Caliburn asked in 科學數學 · 10 years ago

數學 代數

Factorise fully:

2x^2+11x+20

Solve:

3x+4y=13

4x-5y=38

x=?

y=?

Update:

Solve:

6(x+1)>2x+3

2 Answers

Rating
  • 10 years ago
    Favorite Answer

    Factorise

    fully:

    1. 2x^2+11x+20

    Sol

    2x^2+11x+20

    D=11^2-4*2*20=121-160=-39<0

    在實係數下無法再因式分解

    2. Solve

    3x+4y=13

    4x-5y=38

    x=?

    y=?

    Sol

    3x+4y=13--------(1)

    4x-5y=38------(2)

    5*(1)+4*(2)

    5*(3x+4y)+4(4x-5y)=65+152

    15x+16x=217

    x=7---------------(3)代入(1)

    3*7+4y=13

    4y=-8

    y=-2-----------(4)

    3. Solve:

    6(x+1)>2x+3

    Sol

    6x+6>2x+3

    4x>-3

    x>-3/4

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  • Alex
    Lv 5
    10 years ago

    2.

    3x+4y=13.......(1)

    4x-5y=38........(2)

    (1)*5+(2)*4

    31x=217

    x=7

    代入(1)

    y= -2

    Ans:x=7,y= -2

    2010-04-27 15:46:39 補充:

    3.

    6(x+1)>2x+3

    6x+6 > 2x+3

    6x -2x > 3 -6

    4x > -3

    x > -3/4

    2010-04-27 15:58:19 補充:

    1.

    2x^2+11x+20 無法完全因式分解

    題目是否有誤?

    Source(s): alex, alex, alex
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