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anonymous

  • 5 years ago

Solve the following using the elimination method. If the system has an infinite number of solutions, use set-builder notation to write the the solution set. If the system has no solution, state this: 1) 3x - 5y = -3 -9x + 15y=9 2) 7x + 9y = -27 -2x - 7y = 21

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  1. anonymous
    • 5 years ago
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    1) multiply first equation by 3, so your system: 9x-15y=-9 -9x+15y=9 if you add them : 0=0 so your first eq=second one. no solution 2) multiply first eq by 2 & second by 7 14x+18y=-54 -14x-49y=147 add them: 0-31y=93 -31y=93 divide both sides by (-31): y=-3 substitute value of y in any original eq: 9*x-15*(-3)=-9 9x+45=-9 9x=-54 x=-6 solution: x=-6, y=-3

  2. anonymous
    • 5 years ago
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    thank you

  3. anonymous
    • 5 years ago
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    wouldn't you plug y into the equation 14x + 18y = -54 instead of 9x - 15y = -9? I'm confused

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