A fruit stand has to decide what to charge for their produce. They need $10 for 4 apples and 4 oranges. They also need $12 for 6 apples and 6 oranges. We put this information into a system of linear equations.Can we find a unique price for an apple and an orange?Choose 1 answer:A Yes; they should charge $1.00 for an apple and $1.50 for an orange.B Yes; they should charge $1.00 for an apple and $1.00 for an orange.(C) No; the system has many solutions.(D) No; the system has no solution.
Q. A fruit stand has to decide what to charge for their produce. They need $10 for 4 apples and 4 oranges. They also need $12 for 6 apples and 6 oranges. We put this information into a system of linear equations.Can we find a unique price for an apple and an orange?Choose 1 answer:A Yes; they should charge $1.00 for an apple and $1.50 for an orange.B Yes; they should charge $1.00 for an apple and $1.00 for an orange.(C) No; the system has many solutions.(D) No; the system has no solution.
Translate into equations: Let's denote the price of an apple as A and the price of an orange as O. We can then translate the information given into two equations:1. 4A+4O=($)102. 6A+6O=($)12
Simplify equations: We can simplify both equations by dividing them by their common factors to make the coefficients smaller and the equations easier to work with:1. Divide the first equation by 4: A+O=$(2.50)2. Divide the second equation by 6: A+O=$(2)
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