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16 February, 17:59

Given the equation y=-1/3x - 8, write a second linear equation to create a system that: Has exactly one solution. Has no solution. Has infinitely many solutions.

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  1. 16 February, 21:14
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    For two linear equations to have exactly one solution, their gradient/slope must be different. That's the only requirement. For example y=x intersects with - 1/3x - 8 at (-6, - 6). So one that has only one solution could be y=x+5.

    For two linear equations to have no solutions they must have the same gradient/slope but not be the same equation. Hence, y = - 1/3x + 8 would never intersect with y = - 1/3x - 8.

    For two linear equations to have infinite solutions, they must be the same line or simplify to give the same line. I. e. 3y = - x - 8 would have infinitely many solutions with the above graph.
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