Ok i get it now. We solve the equation which can be solved, the one with only one variable, then we plug that value into the other equation. The goal being to find the value of the other variable.
Yes for this situation, but no generally.
If we have two equations with two unknowns, we first use one equation to solve for one unknown in terms of the other unknown (not a numeric solution). Then we reduce the other equation to one unknown by substituting for the unknown that is known in terms of the other unknown. Then we get numeric solutions for the remaining unknown. And finally, we get a numeric solution for the first unknown. This is abstract so here is an example.
\(\displaystyle x^2 - 2xy + y^2 = 81.\)
\(\displaystyle x + y = 5.\)
Step 1: Solve for one unknown in terms of the other.
You can use either equation to do so. In this case, it is obviously easier to use the second equation.
\(\displaystyle y = 5 - x.\)
Step 2: Reduce the other equation to an equation in one unknown by substitution.
\(\displaystyle x^2 - 2xy + y^2 = 81 \implies x^2 - 2x(5 - x) + (5 - x)^2 = 81 \implies x^2 - 10x + 2x^2 + 25 - 10x + x^2 = 81 \implies 4x^2 - 20x - 56 = 0.\)
Step 3: Solve the reduced equation numerically.
\(\displaystyle 4x^2 - 20x - 56 \implies x^2 - 5x - 14 = 0 \implies (x - 7)(x + 2) = 0 \implies x = 7\ or\ x = - 2.\)
Step 4: Use the other equation to solve for the other unknown numerically.
\(\displaystyle x = 7 \implies 7 + y = 5 \implies y = - 2.\)
\(\displaystyle x = - 2 \implies - 2 + y = 5 \implies y = 7.\)
So the solutions are (7, - 2) and (- 2, 7).