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Which describes the system of equations below?\newliney=110x+107y = \frac{1}{10}x + \frac{10}{7}\newliney=110x+107y = \frac{1}{10}x + \frac{10}{7}\newlineChoices:\newline(A) inconsistent\newline(B) consistent and independent\newline(C) consistent and dependent

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Q. Which describes the system of equations below?\newliney=110x+107y = \frac{1}{10}x + \frac{10}{7}\newliney=110x+107y = \frac{1}{10}x + \frac{10}{7}\newlineChoices:\newline(A) inconsistent\newline(B) consistent and independent\newline(C) consistent and dependent
  1. Compare slopes: We have the system of equations:\newliney = (110)x+107(\frac{1}{10})x + \frac{10}{7}\newliney = (110)x+107(\frac{1}{10})x + \frac{10}{7}\newlineFirst, we need to compare the slopes of both equations.\newlineIn y=(110)x+107y = (\frac{1}{10})x + \frac{10}{7}, the slope is 110\frac{1}{10}.\newlineIn y=(110)x+107y = (\frac{1}{10})x + \frac{10}{7}, the slope is also 110\frac{1}{10}.\newlineSince both slopes are equal, we can say that the lines are parallel or the same line.
  2. Compare y-intercepts: Next, we compare the y-intercepts of both equations.\newlineIn y=110x+107y = \frac{1}{10}x + \frac{10}{7}, the y-intercept is 107\frac{10}{7}.\newlineIn y=110x+107y = \frac{1}{10}x + \frac{10}{7}, the y-intercept is also 107\frac{10}{7}.\newlineSince both y-intercepts are equal, we can say that the lines have the same y-intercept.
  3. Determine line relationship: Since both the slope and yy-intercept of the two equations are the same, the lines represented by these equations are the same line. Therefore, every point on one line is also on the other line, which means the system has an infinite number of solutions. This makes the system consistent and dependent.

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