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Customers started to boycott a company to protest unsafe working conditions. In the first month of the boycott, the share price for the company fell $23. In the next month, it fell $17 more. The customers wondered how much more the share price would need to fall for a total decrease of $50. Which of the following equations matches the situation above? Choose 1 answer: (A) 23-17-50= ? (B) -23-17-?=-50 (c) -23-17-50= ?

Customers started to boycott a company to protest unsafe working conditions. In the first month of the boycott, the share price for the company fell $23 \$ 23 . In the next month, it fell $17 \$ 17 more. The customers wondered how much more the share price would need to fall for a total decrease of $50 \$ 50 .\newlineWhich of the following equations matches the situation above?\newlineChoose 11 answer:\newline(A) 23−17−50= 23-17-50= ?\newline(B) −23−17−?=−50 -23-17-?=-50 \newline(C) −23−17−50= -23-17-50= ?

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Q. Customers started to boycott a company to protest unsafe working conditions. In the first month of the boycott, the share price for the company fell $23 \$ 23 . In the next month, it fell $17 \$ 17 more. The customers wondered how much more the share price would need to fall for a total decrease of $50 \$ 50 .\newlineWhich of the following equations matches the situation above?\newlineChoose 11 answer:\newline(A) 23−17−50= 23-17-50= ?\newline(B) −23−17−?=−50 -23-17-?=-50 \newline(C) −23−17−50= -23-17-50= ?
  1. Understand and Set Up Equation: Understand the situation and set up the equation.\newlineThe share price fell $23\$23 in the first month and $17\$17 in the second month. We want to know how much more it needs to fall to reach a total decrease of $50\$50. We can represent the additional decrease needed as a variable, let's call it 'x'. The equation representing the situation is:\newline−23-23 (first month's decrease) −17- 17 (second month's decrease) −x- x (additional decrease) =−50= -50 (total decrease).
  2. Simplify the Equation: Simplify the equation.\newlineCombine the known decreases and set up the equation with the variable xx:\newline−23−17−x=−50-23 - 17 - x = -50\newline−40−x=−50-40 - x = -50
  3. Solve for 'x': Solve for 'x'.\newlineTo find the value of 'x', we need to isolate it on one side of the equation:\newline−40−x=−50-40 - x = -50\newlineAdd 4040 to both sides of the equation to solve for 'x':\newline−40+40−x=−50+40-40 + 40 - x = -50 + 40\newline−x=−10-x = -10\newlineNow, we multiply both sides by −1-1 to get the positive value of 'x':\newline−x×−1=−10×−1-x \times -1 = -10 \times -1\newlinex=10x = 10

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