Solved: "The solution set of the equation (2y−3)13−(4y+5)13=0"

High School
Algebra
Pre-Algebra
Equations, expressions, and inequalitie

ossidianaZ

Answered question

2021-09-16

The solution set of the equation ${(2 y - 3)}^{\frac{1}{3}} - {(4 y + 5)}^{\frac{1}{3}} = 0$

Answer & Explanation

Sally Cresswell

Skilled2021-09-17Added 91 answers

Step 1
Consider the provided equation,
${(2 y - 3)}^{\frac{1}{3}} - {(4 y + 5)}^{\frac{1}{3}} = 0$
Now, add ${(4 y_{5})}^{\frac{1}{3}}$ to the both sides,
${(2 y - 3)}^{\frac{1}{3}} - {(4 y + 5)}^{\frac{1}{3}} = 0$
${(2 t - 3)}^{\frac{1}{3}} - {(4 y + 5)}^{\frac{1}{3}} + {(4 y + 5)}^{\frac{1}{3}} = {(4 y + 5)}^{\frac{1}{3}}$
${(2 y - 3)}^{\frac{1}{3}} = {(4 y + 5)}^{\frac{1}{3}}$
Take cube on both sides of the above equation,
${(2 y - 3)}^{\frac{1}{3}} = {(4 y + 5)}^{\frac{1}{3}}$
${({(2 y - 3)}^{\frac{1}{3}})}^{3} = {({(4 y + 5)}^{\frac{1}{3}})}^{3}$
$2 y - 3 = 4 y + 5$
$- 3 - 5 = 4 y - 2 y$
Simplify the above equation,
$- 3 - 5 = 4 y - 2 y$
$- 8 = 2 y$
$y = - 4$
Check:
Now, substitute the value of $y = - 4$ in the provided equation $(2 y - 3)^{\frac{1}{3}} - (4 y + 5)^{\frac{1}{3}} = 0$
$[2 (- 4) - 3]^{\frac{1}{3}} - [4 (- 4) + 5]^{\frac{1}{3}} \begin{matrix} ? \\ = \end{matrix} 0$
$(- 8 - 3)^{\frac{1}{3}} - (- 16 + 5)^{\frac{1}{3}} \begin{matrix} ? \\ = \end{matrix} 0$
$(- 11)^{\frac{1}{3}} - (- 11)^{\frac{1}{3}} \begin{matrix} ? \\ = \end{matrix} 0$
$0 \begin{matrix} ? \\ = \end{matrix} 0$
Which is true.
Hence, the solution set of the equation ${(2 y - 3)}^{\frac{1}{3}} - {(4 y + 5)}^{\frac{1}{3}} = 0$ is ${- 4}$

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