Question

Solve for x THANK YOU

Answer 1

Step-by-step explanation:

so, I read here

2/3 x + 3x/x = 4x/2 × (3x + 6)

that is

2/3 x + 3 = 4x/2 × (3x + 6)

2x/3 + 3 = 2x × (3x + 6)

2x + 9 = 6x × (3x + 6)

2x + 9 = 18x² + 36x

18x² + 34x - 9 = 0

the general solution to a quadratic equation is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

a = 18

b = 34

c = -9

x = (-34 ± sqrt(34² - 4×18×-9))/(2×18) =

= (-34 ± sqrt(1156 + 648))/36 =

= (-34 ± sqrt(1804))/36 =

= (-34 ± sqrt(4×11×41))/36 =

= (-34 ± 2×sqrt(451))/36

x1 = (-34 + 2×sqrt(451))/36 =

= 0.235375588...

x2 = (-34 - 2×sqrt(451))/36 =

= -2.124264477...

Answer 2

Answer:

$x=\dfrac{-17+ \sqrt{451} }{18}, \quad x=\dfrac{-17- \sqrt{451} }{18}$

Step-by-step explanation:

$\textsf{Given equation}:$

$\dfrac{2}{3}x+\dfrac{3x}{x}=\dfrac{4x}{2}(3x+6)$

$\textsf{Cancel the common factor } x \textsf{ in }\dfrac{3x}{x}:$

$\implies \dfrac{2}{3}x+3=\dfrac{4x}{2}(3x+6)$

$\textsf{Simplify }\dfrac{4x}{2}\textsf{ by dividing the numbers}:$

$\implies \dfrac{2}{3}x+3=2x(3x+6)$

$\textsf{Apply the distributive law}\quad \:a\left(b+c\right)=ab+ac:$

$\implies \dfrac{2}{3}x+3=6x^2+12x$

$\textsf{Multiply both sides by 3 to cancel the denominator of }\dfrac{2}{3}:$

$\implies \dfrac{2}{3}x\cdot 3+3\cdot 3=6x^2\cdot 3+12x \cdot 3$

$\implies 2x+9=18x^2+36x$

$\textsf{Switch sides}:$

$\implies 18x^2+36x=2x+9$

$\textsf{Subtract }2x \textsf{ from both sides}:$

$\implies 18x^2+36x-2x=2x+9-2x$

$\implies 18x^2+34x=9$

$\textsf{Subtract 9 from both sides}:$

$\implies 18x^2+34x-9=9-9$

$\implies 18x^2+34x-9=0$

Solve using the Quadratic Formula:

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$

Define the variables:

$\implies a=18, \quad b=34, \quad c=-9$

Substitute the defined variables into the quadratic formula and solve for x:

$\implies x=\dfrac{-34 \pm \sqrt{34^2-4(18)(-9)} }{2(18)}$

$\implies x=\dfrac{-34 \pm \sqrt{1156+648} }{36}$

$\implies x=\dfrac{-34 \pm \sqrt{1804} }{36}$

$\implies x=\dfrac{-34 \pm \sqrt{4 \cdot 451} }{36}$

$\implies x=\dfrac{-34 \pm \sqrt{4}\sqrt{451} }{36}$

$\implies x=\dfrac{-34 \pm 2\sqrt{451} }{36}$

$\implies x=\dfrac{-17 \pm \sqrt{451} }{18}$

Therefore, the solutions to the given equation are:

$x=\dfrac{-17+ \sqrt{451} }{18}, \quad x=\dfrac{-17- \sqrt{451} }{18}$

Learn more about the Quadratic Formula here:

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