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0.10 Simultaneous equations

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What are the processor rates for processor A and processor B?

Suppose that we define the processing rates (measured in computations/second) for processors A and B to be represented by the variables x and y respectively. With this definition of variables, the number of computations that processor A can perform in 3 seconds is the product (3 x ). Likewise the number of computations that processor B can perform in 2 seconds is the product (2 y ). The total number of computations that must be performed to complete the first algorithm is 17,000,000. Therefore, we obtain the first of two simultaneous equations

3x + 2y = 17 × 10 6 size 12{3x+2y="17" times "10" rSup { size 8{6} } } {}

Let us now consider the implementation of the second algorithm. The number of computations that processor A can make in 2 seconds is equal to the product (2 x ). Similarly, the number of computations that processor B can make in 3 seconds is the product (3 y ). We obtain the second simultaneous equation as

2x + 3y = 15 . 5 × 10 6 size 12{2x+3y="15" "." 5 times "10" rSup { size 8{6} } } {}

Let us use the first equation to solve for x in terms of y

3x = − 2y + 17 × 10 6 size 12{3x= - 2y+"17" times "10" rSup { size 8{6} } } {}

which results in the expression

x = 17 × 10 6 − 2y 3 size 12{x= { {"17" times "10" rSup { size 8{6} } - 2y} over {3} } } {}

Next, we substitute this expression for x into equation ()

2 17 × 10 6 − 2y 3 + 3y = 15 . 5 × 10 6 size 12{2 left ( { {"17" times "10" rSup { size 8{6} } - 2y} over {3} } right )+3y="15" "." 5 times "10" rSup { size 8{6} } } {}

This equation can be simplified as follows

17 × 10 6 − 2y + 3 2 ( 3y ) = 3 2 15 . 5 × 10 6 size 12{"17" times "10" rSup { size 8{6} } - 2y+ { {3} over {2} } \( 3y \) = { {3} over {2} } left ("15" "." 5 times "10" rSup { size 8{6} } right )} {}
17 × 10 6 − 2y + 4 . 5y = 23 . 25 × 10 6 size 12{"17" times "10" rSup { size 8{6} } - 2y+4 "." 5y="23" "." "25" times "10" rSup { size 8{6} } } {}
2 . 5y = 6 . 25 × 10 6 size 12{2 "." 5y=6 "." "25" times "10" rSup { size 8{6} } } {}

This leads to the solution

y = 2 . 5 × 10 6 computations / s size 12{y=2 "." 5 times "10" rSup { size 8{6} } ~ ital "computations"/s} {}

We now may solve for x through the use of equation ()

x = 17 × 10 6 − 2 ( 2 . 5 × 10 6 ) 3 size 12{x= { {"17" times "10" rSup { size 8{6} } - 2` \( 2 "." 5 times "10" rSup { size 8{6} } \) } over {3} } } {}

which lead to the result

x = 12 × 10 6 3 = 4 × 10 6 size 12{x= { {"12" times "10" rSup { size 8{6} } } over {3} } =4 times "10" rSup { size 8{6} } } {}

Thus we conclude that processor A can perform 4 million computations per second while processor B can perform 2.5 million computations per second.

Alloy compositions

Metallurgical engineering involves the study of the physical and chemical behavior of metallic elements and their mixtures, which are called alloys. It is also involves the technology of metals. That is, metallurgical engineering encompasses the way in which the science of metals is applied to produce compounds for practical use.

The following example presents how simultaneous equations can be applied in the study of alloys.

Example 4: Let us consider two alloys comprised of different percentages of copper and zinco. Suppose that the first alloy is made up of a mixture of 70% copper and 30% zinc. The second alloy is comprised of 40% copper and 60% zinc.

By melting the two alloys and re-combining their constituents into one 300 gram casting, the resulting alloy is 60% copper and 40% zinc.

Find the amount of copper and zinc in each of the original alloys.

We begin our solution by establishing the variables A and B as being equal to the mass (in grams) of the first alloy and the second alloy respectively. We know that the final casting has a mass of 300 grams, so our first simultaneous equation is

A + B = 300 size 12{A+B="300"} {}

Because 60% of the final casting is copper, the amount of copper in the final casting is the product 0.60 (300 grams) or 180 grams.

We know that 70% of the first alloy is copper and that 40% of the second alloy is copper. Thus we may write the second simultaneous equation as

0 . 7 A + 0 . 4 B = 180 size 12{0 "." 7`A+0 "." 4`B="180"} {}

The variable A can be expressed as (300 – B) and substituted into the equation

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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Source:  OpenStax, Math 1508 (laboratory) engineering applications of precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11337/1.3
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