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29 June, 23:51

Given the equation 4x-3y=7, 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. 30 June, 00:56
    0
    One way is to solve for y first

    4x-3y=7

    minus 4

    -3y=-4x+7

    divide by - 3

    y=4/3x+7/3

    y=mx+b

    m=sloope

    b=yintercept

    ok, so with 1 solution is different slope and differetn y int

    y=3x+9 would be first

    no solution means same slopt with different y int so they are paralell

    y=4/3x+9

    infinite solutions means same line, but disguised

    2y=8/3x-14/3 (just multiply everybody by 2)

    so

    1 solution: 3x-4y=7

    no solution: 4x-3y=anynumberexcept7 so 4x-3y=9

    infinite solutions simplifies to 4x-3y=7, example 8x-6y=14
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