What is the largest integer $x$ such that $\frac{x}{3}+\frac{4}{5} < \frac{5}{3}$?

Guest May 18, 2019

#1**+3 **

\(\frac{x}{3}+\frac{4}{5} < \frac{5}{3}\)

Subtract \(\frac45\) from both sides of the inequality.

\(\frac{x}{3}+\frac{4}{5}{\color{blue}-\frac45} < \frac{5}{3}{\color{blue}-\frac45}\)

\(\frac{x}{3} < \frac{5}{3}-\frac45\)

Get a common denominator on the right side.

\(\frac{x}{3} < \frac{5}{3}\cdot\frac55-\frac45\cdot\frac33\)

\(\frac{x}{3} < \frac{25}{15}-\frac{12}{15}\)

Combine the fractions on the right side.

\(\frac{x}{3} < \frac{13}{15}\)

Multiply both sides by 3 , a positive number.

\(\frac{x}{3}{\color{blue}\cdot3} < \frac{13}{15}{\color{blue}\cdot3}\)

\(x < \frac{39}{15}\)

And \(\frac{39}{15}\) = 2.6

\(x <2.6\)

What is the largest integer x such that x < 2.6 ?

What is the largest integer less than 2.6 ?

The largest integer less than 2.6 is 2 , so

x = 2

hectictar May 18, 2019