solve both problems and for the question labeled 3 am I to s

solve both problems. and for the question labeled \"3.\" am I to solve for x in all cases like this? rusty in math.y

df Adobe Reader 3. Solve 5x2+3r-2 3x+1 8 3. Solve 5x2 + 3x-2-3x+1=-B 2x2 +15x-27 2x-3 x +9 2 3x + 1 8 =

Solution

3)

[(5x²+3x-2)/(2x²+15x-27)]-[(3x+1)/(2x-3)] =8/(x+9)

[(5x²+5x-2x-2)/(2x²+18x-3x-27)]-[(3x+1)/(2x-3)] =8/(x+9)

[(5x(x+1)-2(x+1))/(2x(x+9)-3(x+9))]-[(3x+1)/(2x-3)] =8/(x+9)

[((5x-2)(x+1))/((2x -3)(x+9))]-[(3x+1)/(2x-3)] =8/(x+9)

[((5x-2)(x+1)) -((3x+1)(x+9))]/[(2x -3)(x+9)] =8/(x+9)

[((5x-2)(x+1)) -((3x+1)(x+9))]=(2x -3)(x+9)8/(x+9)

[((5x-2)(x+1)) -((3x+1)(x+9))]=8(2x -3)

[(5x²+3x-2) -(3x²+28x+9)]=(16x -24)

2x²-25x-11=16x-24

2x²-25x-11-16x+24=0

2x²-41x+13=0

quadratice formula:ax^2 +bx +c =0 ==>x=[-b+(b²-4ac)]/(2a),x=[-b-(b²-4ac)]/(2a)

a =2, b=-41, c=13

x=(41-1577)/4,x=(41+1577)/4

x=0.32,x=20.12