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Created with Fabric.js 1.4.5 VHEMES vhemes = visual + themesstart by dragging a predesigned vheme onto the canvas OBJECTS drag and drop icons,shapes, text or upload your own from our extensive library of artwork your artboard start from scratch[clears the canvas] double click on textto edit or change TEXT Nola was selling tickets at the high school dance. At the end of the evening, she picked up the cash box and noticed a dollar lying on the floor next to it. She said, I wonder whether the dollar belongs inside the cash box or not.The price of tickets for the dance was 1 ticket for $5 (for individuals) or 2 tickets for $8 (for couples). She looked inside the cash box and found $200 and ticket stubs for the 47 students in attendance. Does the dollar belong inside the cash box or not? double click to change this title text! double click to change this header text! double click to changethis text! Drag a cornerto scale proportionally. Let s be the number of tickets sold to individuals and c be the number of tickets sold to couples. We know that 47 tickets were sold so far, so we have s+c=47. Since each individual's ticket is $5, the total amount of money made by selling tickets to individuals is 5s. Similarly, since each ticket sold to couples is $4, the total amount of money made by selling tickets to couples is 4c. The cash box contains $200 total, so we have 5s+4c=200. Thus, we can represent the situation with a system of equations:s+c5s+4c=47=200Does this system of two equations in two unknowns have a positive integer solution? Furthermore, c has to be even, since tickets sold to couples were only sold in sets of 2.Solving the first equation for s we haves=47−c.Substituting for s in the second equation we obtain5(47−c)+4c=200.Distributing and collecting like terms we have235−c=200.Therefore, c=35 and s=12. This means that without the found extra dollar in the cash box, 35 tickets were sold to couples and 12 tickets were sold to individuals. This is not possible, since couples tickets were only sold in sets of 2.First indications are that the $1 on the floor belongs in the cash box. To check this, we change the system of equations so that it reflects the extra $1 in the cash box:s+c5s+4c=47=201The solution to the changed system of equations is s=13 and c=34, so 13 tickets were sold to individuals and 34 tickets were sold to couples. This combination of tickets is indeed possible.Note that it is not really necessary to solve the second system of equations in the standard way. We could just have argued that exchanging one couples' ticket for an individual's ticket would increase the money in the cash box from 200 to 201 and it would result in an even number of couples tickets sold. Alignment: A-CED.AHSAAlgebraDomainHSA-CED: Creating EquationsClusterCreate equations that describe numbers or relationships.Alignment: A-REI.C.6HSAAlgebraDomainHSA-REI: Reasoning with Equations and InequalitiesClusterSolve systems of equations.StandardSolve systems of linear equations exactly and approximately (e.g., with graphs), focusing on pairs of linear equations in two variables.
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