Consider this example: Being that the first equation has the simplest coefficients (1, -1, and 1, for appears in the other two equations: Reducing these two equations to their simplest forms: So far, our efforts have reduced the system from three variables in three equations to two variables in two equations.
Now, we can apply the substitution technique again to the two equations of 2, 4, and 12, respectively, that satisfy all three equations.
What about an example where things aren’t so simple?
Consider the following equation set: We could add these two equations together—this being a completely valid algebraic operation—but it would not profit us in the goal of obtaining values for : The resulting equation still contains two unknown variables, just like the original equations do, and so we’re no further along in obtaining a solution.
The object is to manipulate the two equations so that, when combined, either the x term or the y term is eliminated (hence the name) − the resulting equation with just one unknown can then be solved: Here we will manipulate one of the equations so that when it is combined with the other equation either the x or y terms will drop out.
In this example the x term will drop out giving a solution for y.
The step-by-step example shows how to group like terms and then add or subtract to remove one of the unknowns, to leave one unknown to be solved.
It involves what it says − substitution − using one of the equations to get an expression of the form ‘y = …’ or ‘x = …’ and substituting this into the other equation.
If the total income earned be 0, find the amount invested in A and B separately A trader gains one third of the cost price as profit on a product and one fourth of the cost price as profit on other product. The sum of the cost prices of two products is 0.
If an equation has two unknowns, such as 2y x = 20, it cannot have unique solutions.