Solve the following linear simultaneous equations $$x + y = 16 \qquad\qquad(1)$$ $$x - y = 10 \qquad\qquad(2)$$
Here we have
Solve the following linear simultaneous equations $$\quad y - x = 4 \qquad\qquad(1)$$ $$2y - 2x = 2 \qquad\qquad(2)$$
Here we have two unknowns and two equations.
However, the equations are not consistent. If \(y-x=4\), to be consistent within the system, \(2y-2x\) must be \(8\), not \(2\). Thus, these two equations are inconsistent, and a solution does not exist.
We plot these two equations in Figure 3. The lines are parallel, therefore, never cross. Hence, this system of linear simultaneous equations does not have a solution.
Solve the following linear simultaneous equations $$\quad x + y = 16 \qquad\qquad(1)$$ $$2x + 2y = 32 \qquad\qquad(2)$$
In this system, we have two unknowns and two equations (at least, it looks like that at the beginning).
They are also consistent: \(x+y=16\) and \(2x+2y =32\).
However, equation (2) is a linear transformation of equation (1) as equation (2) can be obtained by multiplying equation (1) by the number \(2\). Lines representing these two equations coincide in the Cartesian plane. Therefore, in this system, there exists an infinite number of \((x,y)\) pairs which satisfy these equations.