Graph the solution set for the of linear inequalities y&lt;13x+1 y≥13x−2

Step 1 We make tables for the auxiliary equations. Step 2 Then we plot the points and graph the lines. then we shade the region depending on the inequalities. (0,0) means x=0 , y=0 satisfies both inequalities. 0&lt;1 and 0≥−2 y&lt;13x+1→0&lt;13(0)+1→0&lt;1 y≥13x−2→0≥13(0)−2→0≥−2 For y&lt;(13)x+1 we draw dotted line because strict inequality. For y≥(13)x−2 we draw solid line because equality holds. For both shaded towards (0,0)

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High School
Algebra
Algebra I
Inequalities systems and graphs

Cabiolab

Answered question

2020-11-02

Graph the solution set for the of linear inequalities
$y < \frac{1}{3} x + 1$
$y \geq \frac{1}{3} x - 2$

Answer & Explanation

Derrick

Skilled2020-11-03Added 94 answers

Step 1
We make tables for the auxiliary equations.
Step 2
Then we plot the points and graph the lines. then we shade the region depending on the inequalities. (0,0) means x=0 , y=0 satisfies both inequalities. 0<1 and $0 \geq - 2$
$y < \frac{1}{3} x + 1 \to 0 < \frac{1}{3} (0) + 1 \to 0 < 1$
$y \geq \frac{1}{3} x - 2 \to 0 \geq \frac{1}{3} (0) - 2 \to 0 \geq - 2$
For $y < (\frac{1}{3}) x + 1$ we draw dotted line because strict inequality.
For $y \geq (\frac{1}{3}) x - 2$ we draw solid line because equality holds.
For both shaded towards (0,0)

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