How Much Math Do You Need To Be A Mathmatician 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’ve seen how to solve a system of linear equations using the substitution method, let’s move on to a more convenient method known as linear combinations. With this method – also known as addition-subtraction – we eliminate one of the variables by adding an appropriate multiple of one of the equations. We can then eliminate one variable and solve for the other. Once done, we use the other equation to solve for the other variable.

This method can be made algorithmic, so the steps to solve a system using linear combinations are listed here:

Step 1: Sorts equations with like terms into columns.

Step 2: Multiply one or both equations to get opposite coefficients for one of the variables.

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

Step 4: Substitute the value obtained in the previous step in one of the equations and solve for the other variable.

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

To illustrate the algorithm, we solve the following system: 4x + 3y = 16 and 2x – 3y = 8. First we arrange these two equations in columns so that like 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 just add the two so that the y terms disappear. Thus, we have 6x = 24. Solving for x, we have x = 4. Substituting this value into the first equation, say, we get 4(4) + 3y = 16 or 16 + 3y = 16 or 3y = 0, oy = 0. Checking by 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 in a 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 its volume to determine that there was fraud in the manufacture of a golden crown. The way it was done is by using the volume displacement principle. You see, if a crown is pure gold, it should displace the same volume as an equal amount of gold. In the problem that follows, the concept of density is also used. By definition, the density of an object is equal to its mass divided by its volume. Gold has a density of 19 grams per cubic centimeter. Silver has a density of 10.5 grams per cubic centimeter. We will use these facts in the problem that follows.

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

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

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

G + S = 46

19G + 10.5S = 714.

Now we multiply the first equation by -19 to get opposite coefficients for 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. So the volume of silver is 19 cubic centimeters, and the percentage of silver in the crown is 19/46 or 41 % to the nearest whole percentage. . Remember this method the next time someone tries to pawn you pure gold, when in reality the reality is something very different. Watch out for the fool’s gold!

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