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Math Problem example - Algebra
Pages 6 (1506 words)
Complete Final Project The system of linear equations may be applied to mixture problems in which items of various types or grades are sold at distinct values. This is particularly useful for starting entrepreneurs who consider establishing business on a small-scale with simple calculations carried out…
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Solution: Let x = the quantity of Arabica(M) (in kg) and y = quantity of Robusta(H) (in kg) Based on the given information, equations may be set up as: 10.50(x) + 9.25(y) = (9.74)(2500) ---? equation (1) x + y = 2500 ---? equation (2) Graphing each equation on the same xy-plane: By applying substitution method (equation (2) into equation (1)): 10.50(x) + 9.25*(2500 – x) = (9.74)(2500) 10.50(x) + 23125 - 9.25(x) = 24350 1.25(x) = 1225 Then dividing each side by 0.8, x = 980 kgs Arabica(M) And 980 + y = 2500 ---? y = 1520 kgs Robusta(H) Thus, the point of intersection is at (980, 1520) and this pertains to the quantities each of the Arabica(M) and the Robusta(H) that must be present in the bean-mixture so that Matthew is able to satisfy the condition of selling a total of 2500-kg mixture where each kilogram is sold for $9.74. Summary of Learning Besides its flexible range of applications, I have learned that there can be alternative methods in solving a system of equations once each equation has been properly set up with correct algebraic expressions in which variables are made to represent unknown amounts of objects either count or non-count by nature. ...
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