Integers only please!

Algebra Level 2

How many integer values of x x satisfy both inequalities below?

{ − 3 x + 10 < 1 8 x + 2 < 50 \large \begin{cases} -3x + 10 < 1 \\ 8x + 2 < 50 \end{cases}

3 4 0 1 2

Denton Young by Denton Young , 47, USA

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2 solutions

Denton Young
Jul 20, 2016

-3x + 10 < 1

-3x < -9

x > 3 (remember, multiplying a negative number flips the inequality)

8x + 2 < 50

8x < 48

x < 6

Since 3 < x < 6 and x can only take integer values, the only possibilities are 4 and 5. So there are two values that work.

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Moderator note:

Simple standard approach.

Nikhil Raj
Jun 1, 2017

Given, { − 3 x + 10 < 1 8 x + 2 < 50 ⟹ { − 3 x < − 9 8 x < 48 ⟹ { x > 3 x < 6 ⟹ 3 < x < 6 ⟹ x = 4 , 5. ∴ No. of solutions = 2 \large \begin{cases} -3x + 10 < 1 \\ 8x + 2 < 50 \end{cases} \\ \implies \large \begin{cases} -3x < -9 \\ 8x < 48 \end{cases} \\ \implies \large \begin{cases} x > 3 \\ x < 6 \end{cases} \\ \implies 3 < x < 6 \implies x = 4,5. \\ \therefore {\text{No. of solutions}} = \color{#3D99F6}{\boxed2}

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