Solve-5x-1-3x -1 - 6x-1-30

Given:5x 1/3(x +1) = 6(x+1/30)⇒ 5x x/3 1/3=6x+6/300ex0ex⇒ 5x x/3 6x=1/5+1/30ex0ex⇒15x x 18x/3=1× 3+1× 5/15⇒ 4x/3=8/150ex0ex⇒ x= 24/60= 2/5Hence, x= 2/5 is the c ...

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Standard X

Mathematics

Types of Inequalities

Solve:5x-1/3x...

Question

Solve: $5 x - \frac{1}{3} (x + 1) = 6 (x + \frac{1}{30})$

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Solution

Given:
$5 x - \frac{1}{3} (x + 1) = 6 (x + \frac{1}{30})$
$\Rightarrow 5 x - \frac{x}{3} - \frac{1}{3} = 6 x + \frac{6}{30} \Rightarrow 5 x - \frac{x}{3} - 6 x = \frac{1}{5} + \frac{1}{3} \Rightarrow \frac{15 x - x - 18 x}{3} = \frac{1 \times 3 + 1 \times 5}{15}$
$\Rightarrow \frac{- 4 x}{3} = \frac{8}{15} \Rightarrow x = \frac{- 24}{60} = \frac{- 2}{5}$
Hence, $x = \frac{- 2}{5}$ is the correct answer.

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Similar questions

Solve :

$5 x - \frac{1}{3} (x + 1) = 6 (x + \frac{1}{30})$

Q. Solve:

5 x - \frac{1}{3} (x + 1) = 6 (x + \frac{1}{30})

Q. Assertion :

Δ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{(1 + a_{1} b_{1}) (1 + a_{1}^{2} b_{1}^{2} - a_{1} b_{1})}{1 + a_{1} b_{1}} & \frac{(1 + a_{1} b_{2}) (1 + a_{1}^{2} b_{2}^{2} - a_{1} b_{2})}{1 + a_{1} b_{2}} & \frac{(1 + a_{1} b_{3}) (1 + a_{1}^{2} b_{3}^{2} - a_{1} b_{3})}{1 + a_{1} b_{3}} \frac{(1 + a_{2} b_{1}) (1 + a_{2}^{2} b_{1}^{2} - a_{2} b_{1})}{1 + a_{2} b_{1}} & \frac{(1 + a_{2} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{2} b_{2})}{1 + a_{2} b_{2}} & \frac{(1 + a_{2} b_{3}) (1 + a_{2}^{2} b_{3}^{2} - a_{2} b_{3})}{1 + a_{2} b_{3}} \frac{(1 + a_{3} b_{1}) (1 + a_{3}^{2} b_{1}^{2} - a_{3} b_{1})}{1 + a_{3} b_{1}} & \frac{(1 + a_{3} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{3} b_{2})}{1 + a_{3} b_{2}} & \frac{(1 + a_{3} b_{3}) (1 + a_{3}^{2} b_{3}^{2} - a_{3} b_{3})}{1 + a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

Δ = 0

Reason:

Δ

can be written as product of two determinants.

Q. If

α

β

γ

are the roots of

x^{3} + a x^{2} + b = 0

b \neq 0

then the determinant

Δ

, where

Δ = ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{1}{α} & \frac{1}{β} & \frac{1}{γ} \frac{1}{β} & \frac{1}{γ} & \frac{1}{α} \frac{1}{γ} & \frac{1}{α} & \frac{1}{β} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣

equals

Determine whether the following numbers are in proportion or not $:$

$\frac{1}{3}, \frac{1}{4}, \frac{1}{6}, \frac{1}{7}$

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Solve:5x-13(x+1)=6(x+130)

Given:5x-13(x+1)=6(x+130)⇒5x-x3-13=6x+630⇒5x-x3-6x=15+13⇒15x-x-18x3=1×3+1×515⇒-4x3=815⇒x=-2460=-25Hence, x=-25 is the correct answer.

Solve: $5 x - \frac{1}{3} (x + 1) = 6 (x + \frac{1}{30})$