SOLVE REGENTS PROBLEMS $\newline$ $10$ ) The value of $x$ which makes true is $\newline$ $\frac{2}{3}\left(\frac{1}{4} x-2\right)=\frac{1}{5}\left(\frac{4}{3} x-1\right)$ $\newline$ A. $-10$ $\newline$ B. $-2$ $\newline$ C. $-9 . \overline{09}$ $\newline$ D. $-11 . \overline{3}$

View step-by-step help

Home

Math Problems

Grade 8

Domain and range of functions

Full solution

Q. SOLVE REGENTS PROBLEMS

\newline

10

) The value of

x

which makes true is

\newline

\frac{2}{3}\left(\frac{1}{4} x-2\right)=\frac{1}{5}\left(\frac{4}{3} x-1\right)

\newline

-10

\newline

-2

\newline

-9 . \overline{09}

\newline

-11 . \overline{3}

Multiply by LCM: First, let's get rid of the fractions by multiplying both sides of the equation by the least common multiple of the denominators, which is $15$ . So, we multiply both sides by $15$ to get: \(15 \times \left(\frac{ $2$ }{ $3$ }\right)\left(\frac{ $1$ }{ $4$ }x - $2$ \right) = $15$ \times \left(\frac{ $1$ }{ $5$ }\right)\left(\frac{ $4$ }{ $3$ }x - $1$ \right)
Simplify both sides: Now, simplify both sides: $15$ \times \left(\frac{$2$}{$3$}\right) \times \left(\frac{$1$}{$4$}\right)x - $15$ \times \left(\frac{$2$}{$3$}\right) \times $2$ = $15$ \times \left(\frac{$1$}{$5$}\right) \times \left(\frac{$4$}{$3$}\right)x - $15$ \times \left(\frac{$1$}{$5$}\right) \times $1$
Further simplification: This simplifies to: $10$ \times \left(\frac{ $1$ }{ $4$ }\right)x - $10$ \times $2$ = $3$ \times \left(\frac{ $4$ }{ $3$ }\right)x - $3$
Simplify $x$ terms: Which further simplifies to: $\newline$ $\frac{10}{4}x - 20 = 4x - 3$
Combine like terms: Now, let's simplify $(\frac{10}{4})x$ to $(\frac{5}{2})x$ : $\newline$ $(\frac{5}{2})x - 20 = 4x - 3$
Isolate x term: Next, we'll get all the x terms on one side and the constants on the other side. Add $20$ to both sides and subtract $(5/2)x$ from both sides: $\newline$ $(5/2)x - (5/2)x - 20 + 20 = 4x - (5/2)x - 3 + 20$
Multiply by reciprocal: This simplifies to: $\newline$ $0 = \frac{8}{2}x - \frac{5}{2}x + 17$
Final solution: Combine like terms: $\newline$ $0 = \frac{3}{2}x + 17$
Final solution: Combine like terms: $\newline$ $0 = \frac{3}{2}x + 17$ Now, subtract $17$ from both sides to isolate the $x$ term: $\newline$ $-17 = \frac{3}{2}x$
Final solution: Combine like terms: $\newline$ $0 = \frac{3}{2}x + 17$ Now, subtract $17$ from both sides to isolate the $x$ term: $\newline$ $-17 = \frac{3}{2}x$ Finally, multiply both sides by the reciprocal of $\frac{3}{2}$ to solve for $x$ : $\newline$ $x = -17 \times \frac{2}{3}$
Final solution: Combine like terms: $\newline$ $0 = \frac{3}{2}x + 17$ Now, subtract $17$ from both sides to isolate the $x$ term: $\newline$ $-17 = \frac{3}{2}x$ Finally, multiply both sides by the reciprocal of $\frac{3}{2}$ to solve for $x$ : $\newline$ $x = -17 \times \frac{2}{3}$ This gives us: $\newline$ $x = -\frac{34}{3}$
Final solution: Combine like terms: $\newline$ $0 = \frac{3}{2}x + 17$ Now, subtract $17$ from both sides to isolate the $x$ term: $\newline$ $-17 = \frac{3}{2}x$ Finally, multiply both sides by the reciprocal of $\frac{3}{2}$ to solve for $x$ : $\newline$ $x = -17 \times \frac{2}{3}$ This gives us: $\newline$ $x = -\frac{34}{3}$ Convert $-\frac{34}{3}$ to a mixed number to match the answer choices: $\newline$ $x = -11 \frac{1}{3}$