Answer :
We have to solve this system by substitution.
This method let us define a varaible in function of the other (or others, if they are more than two) and then replace it in the other equations by this relation.
The system is:
[tex]\begin{gathered} y-4x=3 \\ 2x-3y=21 \end{gathered}[/tex]1. From the first equation, we can easily define y in function as x:
[tex]\begin{gathered} y-4x=3 \\ y=4x+3 \end{gathered}[/tex]2. Now, we substitute y in the second equation:
[tex]\begin{gathered} 2x-3y=21 \\ 2x-3(4x+3)=21 \end{gathered}[/tex]3. We got an equation that is expressed only in function of x, so we can solve it as:
[tex]\begin{gathered} 2x-3(4x+3)=21 \\ 2x-3\cdot4x-3\cdot3=21 \\ 2x-12x-9=21 \\ -10x=21+9 \\ -10x=30 \\ x=\frac{30}{-10} \\ x=-3 \end{gathered}[/tex]4. If we replace x in the equation for y, we get:
[tex]\begin{gathered} y=4x+3 \\ y=4(-3)+3 \\ y=-12+3 \\ y=-9 \end{gathered}[/tex]5. Then, we can write the answer a s a pair (x,y) = (-3,-9)