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- Why Study Math? – Solving Linear Systems by Linear Combinations
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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, let’s move on to a more convenient method called linear combinations. With this method (also known as addition-subtraction), we eliminate one variable by adding an appropriate multiple of one equation. We can then eliminate one variable and solve for the other. Once this is done, we use the second equation to solve for the second variable.

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

** Step 1**: arrange equations with like terms into columns.

** Step 2**: Multiply one or both equations to get coefficients that are opposites of one variable.

** Step 3**: Add the equations from the previous step. Combining like terms eliminates one variable and allows you to solve for another.

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

** Step 5**: Check the solution in each of the original equations.

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

4x + 3y = 16

2x – 3y = 8

Since we see that the coefficients of the y terms are opposites, there is no need to multiply the equations to get this form. So we add the two so the y terms disappear. So we have 6x = 24. Solving for x, we have x = 4. Substituting this value into the first equation, suppose we get 4(4) + 3y = 16 or 16 + 3y = 16 or 3y = 0 or y = 0. Checking , substituting these values into each of the original equations yields a true statement. Therefore, the solution is x = 4 and y = 0, or the point (4, 0) as the point of intersection of these two lines on the coordinate plane.

Let’s see how we can use the method of linear combinations to model a historical problem. According to legend, the famous Greek mathematician Archimedes used the relationship between an object’s weight and volume to determine that fraud had occurred in the manufacture of the golden crown. This was done on the principle of volume displacement. You see, if the crown is pure gold, it should displace the same volume as an equal amount of gold. The following problem also uses the concept of density. 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 next problem.

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

To solve this, we observe that the volume of gold plus the volume of silver must equal the total volume given by the value 46. Since we know the density of both gold and silver, and we know that density times volume equals mass, we have that gold density times gold volume plus silver density times silver volume equals total weight. Let G = volume of gold and S = volume of silver. Now we can translate the problem into mathematics and a linear system.

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

G + S = 46

19G + 10.5S = 714.

Now multiply the first equation by -19 to get the opposite coefficients for G. So it is

-19G + -19S = -874

19G + 10.5S = 714.

Adding the two equations, we get – 8.5S = -160. Dividing both sides by -8.5 gives us 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 krone is 19/46, or 41% with full percent accuracy. . Keep this method in mind the next time someone tries to pawn you pure gold, when the reality is something quite different. Beware of fools gold!

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