}-2.5x+1.5& =& x-1& |-x\\ -3.5x+1.5& =& -1& |-1.5\\ -3.5x& =& -2.5& |:(-3.5)\\ x& =& {\displaystyle \frac{5}{7}}\approx 0.71& \end{array}$

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.

$L=\left\{\left({\displaystyle \frac{5}{7}}|-{\displaystyle \frac{2}{7}}\right)\right\}$

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.

$\begin{array}{ccccc}\mathrm{I}& 5y-3x& =& 10& |+3x\\ \mathrm{I}& 5y& =& 3x+10& |:5\\ \mathrm{I}& y& =& 0.6x+2& \\ \mathrm{II}& 4x+5y& =& 16& |-4x\\ \mathrm{II}& 5y& =& -4x+16& |:5\\ \mathrm{II}& y& =& -0.8x+3.2& \end{array}$

The intersection is at $x\approx 0.86$ and $y\approx 2.51$. Thus the solution set is $L=\left\{\left(0.86|2.51\right)\right\}$.

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.

$\begin{array}{cccc}\mathrm{I}& y& =& 0.6x+2\\ \mathrm{II}& y& =& -0.8x+3.2\end{array}$

Set $\mathrm{I}$ and $\mathrm{II}$ equal and solve for $x$.

$\begin{array}{rccc}0.6x+2& =& -0.8x+3.2& |-2\\ 0.6x& =& -0.8x+1.2& |+0.8x\\ 1.4x& =& 1.2& |:1.4\\ x& =& {\displaystyle \frac{6}{7}}\approx 0.86& \end{array}$

Plug the obtained value for $x$ into one of the equations, e.g. $\mathrm{I}$.

$y=0.6\cdot \frac{6}{7}+2=\frac{88}{35}\approx 2.51$

Now, you can write down the solution set.

$L=\left\{\left({\displaystyle \frac{6}{7}}|{\displaystyle \frac{88}{35}}\right)\right\}$