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

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

${\mathrm{I}}^{\mathrm{\prime }}5y-3(y+1)=1\backslash mathrm\{I\}\text{'}\backslash quad5y-3\backslash left(y+1\backslash right)=1$

Solve for y

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

$L=\left\{\left(x\mid y\right)\right\}=\left\{\left(3\mid 2\right)\right\}L=\backslash left\backslash \{\backslash left(x\backslash ;\backslash left|\backslash ;y\backslash right.\backslash right)\backslash right\backslash \}=\backslash left\backslash \{\backslash left(3\backslash ;\backslash left|\backslash ;2\backslash right.\backslash right)\backslash right\backslash \}$

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

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

${\mathrm{I}}^{\mathrm{\prime }}4x+5\cdot (5x-11)=32\backslash mathrm\{I\}\text{'}\backslash quad\; 4x+5\backslash cdot\; \backslash left(5x-11\backslash right)=32$

Solve for $xx$

Plug $x=3x=3$ into $\mathrm{I}\mathrm{I}\backslash mathrm\{II\}$ and solve for $yy$

$L=\left\{\left(x\mid y\right)\right\}=\left\{\left(3\mid 4\right)\right\}L=\backslash left\backslash \{\backslash left(x\backslash ;\backslash left|\backslash ;y\backslash right.\backslash right)\backslash right\backslash \}=\backslash left\backslash \{\backslash left(3\backslash ;\backslash left|\backslash ;4\backslash right.\backslash right)\backslash right\backslash \}$

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

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

${\mathrm{I}}^{\mathrm{\prime }}15y-4(y+7)=-50\backslash mathrm\{I\}\text{'}\backslash quad15y-4\backslash left(y+7\backslash right)=-50$

Solve for $xx$

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

$x=-2+7x=-2+7$

$x=5x=5$

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

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

Divide $\mathrm{I}\mathrm{I}\backslash mathrm\{II\}$ by 2 and solve for $xx$

$\mathrm{I}\mathrm{I}:2\to {\mathrm{I}\mathrm{I}}^{\mathrm{\prime }}y-5=x\backslash mathrm\{II\}:2\backslash to\backslash mathrm\{II\}\text{'}\backslash quad\; y-5=\; x$

Plug ${\mathrm{I}\mathrm{I}}^{\mathrm{\prime }}\backslash mathrm\{II\}\text{'}$ into $\mathrm{I}\backslash mathrm\{I\}$

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

${\mathrm{I}}^{\mathrm{\prime }}3(y-5)=y+15\backslash mathrm\{I\}\text{'}\backslash quad3\backslash left(y-5\backslash right)=y+15$

Then, solve ${\mathrm{I}}^{\mathrm{\prime }}\backslash mathrm\{I\}\text{'}$ for $yy$

Finally, plug $y=15y=15$ into ${\mathrm{I}\mathrm{I}}^{\mathrm{\prime }}\backslash mathrm\{II\}\text{'}$ and solve for $xx$

$y=15y=15$ plugged into ${\mathrm{I}\mathrm{I}}^{\mathrm{\prime }}\backslash mathrm\{II\}\text{'}$ :

$L=\left\{\left(x\mid y\right)\right\}=\left\{\left(10\mid 15\right)\right\}L=\backslash left\backslash \{\backslash left(x\backslash ;\backslash left|\backslash ;y\backslash right.\backslash right)\backslash right\backslash \}=\backslash left\backslash \{\backslash left(10\backslash ;\backslash left|\backslash ;15\backslash right.\backslash right)\backslash right\backslash \}$

Alternative solution: equating coefficients

Another option is to use the method of equating coefficients, because on the left side of $\mathrm{I}\backslash mathrm\{I\}$ and on the right side of $\mathrm{I}\mathrm{I}\backslash mathrm\{II\}$ there is almost the same term.

Multiply $\mathrm{I}\mathrm{I}\backslash mathrm\{II\}$ by $\frac{3}{2}\backslash frac32$ to get a $3x3x$ on the right hand side.

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

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

$L=\left\{\left(x\mid y\right)\right\}=\left\{\left(10\mid 15\right)\right\}L=\backslash left\backslash \{\backslash left(x\backslash ;\backslash left|\backslash ;y\backslash right.\backslash right)\backslash right\backslash \}=\backslash left\backslash \{\backslash left(10\backslash ;\backslash left|\backslash ;15\backslash right.\backslash right)\backslash right\backslash \}$

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:

Subtract the second equation from the first equation.

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

Plug ${\mathrm{I}}^{\mathrm{\prime }}\backslash mathrm\{I\}\text{'}$ into $\mathrm{I}\mathrm{I}\backslash mathrm\{II\}$ and solve for $yy$.

Plug $y=15y=15$ into ${\mathrm{I}}^{\mathrm{\prime }}\backslash mathrm\{I\}\text{'}$ and solve for $xx$.

$x=-15+25x=-15+25$

$x=10x=10$

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