How do you solve the simultaneous equations \[3x-4y=11\] and \[5x+6y=12?\]

Hint: Simultaneous equations can be solved in one of three ways: by elimination, substitution, or graphing. Elimination is the simplest way to solve this problem. Either x or y must be eliminated by multiplying the equations until both x and y have the same coefficient.

Complete step by step solution:
The first step is to choose a variable to eliminate, and then calculate the least common multiple (LCM) of its coefficients.
Let's say that you want to eliminate \[x\] and solve for \[y~\] first. The two coefficients of \[x\] are 3 and 5, which means that they're LCM will be equal to 15.
So, multiply the first equation by 5 and the second equation by \[-3\]to get,
\[5\cdot (3x-4y) =5(11)\]
\[15x-20y=55\]
And,
\[\begin{align}
  & (-3)\cdot (5x+6y) =-3(12) \\
 & -15x-18y=-36 \\
\end{align}\]
Add the left side and the right side of these two equations separately to get,
\[\begin{align}
  & 15x-20y-15x-18y=55-36 \\
 & -38y=19\Rightarrow y=\dfrac{-19}{38}=-\dfrac{1}{2} \\
\end{align}\]
Now use this value of \[y~\] in one of the two equations to determine the value of\ [x\].
\[\begin{align}
  & 3x-4(-\dfrac {1} {2})=11 \\
 & 3x+2=11\Rightarrow x=\dfrac {11-2} {3}=3 \\
\end{align}\]
The solutions to this system of equations are,
\[\left\{ \begin{align}
  & x=3 \\
 & y=-\dfrac {1}{2} \\
\end{align} \right\}\]

Note: There may be times when one of the variables is not equal to the other (if you follow the steps carefully). This means there is no solution because the system of equations cannot equal each other. When solving fractional equations, be careful; I would multiply each one by the most common denominator to get whole integers, making the equations easier to solve.