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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 practical 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 and here are the steps to follow to solve a system by linear combinations:
Step 1: Arrange equations with like terms in columns.
2nd step: Multiply one or both equations so as to obtain opposite coefficients for one of the variables.
Step 3: Add the equations from the previous step. Combining similar terms will eliminate one of the variables and allow you to solve 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 in each of the original equations.
To illustrate the algorithm, let’s solve the following system: 4x + 3y = 16 and 2x – 3y = 8. We first organize these two equations into columns so that identical variables line up. So we have
4x + 3a = 16
2x – 3a = 8
Since we see that the coefficients of the terms are opposite there, it is not necessary to multiply the equations to obtain this form. Thus, we simply add the two so that the terms disappear there. So, we have 6x = 24. Solving x, we have x = 4. Substituting this value into the first equation, say, gives 4(4) + 3y = 16 or 16 + 3y = 16 or 3y = 0, or y = 0. Verification 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 to 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 making of a gold crown. This was done using the principle of volume displacement. You see, if a crown is pure gold, it should displace the same volume as 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. 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 following problem.
Problem: Suppose a gold crown suspected of containing silver, weighing 714 grams and having a volume of 46 cubic centimeters. What percentage of the crown was silver?
In order to solve this problem, we observe that the volume of gold plus the volume of silver must equal the given total volume of 46. Since we know the densities of gold and silver, and we know that density multiplied by volume equals mass, we have that gold density times gold volume plus silver density multiplied by silver volume equals total weight. Let G = the volume of gold and S = the volume of silver. We can now translate the problem into mathematics and into a linear system.
We have G + S = 46 and 19G + 10.5S = 714. Putting these equations in columns, we have
We now multiply the first equation by -19 to get opposite coefficients for G. So we have
-19G + -19S = -874
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 pure gold on you, when in reality the reality is something quite different. Beware of Fool’s Gold!
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