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## Why Study Math? – Solving Linear Systems by Linear Combinations

Now that we have seen how to solve a system of linear equations using the substitution method, we move on to a more convenient method called linear combination. With this method—also called addition-subtraction—we eliminate one variable by adding the appropriate multiples of one of the equations. Then we can eliminate one variable and solve for the other. Once done, we use the next equation to solve for the next variable.

This method can be made algorithmic and hence the steps to be followed to solve the system by linear combinations are listed here:

** Step 1**: Match the equations with like terms in the columns.

** Step 2**: Multiply one or both equations to obtain the opposite coefficients of one of the variables.

** Step 3**: Add the equations from the previous step. Combining terms like this will eliminate one variable and allow you to solve for the other.

** Step 4**: Substitute the value obtained in the previous step into an equation and solve for the other variable.

** Step 5**: Check the solution to each original equation.

To illustrate the algorithm, let’s solve the following system: 4x + 3y = 16 and 2x – 3y = 8. We first arrange these two equations into columns so that the variables line up like so. Thus we have

4x + 3y = 16

2x – 3y = 8

Since we have seen that the coefficients of the y-terms are opposite, it is not necessary to multiply the equations to obtain this form. Thus we add the two so that the y-terms vanish. Thus we have 6x = 24. Solving for x, we have x = 4. Substituting this value into the first equation, we get, say, 4(4) + 3y = 16 or 16 + 3y = 16 or 3y = 0, or y = 0. Substituting these values into each of the original equations yields a true statement. Thus the solution is x = 4 and y = 0 or the point (4, 0) as the intersection of these two lines on a coordinate plane.

Let’s see how we can use the method of linear combination to model historical problems. According to legend, the famous Greek mathematician Archimedes used the relationship between an object’s weight and its volume to detect fraud in the manufacture of the golden crown. This method was done using the principle of volume displacement. You see, if the crown is pure gold, it must displace an equal amount of gold. In the following problem, the concept of density is also used. By definition the density of an object is equal to its mass divided by its volume. The density of gold is 19 grams per cubic centimeter. The density of silver is 10.5 grams per cubic centimeter. We will use these facts in the following problem.

** the problem**: Suppose a gold crown suspected of containing some silver, weighs 714 grams and has a volume of 46 cubic centimeters. What percentage of the crown was silver?

To solve this we see that the volume of gold plus the volume of silver must equal a total volume of 46. Since we know the density of both gold and silver, and since we know that density properties equal volume, we have gold. Density times the amount of gold and the density of silver times the amount of silver equals the total weight. We let G = amount of gold and S = amount of silver. Now we can translate the problem into mathematics and linear systems.

We have G + S = 46 and 19G + 10.5S = 714. Putting these equations in columns, we have

G + S = 46

19G + 10.5S = 714.

We now multiply the first equation by -19 to get the inverse coefficient of G. Thus we have

-19G + -19S = -874

19G + 10.5S = 714.

Adding the two equations, we have – 8.5S = -160. Dividing both sides by -8.5, we have S = 18.8, which we can round to 19. Thus the volume of silver is 19 cubic centimeters, and the percentage of silver in the crown is 19/46 or 41% of the nearest absolute percentage. . Remember this method the next time someone tries to pawn you off on pure gold, when in fact the reality is something different. Beware of fools!

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