Question

question 10
solve and graph:
13x - 6 < 8x + 4 and -16 + 3x ≤ 12x - 7
-5 -4 -3 -2 -1 0 1 2 3 4 5
clear all draw:
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Explanation:

Step1: Solve the first inequality

Subtract $8x$ from both sides of $13x - 6<8x + 4$:
$13x-8x-6<8x - 8x+4$, which simplifies to $5x-6 < 4$. Then add 6 to both sides: $5x-6 + 6<4 + 6$, so $5x<10$. Divide both sides by 5: $x < 2$.

Step2: Solve the second inequality

Subtract $3x$ from both sides of $-16 + 3x\leq12x-7$:
$-16+3x-3x\leq12x-3x - 7$, which simplifies to $-16\leq9x-7$. Add 7 to both sides: $-16 + 7\leq9x-7 + 7$, so $-9\leq9x$. Divide both sides by 9: $x\geq - 1$.

Step3: Find the intersection

The solution of the compound - inequality is the intersection of $x < 2$ and $x\geq - 1$, which is $-1\leq x<2$.

Step4: Graph the solution

On the number - line, put a closed circle at $x=-1$ (because $x$ can equal $-1$) and an open circle at $x = 2$ (because $x$ cannot equal 2), and draw a line segment between them.

Answer:

The solution is $-1\leq x<2$. On the number - line, use a closed circle at $-1$, an open circle at $2$, and draw a line segment between them.