I am trying to solve this system of three equations that has 4 variables, but without using a matrix.
$$2x+7y+3z+u=6$$ $$3x+5y+2z+2u=4$$ $$9x+4y+z+7u=2$$
How should I proceed, I've never done a system of equations with 4 variables, and looking online I only come to solutions using the matrix.
On
It doesn't matter how many variable there are, you do the same thing. Eliminate $x$ from every equation but the first. Then eliminate $y$ from the third equation. Solve the third equation of $z$ in terms of $u$, the solve the second for $y$ in terms of $u$ and the first for $x$ in terms of $u$. (It may happen that the third equation becomes $0=0$ so that $u$ and $z$ are both free parameters, or you may get $0=1$ o that there are no solutions at all.)
On
$$2x+7y+3z+u=6$$ $$3x+5y+2z+2u=4$$ $$9x+4y+z+7u=2$$ $$ $$ $$2x+7y+3z+u=6$$ $$-x-9y-4z=-8$$ $$-5x-45y-20z=-40$$ $$ $$ $$2x+7y+3z+u=6$$ $$-x-9y-4z=-8$$ $$x+9y+4z=8$$ $$ $$ $$2x+7y+3z+u=6$$ $$x+9y+4z=8$$ $$ $$ $$-11y-5z+u=-10$$ $$x+9y+4z=8$$
so $y,z$ can be whatever you like. If they are integers, so are $x,u$
You can use substitution. First, you get that $$u=6-2x-7y-3z$$ from the first equation. Now you replace it the second an third and get the new 2 equation system (it has been simplyfied): $$x+9y+4z=8,$$ $$3x+45y+20z=40.$$ Now, do the same on the first of the above equations for $x$ to get that $$x=8-9y-4z$$ and replacce it in the last one to get the equation (it has been simplyfied) $$7y+2z=4.$$ Finally, you now know that $y=\frac{4-2z}{7}$. You replace now this value in $x$ equation to get its value in function of $z$ and finally the same for $u$. Now, it's finished, obtaining infinite solutions determined by the value of $z$.