Question

question 12
solve and graph:
9x + 3 < 2x + 10 and - 5 + 3x ≤ 9x + 1
-5 -4 -3 -2 -1 0 1 2 3 4 5
clear all draw:

Explanation:

Step1: Solve the first inequality

Subtract $2x$ and $3$ from both sides of $9x + 3<2x + 10$.
$9x-2x<10 - 3$
$7x<7$
$x < 1$

Step2: Solve the second inequality

Subtract $3x$ and $1$ from both sides of $-5 + 3x\leq9x+1$.
$-5 - 1\leq9x-3x$
$-6\leq6x$
Divide both sides by $6$: $x\geq - 1$

Step3: Find the intersection

The solution of the compound - inequality is $-1\leq x<1$.
To graph:

• Place a closed circle at $x = - 1$ (because $x$ can equal $-1$) and an open circle at $x = 1$ (because $x$ cannot equal $1$).
• Draw a line segment between them.

Answer:

The solution is $-1\leq x<1$. On the number - line, place a closed circle at $-1$, an open circle at $1$, and draw a line segment connecting them.