System of two equations is generally given as[br][br]   [math] \large \begin{cases}[br]\textcolor{blue}{a_{11}x_1+a_{12}x_2=b_1}\\[br]\textcolor{blue}{a_{21}x_1+a_{22}x_2=b_2}[br]\end{cases}.[/math][br][br]It has [br][list][*]a unique solution, if lines are not parallel.[/*][*]no solution, if lines are parallel but not equal.[/*][*]infinite number of solutions, if line are exactly the same.[/*][/list][br]The easiest way to solve system of two equations is substituting method. It is not a reasonable method, if the number of variables exceeds 2, as expressions are long and errors are very probable. [br][br]The elimination method of solving a system of linear equations algebraically is the most widely used method.[br][br][color=#0000ff][u]Substituting method:[/u][/color][br] [list][*]Solve one variable from the first equations. [/*][*]Substitute the solution to the unknown in the second equation. Now, there is normal linear equation to be solved.[br][/*][*]Solve the linear equation.[/*][*]Substitute the solution to the first equations in order to find solution for the second variable.[/*][*]Check your solution by substituting bot solution to [u]both original equations[/u] (not only to one equation).[/*][/list]

[color=#0000ff][u]Elimination method:[/u][/color][br] [list][*]Multiply or divide both the linear equations with a non-zero number to get a common opposite coefficient of any one of the variables in both equations.[/*][*]Add both the equations such that the same terms will get eliminated.[/*][*]Solve the linear equation.[/*][*]Substitute this value in any of the given equations to find the value of the other given variable.[/*][*]Check your solution by substituting bot solution to [u]both original equations[/u] (not only to one equation).[/*][/list]

Solve[br][br]  [math] \begin{cases}\begin{align}[br]3x+4y&=-5\\[br]-2x+5y&=-12\end{align}[br]\end{cases}[/math] [br] [br]Multiply equations so that multipliers of one variable are opposite numbers:[br][br][math]\begin{cases}\begin{align}[br]3x+4y&=-5&|\cdot 2\\[br]-2x+5y&=-12&|\cdot 3\end{align}[br]\end{cases}[br]\;\;\Leftrightarrow\;\;[br]\begin{cases}\begin{align}[br]6x+8y&=-10&\\[br]-6x+15y&=-36\end{align}[br]\end{cases}[/math][br] [br]Add equations so, that only one variable is left:[br][br]  [math]\begin{array}{rcll} 23y&=&-46&|:23 \\[br]y&=&-2\end{array}[/math][br] [br]After this, the second variable can be solved from any equation:[br][br][math]\begin{array}{rcll} [br]3x+4\cdot (-2)&=&-5\\[br]3x-8&=&-5&|+8\\[br]3x&=&-5+8\\[br]3x&=&3&|:3\\[br]x&=&1[br]\end{array}[/math][br] [br]The solution is check by substituting to the original equations:[br][br]  [math] -2\cdot 1+5\cdot (-2)=-2-10=-12.[/math][br] [br]As solutions in the second equation gives the same value as given at the right side, solution seems to be fine.[br][br]