# Solving Systems of Linear Inequalities

How do you solve a system of linear inequalities? Aw, man...and we thought solving a problem like Maria was tough...

 Algebra Represent and solve equations and inequalities graphicallySystems of Equations Language English Language Mathematics and Statistics Assessment Linear Equations, Inequalities, and Systems

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00:26

hint. y is greater than x plus 3, and y less than

00:31

or equal to negative 3/2 x plus 1. Thankfully, Black Beard paid attention in

00:37

math class… … and sees that the hint gives them a rough

00:40

idea of where on the map to find that treasure chest.

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Let’s start by graphing the first line: y equals x plus 3.

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The equation is in slope-intercept form, so we can plot 3 on the y-axis as the y-intercept.

01:05

The slope of the line is 1. We know that slope is rise over run, so we can rise 1 units on

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the y-axis, and run over 1 unit to the right. Since the inequality doesn’t include equals

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to, we know the line will be dotted and not solid.

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To see which side of the line we should shade for the inequality, we can test a point on

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one side of the line. For example…(0, 4) If we plug in 0 for x,

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we get that 0 plus 3 is 3. Since our point’s y-value of 4 is greater

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than 3, we know that we can shade the upper half of the line.

01:47

For the second line, we have y is less than or equal to negative 3/2 x plus 1, so we can

01:59

graph y equals to negative 3/2 x plus 1 first. The equation is in slope-intercept form, so

02:07

we can plot 1 on the y-axis as the y-intercept. The slope of the line is -3/2. We know that

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slope is rise over run, so we can go down 3 units on the y-axis, and run over 2 units

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to the right. Since the inequality includes equals to, we

02:26

know the line will be solid. To see which side of the line we should shade

02:31

for the inequality, we can test a point on one side of the line.

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For example…(-2, 0). If we plug in -2 for x, we get that negative 3/2 times -2 equals

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3…3 plus 1 is 4. Since our original y-value of 0 is less than

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4, we know we can shade the lower half of the line.

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Seeing where the two shaded regions intersect, we get our solution.

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Now Black Beard and Red Beard know where to start looking for that treasure.

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Hopefully they can track it down before they’re both White Beard.