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

What system of linear equations is represented by the lines shown?

$8 x + 4 y = 32$

$- 10 x - 4 y = - 64$

$8 x - 4 y = 32$

$- 10 x + 4 y = - 64$

$4 x - 10 y = 32$

$- 8 x + 10 y = - 64$

$4 x + 10 y = 32$

$- 8 x - 10 y = - 64$

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Explanation

Choice D is correct. A line in the xy-plane that passes through the points $left parenthesis x 1 comma y 1 right parenthesis$ and $left parenthesis x 2 comma y 2 right parenthesis$ has slope $m$ , where $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ , and can be defined by an equation of the form $y - y_{1} = m (x - x_{1})$ . One of the lines shown in the graph passes through the points $left parenthesis 8 comma 0 right parenthesis$ and $left parenthesis 3 comma 4 right parenthesis$ . Substituting $8$ for $x 1$ , $0$ for $y 1$ , $3$ for $x 2$ , and $4$ for $y 2$ in the equation $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ yields $m = \frac{4 - 0}{3 - 8}$ , or $m = - \frac{4}{5}$ . Substituting $- \frac{4}{5}$ for $m$ , $8$ for $x 1$ and $0$ for $y 1$ in the equation $y - y_{1} = m (x - x_{1})$ yields $y - 0 = - \frac{4}{5} (x - 8)$ , which is equivalent to $y = - \frac{4}{5} x + \frac{32}{5}$ . Adding $\frac{4}{5} x$ to both sides of this equation yields $\frac{4}{5} x + y = \frac{32}{5}$ . Multiplying both sides of this equation by $negative 10$ yields $- 8 x - 10 y = - 64$ . Therefore, an equation of this line is $- 8 x - 10 y = - 64$ . Similarly, the other line shown in the graph passes through the points $left parenthesis 8 comma 0 right parenthesis$ and $left parenthesis 3 comma 2 right parenthesis$ . Substituting $8$ for $x 1$ , $0$ for $y 1$ , $3$ for $x 2$ , and $2$ for $y 2$ in the equation $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ yields $m = \frac{2 - 0}{3 - 8}$ , or $m = - \frac{2}{5}$ . Substituting $- \frac{2}{5}$ for $m$ , $8$ for $x 1$ , and $0$ for $y 1$ in the equation $y - y_{1} = m (x - x_{1})$ yields $y - 0 = - \frac{2}{5} (x - 8)$ , which is equivalent to $y = - \frac{2}{5} x + \frac{16}{5}$ . Adding $\frac{2}{5} x$ to both sides of this equation yields $\frac{2}{5} x + y = \frac{16}{5}$ . Multiplying both sides of this equation by $10$ yields $4 x + 10 y = 32$ . Therefore, an equation of this line is $4 x + 10 y = 32$ . So, the system of linear equations represented by the lines shown is $4 x + 10 y = 32$ and $- 8 x - 10 y = - 64$ .

Choice A is incorrect and may result from conceptual or calculation errors.

Choice B is incorrect and may result from conceptual or calculation errors.

Choice C is incorrect and may result from conceptual or calculation errors.