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Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks.\newlineFor a project in statistics class, a pair of students decided to invest in two companies, one that produces software and one that does biotechnology research. Jill purchased 8282 shares in the software company and 8282 shares in the biotech firm, which cost a total of $5,822\$5,822. At the same time, Kate invested a total of $5,978\$5,978 in 9595 shares in the software company and 8282 shares in the biotech firm. How much did each share cost?\newlineEach share in the software company cost $_\$\_, and each share in the biotech firm cost $_\$\_.

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Q. Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks.\newlineFor a project in statistics class, a pair of students decided to invest in two companies, one that produces software and one that does biotechnology research. Jill purchased 8282 shares in the software company and 8282 shares in the biotech firm, which cost a total of $5,822\$5,822. At the same time, Kate invested a total of $5,978\$5,978 in 9595 shares in the software company and 8282 shares in the biotech firm. How much did each share cost?\newlineEach share in the software company cost $_\$\_, and each share in the biotech firm cost $_\$\_.
  1. Equations setup: Let's denote the cost of one share in the software company as SS dollars and the cost of one share in the biotech firm as BB dollars. We can write two equations based on the information given:\newline11. For Jill's investment: 82S+82B=582282S + 82B = 5822\newline22. For Kate's investment: 95S+82B=597895S + 82B = 5978
  2. Elimination method: To solve this system using elimination, we need to eliminate one of the variables. We can do this by multiplying the first equation by −1-1 to get the coefficients of BB to be opposites:\newline−1×(82S+82B)=−1×5822-1 \times (82S + 82B) = -1 \times 5822\newlineThis gives us: −82S−82B=−5822-82S - 82B = -5822
  3. Solving for S: Now we add this new equation to the second equation to eliminate B:\newline(−82S−82B)+(95S+82B)=(−5822)+5978(-82S - 82B) + (95S + 82B) = (-5822) + 5978\newlineThis simplifies to: 13S=15613S = 156
  4. Substitute SS into equation: We divide both sides of the equation by 1313 to solve for SS:13S13=15613\frac{13S}{13} = \frac{156}{13}S=12S = 12
  5. Calculate value of B: Now that we have the value of S, we can substitute it back into one of the original equations to solve for B. We'll use the first equation:\newline82S+82B=582282S + 82B = 5822\newline82(12)+82B=582282(12) + 82B = 5822
  6. Final solution for B: We calculate the value of 82×1282 \times 12 and subtract it from 58225822 to solve for B:\newline984+82B=5822984 + 82B = 5822\newline82B=5822−98482B = 5822 - 984\newline82B=483882B = 4838
  7. Final solution for B: We calculate the value of 82×1282 \times 12 and subtract it from 58225822 to solve for B:\newline984+82B=5822984 + 82B = 5822\newline82B=5822−98482B = 5822 - 984\newline82B=483882B = 4838Finally, we divide both sides of the equation by 8282 to find the value of B:\newline82B82=483882\frac{82B}{82} = \frac{4838}{82}\newlineB=59B = 59

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