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The tables represent some points on the graphs of lines m and n. Which system of equations is represented by lines m and n? A) 6x − y = 8 3x 2y = 9 B) 6x y = −8 3x − 2y = 9 C) −6x − y = 8 −3x 2y = 9 D) 6x − y = −8 −3x − 2y = 9.

Sagot :

A system of equation is a collection of equations

The system of equation is: [tex]\mathbf{6x -y = 8}[/tex] and [tex]\mathbf{3x + 2y = 3 }[/tex]

The points are given as:

[tex]\mathbf{Line\ m = (0,-8)(1,-2)}[/tex]

[tex]\mathbf{Line\ n = (1,3)(3,0)}[/tex]

For line m

Calculate the slope

[tex]\mathbf{m = \frac{y_2 - y_1}{x_2 - x_1}}[/tex]

So, we have:

[tex]\mathbf{m = \frac{-2 + 8}{1 - 0}}[/tex]

[tex]\mathbf{m = \frac{6}{1}}[/tex]

[tex]\mathbf{m = 6}[/tex]

The equation is then calculated as:

[tex]\mathbf{y = m(x - x_1) + y_1}[/tex]

This gives

[tex]\mathbf{y = 6(x - 0) -8}[/tex]

[tex]\mathbf{y = 6x -8}[/tex]

Rewrite as:

[tex]\mathbf{6x -y = 8}[/tex]

For line n

Calculate the slope

[tex]\mathbf{m = \frac{y_2 - y_1}{x_2 - x_1}}[/tex]

So, we have:

[tex]\mathbf{m = \frac{0 - 3}{3 - 1}}[/tex]

[tex]\mathbf{m =- \frac{3}{2}}[/tex]

The equation is then calculated as:

[tex]\mathbf{y = m(x - x_1) + y_1}[/tex]

This gives

[tex]\mathbf{y =- \frac{3}{2}(x -0) + 3 }[/tex]

[tex]\mathbf{y =- \frac{3}{2}x + 3 }[/tex]

Multiply through by 2

[tex]\mathbf{2y =- 3x + 3 }[/tex]

Rewrite as:

[tex]\mathbf{3x + 2y = 3 }[/tex]

Hence, the system of equation is:

[tex]\mathbf{6x -y = 8}[/tex] and [tex]\mathbf{3x + 2y = 3 }[/tex]

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