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inequalities
- Thread starter sri340
- Start date
sri340 said:If x and y are integers, 7 < y < 16, and x/y = 2/5 how many possible values are there for x ?
(A)One (B) Two (C) Three (D) Four (E) Five
Hi sri340,
\(\displaystyle \frac{x}{y}=\frac{2}{5}=\frac{4}{10}=\frac{6}{15}\).
How many values of x did we find when \(\displaystyle 7 < y < 16\)?
One possible way of finding how many possible values there are for x is to solve for y in the equation and substitute it into the inequality.
From \(\displaystyle \frac{x}{y}=\frac{2}{5}\)
you get \(\displaystyle y=\frac{5}{2}x\)
Pugging that into the inequality gives you
\(\displaystyle 7<\frac{5}{2}x<16\)
Then solve for x
Any Questions?
From \(\displaystyle \frac{x}{y}=\frac{2}{5}\)
you get \(\displaystyle y=\frac{5}{2}x\)
Pugging that into the inequality gives you
\(\displaystyle 7<\frac{5}{2}x<16\)
Then solve for x
Any Questions?
Hello, sri340!
You can talk your way through this . . .
\(\displaystyle \text{We are told that }y\text{ is between 7 and 16: }\;y \,\in\,\{8,9,10,11,12,13,14,15\}\)
\(\displaystyle \text{We are told that: }\:\frac{x}{y} \,=\,\frac{2}{5} \quad\Rightarrow\quad x \:=\:\frac{2}{5}y\)
\(\displaystyle \text{Since }x\text{ and }y\text{ are integers, }y\text{ must be a multiple of 5.}\)
\(\displaystyle \text{Hence: }\:y \:=\:10,15\quad\hdots\;\text{There are two values for }y.\)
. . \(\displaystyle \text{Therefore: }\:x \:=\:4,6 \quad\hdots\;\text{There are }two\text{ values for }x.\)
You can talk your way through this . . .
\(\displaystyle \text{If }x\text{ and }y\text{ are integers, and: }\:7 \,<\,y\,<\,16\,\text{ and }\,\frac{x}{y} \,=\,\frac{2}{5}\)
\(\displaystyle \text{how many possible values are there for }x\text{ ?}\)
. . . \(\displaystyle (A)\;\text{ One}\qquad (B)\;\text{Two} \qquad (C)\;\text{Three} \qquad (D)\;\text{Four} \qquad (E)\;\text{Five}\)
\(\displaystyle \text{We are told that }y\text{ is between 7 and 16: }\;y \,\in\,\{8,9,10,11,12,13,14,15\}\)
\(\displaystyle \text{We are told that: }\:\frac{x}{y} \,=\,\frac{2}{5} \quad\Rightarrow\quad x \:=\:\frac{2}{5}y\)
\(\displaystyle \text{Since }x\text{ and }y\text{ are integers, }y\text{ must be a multiple of 5.}\)
\(\displaystyle \text{Hence: }\:y \:=\:10,15\quad\hdots\;\text{There are two values for }y.\)
. . \(\displaystyle \text{Therefore: }\:x \:=\:4,6 \quad\hdots\;\text{There are }two\text{ values for }x.\)