Further Exercises: Linear systems

Solve equation $(\mathrm{I})$ for $x_2$:

$x_2={\displaystyle \frac{1}{2}}\cdot x_1+1$

The line has slope $m=\frac{1}{2}\text{}$ and $x_2$-axis intercept $1$.

Draw this line into a coordinate system. Below, this is the $\text{purple line}$.

Solve equation $(\mathrm{I}\mathrm{I})$ for $x_2$:

$x_2=2\cdot x_1-2$

The line has slope $m=2$ and $x_2$-axis intercept $-2$.

Draw this line into the same coordinate system. Below, this is the $\text{red line}$.

The two lines intersect at the point $S(2|2)$.

Solution by equating coefficients

You may equivalently write equations $(\mathrm{I})$ and $(\mathrm{I}\mathrm{I})$ as $x_2={\displaystyle \frac{1}{2}}\cdot x_1+1$ and $x_2=2\cdot x_1-2$ . Setting both equations equal renders:

Plug $x_1=2$ into one of the two equations and calculate $x_2$.

$x_2={\displaystyle \frac{1}{2}}\cdot x_1+1\Rightarrow x_2={\displaystyle \frac{1}{2}}\cdot 2+1=2$

Result: The solution of the linear system is the point $(2|2)$.

Solution by substitution method

You may equivalently write equation $(\mathrm{I})$ as $x_2={\displaystyle \frac{1}{2}}\cdot x_1+1$. Now, plug this expression for $x_2$ into equation $(\mathrm{I}\mathrm{I})$:

Plug $x_1=2$ into one of the two equations and calculate $x_2$.

$x_2={\displaystyle \frac{1}{2}}\cdot x_1+1\Rightarrow x_2={\displaystyle \frac{1}{2}}\cdot 2+1=2$

Result: The solution of the linear system is the point $(2|2)$.

Solution by addition method

$\begin{array}{cccccc}(\mathrm{I})& -x_1& +& 2x_2& =& 2\\ (\mathrm{I}\mathrm{I})& 2x_1& -& x_2& =& 2\end{array}$

Eliminate e.g. the variable $x_1$ by calculating: $2\cdot (\mathrm{I})+(\mathrm{I}\mathrm{I})$

$\Rightarrow 4x_2-x_2=4+2\Rightarrow 3x_2=6\Rightarrow x_2=2$

Plug the calculated value $x_2=2$ into equation $(\mathrm{I}\mathrm{I})$:

$\Rightarrow 2x_1-2=2\Rightarrow 2x_1=4\Rightarrow x_1=2$

Result: The solution of the linear system is the point $(2|2)$.