Lesson Using systems of equations to solve problems on investment
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<H3>Using systems of equations to solve problems on investment</H3> <H3>Problem 1</H3>Madison invested a total of $30,000 at two different banks. At one bank she earned 3.5% on her investment and at the other bank she earned 4.5%. If her total earning per year is $1,320 then how much did she invest at each bank? <B>Solution</B> <pre> Let x be the amount of money Madison invested at 3.5%, and let y be the amount of money Madison invested at 4.5% At the 3.5% account Madison earned 0.035x dollars per year (0.035 = 3.5% !) At the 4.5% account Madison earned 0.045x dollars per year (0.045 = 4.5% !) You are given that 0.035x + 0.045y = 1320 dollars. You also know that x + y = 30000 dollars. So, you have this system of two equations x + y = 30000, (1) 0.035x + 0.045y = 1320. (2) To solve it, express x = 30000 - y from (1) and sibstitute it into equation (2). You will get 0.035*(30000-y) + 0.045y = 1320, or 1050 - 0.035y + 0.045y = 1320, 0.01y = 1320 - 1050 ====> 0.01y = 270 ====> y = {{{270/0.01}}} = 27000. Thus you found that Madison invested $27000 at 4.5%. Then the investment at 3.5% is 30000 - 27000 = 3000 dollars. <U>Check</U>. 0.035*3000 + 0.045*27000 = 1320. ! Correct !. <U>Answer</U>. $3000 was invested at 3.5% and $27000 was invested at 4.5%. </pre> The method I used to solve the system of equations in this problem was the <U>Substitution method</U>. <H3>Problem 2</H3>Sue has $80,000 to invest in a savings account, which pays 7% and a certificate of deposit which pays 8.4%. Sue would like to receive $6,300 as interest income. How much should she invest in each? <B>Solution</B> <pre> From the condition, you have these two equations A + B = 80000, (1) 0.07A + 0.084B = 6300. (2) To solve this system, multiply the second equation by 1000 (both sides). You will get the two equations in this form A + B = 80000, (3) 70A + 84B = 6300000. (4) Next, multiply equation (3) by 70 (both sides). You will get the two equations in this form 70A + 70B = 5600000, (5) 70A + 84B = 6300000. (6) Subtract from (5) from equation (6) to eliminate A: 70A + 84B - 70A - 70B = 6300000-5600000, 14B = 700000 ====> B = {{{700000/14}}} = 50000. Thus the amount of money for certificate of deposit is $50000. Then the amount of money for saving account is 80000 - 50000 = 30000 dollars. <U>Check</U>. 0.07*30000 + 0.084*50000 = 6300 dollars. ! Correct ! <U>Answer</U>. $30000 for the saving account at 7% and $50000 for the certificate of deposit at 8.4%. </pre> The method I used to solve the system of equations in this problem was the <U>Elimination method</U>. <H3>Problem 3</H3>$15,074 is invested part at 14% and the rest at 5%. If the interest earned from the amount invested at 14% exceeds the interest earned from the amount invested at 5% by $1694.07, how much is invested at each rate? <B>Solution</B> <pre> From the condition, you have THIS system of 2 equations in 2 unknowns x + y = 15074 (1) (counting investments) 0.14x - 0.05y = 1694.07 (2) (the difference between earnings) To make your calculations easier, multiply eq(2) by 100 (both sides). The modified system is x + y = 15074 (3) 14x - 5y = 169407 (4) Now multiply eq(3) by 5. The modified system takes the form 5x + 5y = 5*15074 (5) 14x - 5y = 169407 (4) Next add the equations (5) and (6) (both sides). The terms with 5y will cancel each other, and you will get a single equation for x: 19x = 5*15074 + 169407 = 244777. x = {{{244777/19}}} = 12883. <U>Answer</U>. It was invested $12883 at %14. The rest 15074 - 12883 = 2191 was invested at 5%. <U>Check</U>. 0.14*12883 - 0.05*2191 = 1694.07. ! Correct ! </pre> <H3>Problem 4</H3>Jack inherited 250000 pesos and invested money in SM, Meralco, and Manila Water. After a year, he got a small return of 16200 pesos from the three investments. SM returned 6%, Meralco returned 7%, and Manila Water returned 8%. There was 60000 more pesos invested in Meralco than in Manila Water. How much did he invest in SM, Meralco, and Manila Water? <B>Solution</B> <pre> I will solve the problem by reducing to two equations in two unknown. Let S be the amount invested in SM, and let W be the amount invested in Manila Water. Then investment in Meralco is (W + 60000), according to the condition. First equation is for the total investment S + (W + 60000) + W = 250000, or, equivalently, S + 2W = 250000 - 60000 = 190000. (1) Second equation is for earning 0.06*S + 0.07*(W + 60000) + 0.08*W = 16200. Multiplying by 100 and simplifying, you can get this form 5*S + 15W = 1620000 - 420000, or, equivalently, 5S + 15W = 1200000. (2) Now, to make calculations easier, I will take off 3 zeroes in the right side of equations (1) and (2). Later I simply will multiply the solution by 1000. Thus I get these two equations S + 2W = 190 (3) ( instead of (1) ) and 6S + 15W = 1200 (4) ( instead of (2) ) Now the solution is an EASY EXERCISE. Multiply eq(3) by 6 (both sides) and then subtract from eq(4). You will get 15W - 12W = 1200 - 6*190 = 60, or 3W = 60 ====> W = {{{60/3}}} = 20. Then from (3) S = 190 - 2W = 190 - 2*20 = 150. Thus it was invested 20000 in Manila Water; 150000 in SM; and 20000+60000 = 80000 in Meralco. <U>Check</U>. 20000 + 150000 + 80000 = 250000 (total). ! Correct ! 0.06*150000 + 0.07*80000 + 0.08*20000 = 16200 (earning) ! Correct ! </pre> <H3>Problem 5</H3>Joseph invested a total of $10,000 in two accounts. After a year, one account lost 7.8%, while the other account gained 6.7%. In total, Joseph lost $562.50. How much money did Joseph invest in each account? <B>Solution</B> <pre> Let x be the amount which was invested and gained 6.7%. Let y be the amount which was invested and lost 7.8%. Then you can write two equations. First equation is OBVIOUS: it is x + y = 10000 dollars. (1) It says that the total of the two accounts is $10,000, exactly as the problem states. Second equation is about the net loss. It is also obvious: 0.078y - 0.067x = 562.50 dollars. (2) Indeed, 0.078y is the lost amount, while 0.067x is the gained amount. We take 0.078y with the sign "+", since in equation (2) we consider the lost amount as positive. You may consider the loss as negative value, but then you should be consistent and write all the terms of equation (2) with opposite signs. So, you either write the "loss" equation in the form (2), or you change all the terms in this equation to opposite. It is your choice to write (2) as is or with opposite signs - the meaning of the equation (2) will be the same. So, for what follows, let's write these two equations as x + y = 10000 (1) -0.067x + 0.078y = 562.50 (2) So, we consider the loss in (2) with the sign "+". As soon as you wrote these equations, the setup is completed. To solve the system, let's multiply equation (1) by 0.067 (all the terms). Keep equation (2) as is. You will get then 0.067x + 0.067y = 670.00 (1') -0.067x + 0.078y = 562.50 (2') Now add equations (1') and (2'). The terms 0.067x and -0.067x will kill each other, and you will get a single equation with unknown y 0.067y + 0.078y = 670 + 562.50, or 0.145y = 1232.5, which gives y = 1232.5/0.145 = 8500. Then from equation (1), x = 10000 - 8500 = 1500. So, the invested amount to the account which gained 6.7% was 1500 dollars. the invested amount to the account which lost 7.8% was 8500 dollars. <U>CHECK</U>. -0.067*1500 + 0.078*8500 = (I use my calculator) = 562.5 dollars, the net loss. ! correct ! </pre> My other lessons in this site on solving systems of two linear equations in two unknowns (Algebra-I curriculum) are - <A HREF = http://www.algebra.com/algebra/homework/coordinate/lessons/Solution-of-the-lin-system-of-two-eqns-by-the-Subst-method.lesson>Solution of the linear system of two equations in two unknowns by the Substitution method</A> - <A HREF = http://www.algebra.com/algebra/homework/coordinate/lessons/Solution-of-the-lin-syst-of-two-eqns-with-two-unknowns-Elimination-method.lesson>Solution of the linear system of two equations in two unknowns by the Elimination method</A> - <A HREF =http://www.algebra.com/algebra/homework/coordinate/lessons/Solution-of-the-lin-syst-of-two-eqns-with-two-unknowns-using-det.lesson>Solution of the linear system of two equations in two unknowns using determinant</A> - 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