The sum of 11 and three-fourths of a number is less than 112. What are all the possible values of the number? Write an inequality that could be used to solve this problem. Use the letter x as the variable, and write the inequality so that the a term comes first. Where necessary, write numbers as fractions (rather than decimals). Do not solve the inequality.
To find all the possible values of the number, we can set up an inequality.
Let's first find the expression that represents "three-fourths of a number." We can represent this expression as $\frac{3}{4}x$.
The sum of 11 and three-fourths of a number can be written as $11 + \frac{3}{4}x$.
The problem states that this sum is less than 112. Therefore, we can write the inequality as:
$11 + \frac{3}{4}x < 112$
So, the inequality that could be used to solve this problem is:
$\frac{3}{4}x + 11 < 112$
can u solve it again
Certainly! We have the inequality:
$\frac{3}{4}x + 11 < 112$
To solve for $x$, let's first subtract 11 from both sides of the inequality:
$\frac{3}{4}x < 112 - 11$
Simplifying the right side:
$\frac{3}{4}x < 101$
To isolate $x$, we'll divide both sides of the inequality by $\frac{3}{4}$:
$x < \frac{101}{\frac{3}{4}}$
Invert and multiply to simplify the right side:
$x < \frac{101}{\frac{3}{4}} \cdot \frac{4}{1}$
$x < \frac{404}{3}$
Therefore, the possible values of $x$ are all numbers less than $\frac{404}{3}$, or $\frac{404}{3}$ non-inclusive.