How do you solve the following linear system? \[3s+4=-4t\],\[7s+6t+11=0\]?

Hint: From the question we have been given two linear equations and we are asked to find the solution. For solving these linear equations we will make one variable subject by taking one equation and after making the subject we will substitute that variable value in another given equation and solve the question.

Complete step-by-step solution:
Firstly, as we said before we will take one given equation and make the variable \[s\] as subject and write it in terms of another variable. So, the equation will be reduced as follows.
We will take the equation \[3s+4=-4t\] and make the variable \[s\] as a subject.
\[\Rightarrow 3s+4=-4t\]
\[\Rightarrow 3s=-4t-4\]
\[\Rightarrow s=\dfrac{-4t-4}{3}\]
Here we got the value of \[s\] as follows. Now we will substitute this value in the other given equation.
So, we get the equation reduced as follows.
\[\Rightarrow 7s+6t+11=0\]
\[\Rightarrow 7\left( \dfrac{-4t-4}{3} \right)+6t+11=0\]
Now, we will multiple the whole equation three on both sides. So, we get the equation reduced as follows.
\[\Rightarrow 7\left( -4t-4 \right)+3\left( 6t+11 \right)=0\]
Now, we will expand the brackets in the equation and simplify the equation using basic mathematical operations like addition and subtraction. So, we get the equation reduced as follows.
\[\Rightarrow -28t-28+18t+33=0\]
\[\Rightarrow -10t+5=0\]
\[\Rightarrow -10t=-5\]
\[\Rightarrow t=\dfrac{1}{2}\]
Now, we will substitute this \[\Rightarrow t=\dfrac{1}{2}\] in any one of the given equations in the question and find the value of the other variable.
We will substitute this \[\Rightarrow t=\dfrac{1}{2}\] in the equation \[\Rightarrow 3s+4=-4t\]. So, we get the equation reduced as follows.
\[\Rightarrow 3s+4=-4t\]
\[\Rightarrow 3s+4=-4\left( \dfrac{1}{2} \right)\]
\[\Rightarrow 3s+4=-2\]
\[\Rightarrow 3s=-2-4\]
\[\Rightarrow 3s=-6\]
\[\Rightarrow s=-2\]
Therefore, the solution is \[\Rightarrow s=-2\] and \[\Rightarrow t=\dfrac{1}{2}\].

Note: Students must be very careful in doing the calculations. Students must have good knowledge in substitution methods and basic operations of mathematics like addition and subtraction. We must not do mistakes in calculation like for example after this step \[\Rightarrow 3s+4=-2\] if we write the next step as \[\Rightarrow 3s=-2+4\] instead of \[\Rightarrow 3s=-2-4\] our solution will be wrong