Lesson Using systems of equations to solve standard (typical) problems on shares
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<H2>Using systems of equations to solve standard (typical) problems on shares</H2> <H3>Problem 1</H3>Mary spends $2,860 to buy stock in two companies. She pays $13 a share to one of the companies and $26 a share to the other. If she ends up with a total of 150 shares, how many shares did she buy at $13 a share and how many did she buy at $26 a share? It is a typical and standard word problem on shares. There are different methods to solve such problems. I will show you 3 (three) basic methods in this lesson. <B>Solution 1</B> (the system of 2 equations approach) <pre> Let x be the number of shares at $13 and let y be the number of shares at $26. Then you have these two equations x + y = 150, (1) (counting shares) 13x + 26y = 2860 dollars (2) (counting dollars) These are your basic equations, and as soon as you got this system, the setup is completed. There are different methods of solving such system (Substitution, Elimination, using determinants). I will use the Elimination method here. Divide equation (2) by 13 (both sides, You will get this equation in this equivalent form x + 2y = 220. (3) Now subtract eq(3) from eq(1). x-terms will cancel each other, and you will get an expression for unknown y y = 220-150 = 70. Thus, you found number of shares at $26. It is 70. Hence the number of shares at $13 was 150 - 70 = 80. <B>Check</B>. 80*13 + 26*70 = 2860 dollars. ! Correct. ! <B>Answer</B>. 80 shares at $13 and 70 shares at $26. </pre> <B>Solution 2</B> (one equation approach) <pre> Let x be the number of shares at $13. Then the number of shares at $6 is (15-x), according to the condition. It was payed 13*x dollars for x 13-dollar shares. It was payed 26*(150-x) dollars for (150-x) 26-dollar shares. Summing up these money, you get the "money" equation 13x + 26*(150-x) = 2860. (4) It is your basic equation in the frame of this approach, and as soon as you got this equation, the setup is completed. To solve equation (4), simplify it step by step. 13x + 3900 - 26x = 2860 ====> -13x = 2860 - 3900 = -1040 ====> x = {{{(-1040)/(-13)}}} = 80. Thus you get 80 shares at $13 dollars. Surely, you got the same answer. </pre> <B>Solution 3</B> (Logical analysis) <pre> Let assume for a minute that all shares are at $13. Then their total cost would be 150*13 = 1950 dollars. But it is by 2860 - 1950 = 910 dollars less than the given total cost. Why did we get the difference ? - But of course, because we account 26-dollar shares as 13-dollar shares. So, we should return these 26-dollar shares back. Returning each, we diminish the difference of $910 by 26-13 = 13 dollars. Then it is clear that the number of 26-dollar shares was {{{910/13}}} = 70, exactly as in the solutions 1 and 2 above. Hence, the number of $13 shares was 150 - 70 = 80, and you got the same answer. </pre> -------------- Congratulations ! You are now familiar with 3 methods for shares problems solution. I suggest that algebraic methods will be your basic methods for such problems, and the logical analysis method will allow you to solve the problems MENTALLY without using equations. I will be happy if it will make your horizon wider. -------------- This share problem is very similar to traditional ticket problems. To see similar solved problems on tickets, look into the lessons - <A HREF=https://www.algebra.com/algebra/homework/coordinate/lessons/Using-systems-of-equations-to-solve-problems-on-Tickets.lesson>Using systems of equations to solve problems on tickets</A> - <A HREF=https://www.algebra.com/algebra/homework/coordinate/lessons/Three-methods-for-solving-standard-typical-problem-on-tickets.lesson>Three methods for solving standard (typical) problems on tickets</A> in this site. To see how the logical method works for other similar problems, look into the lessons - <A HREF=http://www.algebra.com/algebra/homework/word/misc/Problem-on-two-wheels-and-three-wheels-bicycles-in-a-sale.lesson>Problem on two-wheel and three-wheel bicycles</A> - <A HREF=http://www.algebra.com/algebra/homework/word/misc/Problem-on-animals-at-a-farm.lesson>Problem on animals at a farm</A> - <A HREF=http://www.algebra.com/algebra/homework/word/misc/Problem-on-tablets-in-containers.lesson>Problem on pills in containers</A> - <A HREF=http://www.algebra.com/algebra/homework/word/misc/What-type-of-problems-are-these.lesson>What type of problems are these?</A> in this site. 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