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anonymous

  • one year ago

how do i know if there is 1 solution, no solution, or infinite solutions?

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  1. anonymous
    • one year ago
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    i need to know this but i don't

  2. anonymous
    • one year ago
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    1 solution: Different slopes No solution: Same slope, different y-intercept Infinite solution: Same slope and same y-intercept I'm writing this on the assumption that you are talking about lines, and not parabolas and such.

  3. anonymous
    • one year ago
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    If you can solve for x, there is one solution. If both sides of the equation are equal, for example 9x+3=9x+3, there are infinitely many solutions. If the solution is untrue, for example 7=8, 9=3, etc, there are no solutions.

  4. anonymous
    • one year ago
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    Sorry, it doesn't necessarily have to be x. It can be any variable.

  5. anonymous
    • one year ago
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    how about one solution

  6. anonymous
    • one year ago
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    If you can work out the equation and find the value of the variable normally, it has one solution.

  7. anonymous
    • one year ago
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    if like 6=6 does that have 1 solution

  8. anonymous
    • one year ago
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    That would have infinitely many solutions, because both sides of the equation are the same.

  9. anonymous
    • one year ago
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    how about 2=3 or 1=7

  10. anonymous
    • one year ago
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    2 does not equal 3, nor does 1 equal 7. Both of the equations are false, so they have no solution.

  11. anonymous
    • one year ago
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    how about this prob.5(6x+2)=3(10x−2)−2x

  12. anonymous
    • one year ago
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    First you would use the distributive property to solve the parenthesis on each side. Then you would continue working out the equation until you come to a solution, whether it be finding the value of x or finding that there are an infinite amount of solutions or no solutions.

  13. anonymous
    • one year ago
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    how many solutions is there/

  14. anonymous
    • one year ago
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    1

  15. anonymous
    • one year ago
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    \[5(6x+2)=3(10x-2)-2x\]Distributive property \[5(6x)+5(2)=3(10x)+3(-2)-2x\]Multiply \[30x+10=30x-6-2x\]Combine like terms on right side to get the two equations\[30x+10=28x-6\]Get the x's on the left side\[2x+10=-6\] Subtract the 10 over\[2x=-16\] Divide to get\[x=-8\] If you want to find the point where they meet, plug the x into one of the original equations.\[y=30(-8)+10\]\[y=-240+10\]\[y=-230\] So the point they meet would be \[(-8,-230)\]

  16. anonymous
    • one year ago
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    LegendarySadist was faster. But anyway, obviously it has one solution.

  17. anonymous
    • one year ago
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    He left before I got mine in =(

  18. anonymous
    • one year ago
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    Shame.

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