Exercises: systems with two variables

In this case, the substitution method is useful, since the second equation is already solved for one variable.

$\begin{array}{cccccc}\mathrm{I}& 5y& -& 3x& =& 1\\ \mathrm{II}& x& =& y& +& 1\end{array}$

Plug equation $\mathrm{II}$ into $\mathrm{I}$

${\mathrm{I}}^{\prime }5y-3(y+1)=1$

Solve for y

$\begin{array}{rccc}5y-3y-3& =& 1& \\ 2y-3& =& 1& |+3\\ 2y& =& 4& |:2\\ y& =& 2& \end{array}$

Plug $y=2$ into $\mathrm{II}$ and solve for x

$\begin{array}{rccc}5\cdot 2-3x& =& 1& |-10\\ -3x& =& -9& |:(-3)\\ x& =& 3& \end{array}$

$L=\left\{\left(x|y\right)\right\}=\left\{\left(3|2\right)\right\}$

In this case, the substitution method is useful, since the second equation is already solved for one variable.

$\begin{array}{cccccc}\mathrm{I}& 4x& +& 5y& =& 32\\ \mathrm{II}& y& =& 5x& -& 11\end{array}$

Plug equation $\mathrm{II}$ into $\mathrm{I}$

${\mathrm{I}}^{\prime }4x+5\cdot (5x-11)=32$

Solve for $x$

$\begin{array}{rccc}4x+25x-55& =& 32& \\ 29x-55& =& 32& |+55\\ 29x& =& 87& & |:29\\ x& =& 3& \end{array}$

Plug $x=3$ into $\mathrm{II}$ and solve for $y$

$\begin{array}{rcl}y& =& 5\cdot 3-11\\ y& =& 4\end{array}$

$L=\left\{\left(x|y\right)\right\}=\left\{\left(3|4\right)\right\}$

In this case, the substitution method is useful, since the second equation is already solved for one variable.

$\begin{array}{cccccc}\mathrm{I}& 15y& -& 4x& =& -50\\ \mathrm{II}& x& =& y& +& 7\end{array}$

Plug equation $\mathrm{II}$ into $\mathrm{I}$

${\mathrm{I}}^{\prime }15y-4(y+7)=-50$

Solve for $x$

$\begin{array}{rcll}15y-4y-28& =& -50& \\ 11y-28& =& -50& |+28\\ 11y& =& -22& |:11\\ y& =& -2& \end{array}$

Plug $y=-2$ into $\mathrm{II}$ and solve for x

$x=-2+7$

$x=5$

$L=\left\{\left(x|y\right)\right\}=\left\{\left(5|-2\right)\right\}$

In this case, the substitution method is useful, since the second equation is already solved for one variable.

$\begin{array}{cccccc}\mathrm{I}& 3x& =& y& +& 15\\ \mathrm{II}& 2y& -& 10& =& 2x\end{array}$

Divide $\mathrm{II}$ by 2 and solve for $x$

$\mathrm{II}:2\to {\mathrm{II}}^{\prime }y-5=x$

Plug ${\mathrm{II}}^{\prime }$ into $\mathrm{I}$

${\mathrm{II}}^{\prime }$ plugged into $\mathrm{I}$ :

${\mathrm{I}}^{\prime }3(y-5)=y+15$

Then, solve ${\mathrm{I}}^{\prime }$ for $y$

$\begin{array}{rcll}3y-15& =& y+15& |-y;+15\\ 2y& =& 30& |:2\\ y& =& 15& \end{array}$

Finally, plug $y=15$ into ${\mathrm{II}}^{\prime }$ and solve for $x$

$y=15$ plugged into ${\mathrm{II}}^{\prime }$ :

$\begin{array}{rcl}15-5& =& x\\ 10& =& x\end{array}$

$L=\left\{\left(x|y\right)\right\}=\left\{\left(10|15\right)\right\}$

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.

$\begin{array}{cccccc}\mathrm{I}& 3x& =& y& +& 15\\ \mathrm{II}& 2y& -& 10& =& 2x\end{array}$

Multiply $\mathrm{II}$ by $\frac{3}{2}$ to get a $3x$ on the right hand side.

$\begin{array}{cccccc}\mathrm{I}& 3x& =& y& +& 15\\ {\mathrm{II}}^{\prime }& 3y& -& 15& =& 3x\end{array}$

Set the right side of $\mathrm{I}$ equal to the left side of ${\mathrm{II}}^{\prime }$ and solve for $x$.

$\begin{array}{rcll}3x-15& =& x+5& |-x;+15\\ 2x& =& 20& |:2\\ x& =& 10& \end{array}$

Plug $x=10$ into $\mathrm{I}$ (or $\mathrm{II}$) and solve for $y$.

$\begin{array}{rcll}3\cdot 10& =& y+15& \\ 30& =& y+15& |-15\\ 15& =& y& \end{array}$

$L=\left\{\left(x|y\right)\right\}=\left\{\left(10|15\right)\right\}$

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:

$\begin{array}{cccccc}\mathrm{I}& 3x& =& y& +& 15\\ \mathrm{II}& 2x& =& 2y& -& 10\end{array}$

Subtract the second equation from the first equation.

$\begin{array}{cccccc}{\mathrm{I}}^{\prime }& x& =& -y& +& 25\\ \mathrm{II}& 2x& =& 2y& -& 10\end{array}$

Since the first equation is now solved for $x$, we can use the substitution method again.

Plug ${\mathrm{I}}^{\prime }$ into $\mathrm{II}$ and solve for $y$.

$\begin{array}{crcll}{\mathrm{II}}^{\prime }& 2\cdot (-y+25)& =& 2y-10& \\ & -2y+50& =& 2y-10& |-2y|-50\\ & -4y& =& -60& |:(-4)\\ & y& =& 15& \end{array}$

Plug $y=15$ into ${\mathrm{I}}^{\prime }$ and solve for $x$.

$x=-15+25$

$x=10$

$L=\{(x|y)\}=\{(10|15)\}$