A furniture shop refinishes tables. Employees use two methods to refinish tables. Method I takes 0.5 hours and the material costs $8 . Method II takes 2.0 hours, and the material costs $6 . Next week, they plan to spend 94 hours in labor and $724 in material for refinishing tables. How many tables should they plan to refinish with each method? (Round to two decimal places if necessary.)
x = number of tables refinished using method I
y = number of tables refinished using method II Labor hours 0.5x + 2y = 94 Cost of materials 8x + 6y = 724 Solve the system: 0.5x + 2y = 94 8x + 6y = 724 0.5x + 2y = 94 {first equation} 0.5x = 2y + 94 {subtracted 2y from each side} x = 4y + 188 {multiplied each side by 2} Substitute 4y + 188, in for x, into second equation. 8x + 6y = 724 {second equation} 8(4y + 188) + 6y = 724 {substituted} 32y + 1504 + 6y = 724 {used distributive property} 26y + 1504 = 724 {combined like terms} 26y = 780 {subtracted 1504 from each side} y = 30 {divided each side by 26} Substitute 30, in for y, into x = 4y + 188. x = 4y + 188 x = 4(30) + 188 {substituted} x = 120 + 188 {multiplied} x = 68 {added} x = 68 and y = 30 They should plan on finishing: 68 tables using method I and 30 tables using method II Ask Algebra House
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