Answer :
Sure, let's solve these equations step-by-step.
### 1) [tex]\( 16x - \frac{10}{x} = 27 \)[/tex]
First, multiply both sides of the equation by [tex]\( x \)[/tex] to get rid of the fraction:
[tex]\[ 16x^2 - 10 = 27x \][/tex]
Rearrange the equation to form a standard quadratic equation:
[tex]\[ 16x^2 - 27x - 10 = 0 \][/tex]
Solve this quadratic equation using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 16 \)[/tex], [tex]\( b = -27 \)[/tex], and [tex]\( c = -10 \)[/tex]:
[tex]\[ x = \frac{27 \pm \sqrt{(-27)^2 - 4 \cdot 16 \cdot (-10)}}{2 \cdot 16} \][/tex]
[tex]\[ x = \frac{27 \pm \sqrt{729 + 640}}{32} \][/tex]
[tex]\[ x = \frac{27 \pm \sqrt{1369}}{32} \][/tex]
[tex]\[ x = \frac{27 \pm 37}{32} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{64}{32} = 2 \][/tex]
[tex]\[ x = \frac{-10}{32} = -\frac{5}{16} \][/tex]
So, the solutions to the first equation are:
[tex]\[ x = 2 \quad \text{or} \quad x = -\frac{5}{16} \][/tex]
### 2) [tex]\( 9x^2 - 3x - 2 = 0 \)[/tex]
This is already a quadratic equation in standard form. Let's use the quadratic formula again: [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 9 \)[/tex], [tex]\( b = -3 \)[/tex], and [tex]\( c = -2 \)[/tex]:
[tex]\[ x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 9 \cdot (-2)}}{2 \cdot 9} \][/tex]
[tex]\[ x = \frac{3 \pm \sqrt{9 + 72}}{18} \][/tex]
[tex]\[ x = \frac{3 \pm \sqrt{81}}{18} \][/tex]
[tex]\[ x = \frac{3 \pm 9}{18} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{12}{18} = \frac{2}{3} \][/tex]
[tex]\[ x = \frac{-6}{18} = -\frac{1}{3} \][/tex]
So, the solutions to the second equation are:
[tex]\[ x = \frac{2}{3} \quad \text{or} \quad x = -\frac{1}{3} \][/tex]
### 3) [tex]\( \frac{1}{x-3} + \frac{2}{x-2} = \frac{8}{x} \)[/tex]
To solve this, first find a common denominator:
[tex]\[ \frac{(x-2) + 2(x-3)}{(x-3)(x-2)} = \frac{8}{x} \][/tex]
[tex]\[ \frac{x - 2 + 2x - 6}{(x-3)(x-2)} = \frac{8}{x} \][/tex]
[tex]\[ \frac{3x - 8}{(x-3)(x-2)} = \frac{8}{x} \][/tex]
Cross-multiply to clear the fractions:
[tex]\[ x(3x - 8) = 8(x-3)(x-2) \][/tex]
[tex]\[ 3x^2 - 8x = 8(x^2 - 5x + 6) \][/tex]
[tex]\[ 3x^2 - 8x = 8x^2 - 40x + 48 \][/tex]
Rearrange the equation:
[tex]\[ 3x^2 - 8x - 8x^2 + 40x - 48 = 0 \][/tex]
[tex]\[ -5x^2 + 32x - 48 = 0 \][/tex]
Solve using the quadratic formula: [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = -5 \)[/tex], [tex]\( b = 32 \)[/tex], and [tex]\( c = -48 \)[/tex]:
[tex]\[ x = \frac{-32 \pm \sqrt{32^2 - 4 \cdot (-5) \cdot (-48)}}{2 \cdot (-5)} \][/tex]
[tex]\[ x = \frac{-32 \pm \sqrt{1024 - 960}}{-10} \][/tex]
[tex]\[ x = \frac{-32 \pm \sqrt{64}}{-10} \][/tex]
[tex]\[ x = \frac{-32 \pm 8}{-10} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{-24}{-10} = \frac{12}{5} \][/tex]
[tex]\[ x = \frac{-40}{-10} = 4 \][/tex]
So, the solutions to the third equation are:
[tex]\[ x = \frac{12}{5} \quad \text{or} \quad x = 4 \][/tex]
Thus, the solutions are:
1) [tex]\( x = 2 \)[/tex] or [tex]\( x = -\frac{5}{16} \)[/tex]
2) [tex]\( x = \frac{2}{3} \)[/tex] or [tex]\( x = -\frac{1}{3} \)[/tex]
3) [tex]\( x = \frac{12}{5} \)[/tex] or [tex]\( x = 4 \)[/tex]
### 1) [tex]\( 16x - \frac{10}{x} = 27 \)[/tex]
First, multiply both sides of the equation by [tex]\( x \)[/tex] to get rid of the fraction:
[tex]\[ 16x^2 - 10 = 27x \][/tex]
Rearrange the equation to form a standard quadratic equation:
[tex]\[ 16x^2 - 27x - 10 = 0 \][/tex]
Solve this quadratic equation using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 16 \)[/tex], [tex]\( b = -27 \)[/tex], and [tex]\( c = -10 \)[/tex]:
[tex]\[ x = \frac{27 \pm \sqrt{(-27)^2 - 4 \cdot 16 \cdot (-10)}}{2 \cdot 16} \][/tex]
[tex]\[ x = \frac{27 \pm \sqrt{729 + 640}}{32} \][/tex]
[tex]\[ x = \frac{27 \pm \sqrt{1369}}{32} \][/tex]
[tex]\[ x = \frac{27 \pm 37}{32} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{64}{32} = 2 \][/tex]
[tex]\[ x = \frac{-10}{32} = -\frac{5}{16} \][/tex]
So, the solutions to the first equation are:
[tex]\[ x = 2 \quad \text{or} \quad x = -\frac{5}{16} \][/tex]
### 2) [tex]\( 9x^2 - 3x - 2 = 0 \)[/tex]
This is already a quadratic equation in standard form. Let's use the quadratic formula again: [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 9 \)[/tex], [tex]\( b = -3 \)[/tex], and [tex]\( c = -2 \)[/tex]:
[tex]\[ x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 9 \cdot (-2)}}{2 \cdot 9} \][/tex]
[tex]\[ x = \frac{3 \pm \sqrt{9 + 72}}{18} \][/tex]
[tex]\[ x = \frac{3 \pm \sqrt{81}}{18} \][/tex]
[tex]\[ x = \frac{3 \pm 9}{18} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{12}{18} = \frac{2}{3} \][/tex]
[tex]\[ x = \frac{-6}{18} = -\frac{1}{3} \][/tex]
So, the solutions to the second equation are:
[tex]\[ x = \frac{2}{3} \quad \text{or} \quad x = -\frac{1}{3} \][/tex]
### 3) [tex]\( \frac{1}{x-3} + \frac{2}{x-2} = \frac{8}{x} \)[/tex]
To solve this, first find a common denominator:
[tex]\[ \frac{(x-2) + 2(x-3)}{(x-3)(x-2)} = \frac{8}{x} \][/tex]
[tex]\[ \frac{x - 2 + 2x - 6}{(x-3)(x-2)} = \frac{8}{x} \][/tex]
[tex]\[ \frac{3x - 8}{(x-3)(x-2)} = \frac{8}{x} \][/tex]
Cross-multiply to clear the fractions:
[tex]\[ x(3x - 8) = 8(x-3)(x-2) \][/tex]
[tex]\[ 3x^2 - 8x = 8(x^2 - 5x + 6) \][/tex]
[tex]\[ 3x^2 - 8x = 8x^2 - 40x + 48 \][/tex]
Rearrange the equation:
[tex]\[ 3x^2 - 8x - 8x^2 + 40x - 48 = 0 \][/tex]
[tex]\[ -5x^2 + 32x - 48 = 0 \][/tex]
Solve using the quadratic formula: [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = -5 \)[/tex], [tex]\( b = 32 \)[/tex], and [tex]\( c = -48 \)[/tex]:
[tex]\[ x = \frac{-32 \pm \sqrt{32^2 - 4 \cdot (-5) \cdot (-48)}}{2 \cdot (-5)} \][/tex]
[tex]\[ x = \frac{-32 \pm \sqrt{1024 - 960}}{-10} \][/tex]
[tex]\[ x = \frac{-32 \pm \sqrt{64}}{-10} \][/tex]
[tex]\[ x = \frac{-32 \pm 8}{-10} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{-24}{-10} = \frac{12}{5} \][/tex]
[tex]\[ x = \frac{-40}{-10} = 4 \][/tex]
So, the solutions to the third equation are:
[tex]\[ x = \frac{12}{5} \quad \text{or} \quad x = 4 \][/tex]
Thus, the solutions are:
1) [tex]\( x = 2 \)[/tex] or [tex]\( x = -\frac{5}{16} \)[/tex]
2) [tex]\( x = \frac{2}{3} \)[/tex] or [tex]\( x = -\frac{1}{3} \)[/tex]
3) [tex]\( x = \frac{12}{5} \)[/tex] or [tex]\( x = 4 \)[/tex]
