Solution:
Given:
[tex]\begin{gathered} 4x-y=-38 \\ x+y=3 \end{gathered}[/tex]
Solving the system simultaneously by substitution method;
[tex]\begin{gathered} 4x-y=-38\ldots\ldots\ldots\ldots\ldots(1) \\ x+y=3\ldots\ldots\ldots\ldots\ldots\ldots.(2) \\ \\ \text{making y the subject of the formula from equation (2);} \\ x+y=3 \\ y=3-x\ldots\ldots\ldots\text{.}\mathrm{}(3) \\ \\ \text{Substituting equation (3) into equation (1);} \\ 4x-y=-38 \\ 4x-(3-x)=-38 \\ 4x-3+x=-38 \\ 4x+x=-38+3 \\ 5x=-35 \\ \text{Dividing both sides by 5;} \\ x=-\frac{35}{5} \\ x=-7 \\ \\ \text{Substituting the value of x into equation (3) to get y;} \\ y=3-x \\ y=3-(-7) \\ y=3+7 \\ y=10 \end{gathered}[/tex]
Therefore, the solution of the system is;
[tex](x,y)=(-7,10)[/tex]
