Exercises: systems with two variables

Solving linear systems by equating coefficients

$\begin{array}{rrcl}\text{I})& 3x+4& =& 2y\\ \text{II})& 4y& =& 2x+10\end{array}$

1. solve both equations for y

Solve both equations for one variable. In this case, $y$ is already singled out, so it is easier to solve for $y$.

$\begin{array}{rrcll}\text{I})& 3x+4& =& 2y& |:2\\ \Rightarrow {\text{I}}^{\prime })& 1.5x+2& =& y& \end{array}$

$\begin{array}{rrcll}\text{II})& 4y& =& 2x+10& |:4\\ \Rightarrow {\text{ II}}^{\prime })& y& =& 0.5x+2.5& \end{array}$

2. Set equal

Equate ${\mathrm{I}}^{\prime }$ and $\mathrm{I}{\mathrm{I}}^{\prime }$ .

$\Rightarrow 1.5x+2=0.5x+2.5$

3. Solve equation for x

4. Plug in x to find y

Plug $x$ into ${\mathrm{I}}^{\prime }$ or $\mathrm{I}{\mathrm{I}}^{\prime }$ .

$\begin{array}{l}y=0.5\cdot 0.5+2.5\\ =0.25+2.5\\ =2.75\end{array}$

Write down the solution set.

$${\displaystyle L=\left\{(0.5;2.75)\right\}}$$

Solving linear systems by equating coefficients

$\begin{array}{lrcl}\text{I})& y-1& =& 2x+3\\ \text{II})& 2y-2& =& 5x-1\end{array}$

1. Solve both equations for x

Solve both equations for one variable. For example, for $x$.

$\begin{array}{lrcll}\text{I})& y-1& =& 2x+3& |-3\\ {\text{I}}^{\prime })& y-4& =& 2x& |:2\\ {\text{I}}^{\prime \prime })& 0.5y-2& =& x& \end{array}$

$\begin{array}{lrcll}\text{II})& 2y-2& =& 5x-1& |+1\\ {\text{II}}^{\prime })& 2y-1& =& 5x& |:5\\ {\text{II}}^{\prime \prime })& 0.4y-0.2& =& x& \end{array}$

2. Set equal

Set equal ${\mathrm{I}}^{\prime \prime }$ and $\mathrm{I}{\mathrm{I}}^{\prime \prime }$ .

$\Rightarrow 0.5y-2=0.4y-0.2$

3. Solve for y

4. Plug in y to find x

plug $y$ into ${\mathrm{I}}^{\prime \prime }$

$0.5\cdot 18-2=x=9-2=7$

Write down the solution set

$L=\{(\mathrm{7,18})\}$

Solving linear systems by equating coefficients

$\phantom{\mathrm{I}}\mathrm{I})2x+3y=4x-5$

$\mathrm{II})3x-2y=2y+8$

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

$$$\begin{array}{lrll}\phantom{\mathrm{I}}\mathrm{I})& 2x+3y& =& 4x-5|-2x\\ \phantom{\mathrm{I}}{\mathrm{I}}^{\prime })& 3y& =& 2x-5|:3\\ \phantom{\mathrm{I}}{\mathrm{I}}^{\prime \prime })& y& =& \frac{2}{3}x-\frac{5}{3}\end{array}$

$\begin{array}{lrcll}\mathrm{II})& 3x-2y& =& 2y+8& |+2y-8\\ {\mathrm{II}}^{\prime })& 3x-8& =& 4y& |:4\\ {\mathrm{II}}^{\prime \prime })& \frac{3}{4}x-2& =& y& \end{array}$

2. Set equal

Set equal ${\mathrm{I}}^{\prime \prime }$ and $\mathrm{I}{\mathrm{I}}^{\prime \prime }$ .

$\begin{array}{lrll}\Rightarrow & \frac{3}{4}x-2& =& \frac{2}{3}x-\frac{5}{3}\end{array}$

3. Solve for that one variable

Solve for $x$.

$\begin{array}{rcll}\frac{3}{4}x-2& =& \frac{2}{3}x-\frac{5}{3}& |-\frac{2}{3}x\\ \frac{3}{4}x-\frac{2}{3}x-2& =& -\frac{5}{3}& |+2\\ \frac{9}{12}x-\frac{8}{12}x& =& -\frac{5}{3}+2& \\ \frac{1}{12}x& =& \frac{1}{3}& |\cdot 12\\ x& =& 4& \end{array}$

4. Plug into one equation to find the other variable

Plug $x$ into $\mathrm{I}{\mathrm{I}}^{\prime }$.

$\begin{array}{rll}\frac{3}{4}\cdot (4)-2& =& y\\ 3-2& =& y\\ y& =& 1\end{array}$

Now, you can write down the solution set again

$L=\{(\mathrm{4,1})\}$