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Sagot :
Convert all the equations to slope-intercept form:
[tex]2y-4x=3 \ \ \ |+4x \\ 2y=4x+3 \ \ \ |\div 2 \\ y=2x+\frac{3}{2}[/tex]
[tex]A. \\ -2y+4x=1 \ \ \ |-4x \\ -2y=-4x+1 \ \ \ |\div (-2) \\ y=2x-\frac{1}{2} \\ \\ B. \\ 3y-6x=6.5 \ \ \ |+6x \\ 3y=6x+6.5 \ \ \ |\div 3 \\ y=2x+\frac{13}{6}[/tex]
[tex]C. \\ y-2x=3 \ \ \ |+2x \\ y=2x+3 \\ \\ D. \\ -2y-4x=3 \ \ \ |+4x \\ -2y=4x+3 \ \ \ |\div (-2) \\ y=-2x-\frac{3}{2}[/tex]
Lines A, B and C have the same slope as the line 2y-4x=3, so they are parallel and don't intersect it.
Line D has another slope, so it intersects the line 2y-4x=3.
The answer is D.
[tex]2y-4x=3 \ \ \ |+4x \\ 2y=4x+3 \ \ \ |\div 2 \\ y=2x+\frac{3}{2}[/tex]
[tex]A. \\ -2y+4x=1 \ \ \ |-4x \\ -2y=-4x+1 \ \ \ |\div (-2) \\ y=2x-\frac{1}{2} \\ \\ B. \\ 3y-6x=6.5 \ \ \ |+6x \\ 3y=6x+6.5 \ \ \ |\div 3 \\ y=2x+\frac{13}{6}[/tex]
[tex]C. \\ y-2x=3 \ \ \ |+2x \\ y=2x+3 \\ \\ D. \\ -2y-4x=3 \ \ \ |+4x \\ -2y=4x+3 \ \ \ |\div (-2) \\ y=-2x-\frac{3}{2}[/tex]
Lines A, B and C have the same slope as the line 2y-4x=3, so they are parallel and don't intersect it.
Line D has another slope, so it intersects the line 2y-4x=3.
The answer is D.
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