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Algebra / Systems of two linear equations in two variables Difficulty: Medium

Which of the following systems of linear equations has no solution?

$y = 6 x + 3$

$y = 6 x + 9$

$y equals 10$

$y = 10 x + 10$

$y = 14 x + 14$

$y = 10 x + 14$

$x equals 3$

$y equals 10$

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Explanation

Choice A is correct. A system of two linear equations in two variables, $x$ and $y$ , has no solution if the graphs of the lines represented by the equations in the xy-plane are distinct and parallel. The graphs of two lines in the xy-plane represented by equations in slope-intercept form, $y = m x + b$ , where $m$ and $b$ are constants, are parallel if their slopes, $m$ , are the same and are distinct if their y-coordinates of the y-intercepts, $b$ , are different. In the equations $y = 6 x + 3$ and $y = 6 x + 9$ , the values of $m$ are each $6$ , and the values of $b$ are $3$ and $9$ , respectively. Since the slopes of these lines are the same and the y-coordinates of the y-intercepts are different, it follows that the system of linear equations in choice A has no solution.

Choice B is incorrect. The two lines represented by these equations are a horizontal line and a line with a slope of $10$ that have the same y-coordinate of the y-intercept. Therefore, this system has a solution, $left parenthesis 0 comma 10 right parenthesis$ , rather than no solution.

Choice C is incorrect. The two lines represented by these equations have different slopes and the same y-coordinate of the y-intercept. Therefore, this system has a solution, $left parenthesis 0 comma 14 right parenthesis$ , rather than no solution.

Choice D is incorrect. The two lines represented by these equations are a vertical line and a horizontal line. Therefore, this system has a solution, $left parenthesis 3 comma 10 right parenthesis$ , rather than no solution.