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Holt Algebra 2
2-5 Linear Inequalities in Two Variables
Graph linear inequalities on the coordinate plane.Solve problems using linear inequalities.
Objectives
Linear functions form the basis of linear inequalities. A linear inequality in two variables relates two variables using an inequality symbol, such as y > 2x – 4. Its graph is a region of the coordinate plane bounded by a line. The line is a boundary line, which divides the coordinate plane into two regions.
Holt Algebra 2
2-5 Linear Inequalities in Two Variables
For example, the line y = 2x – 4, shown at right, divides the coordinate plane into two parts: one where y > 2x – 4 and one where y < 2x – 4. In the coordinate plane higher points have larger y values, so the region where y > 2x – 4 is above the boundary line where y = 2x – 4.
Holt Algebra 2
2-5 Linear Inequalities in Two Variables
To graph y ≥ 2x – 4, make the boundary line solid, and shade the region above the line. To graph y > 2x – 4, make the boundary line dashed because y-values equal to 2x – 4 are not included.
Think of the underlines in the symbols ≤ and ≥ as representing solid lines on the graph.
Helpful Hint
Holt Algebra 2
2-5 Linear Inequalities in Two Variables
Graph the inequality .
The boundary line is which has a
y-intercept of 2 and a slope of .
Draw the boundary line dashed because it is not part of the solution.
Then shade the region above the boundary line to show
.
Holt Algebra 2
2-5 Linear Inequalities in Two Variables
Example 1A Continued
Check Choose a point in the solution region, such as (3, 2) and test it in the inequality.
The test point satisfies the inequality, so the solution region appears to be correct.
?
2 > 1 ?
Holt Algebra 2
2-5 Linear Inequalities in Two Variables
Graph the inequality y ≤ –1.
Recall that y= –1 is a horizontal line.
Check The point (0, –2) is a solution because –2 ≤ –1. Note that any point on or below y = –1 is a solution, regardless of the value of x.
Holt Algebra 2
2-5 Linear Inequalities in Two Variables
The boundary line is y = 3x – 2 which has a
y–intercept of –2 and a slope of 3.
Draw a solid line because it is part of the solution.
Then shade the region above the boundary line to show y > 3x – 2.
Graph the inequality y ≥ 3x –2.
Check Choose a point in the solution region, such as (0, 0) and test it in the inequality.
Holt Algebra 2
2-5 Linear Inequalities in Two Variables
If the equation of the boundary line is not in slope-intercept form, you can choose a test point that is not on the line to determine which region to shade. If the point satisfies the inequality, then shade the region containing that point. Otherwise, shade the other region.
The point (0, 0) is the easiest point to test if it is not on the boundary line.
Helpful Hint
Holt Algebra 2
2-5 Linear Inequalities in Two Variables
Graph 3x + 4y ≤ 12 using intercepts.
Step 1 Find the intercepts.
Substitute x = 0 and y = 0 into 3x + 4y = 12 to find the intercepts of the boundary line.
y-intercept x-intercept
3x + 4y = 12
3(0) + 4y = 12 3x + 4(0) = 12
4y = 12
3x + 4y = 12
y = 3
3x = 12
x = 4
Holt Algebra 2
2-5 Linear Inequalities in Two Variables
Step 2 Draw the boundary line.The line goes through (0, 3) and (4, 0). Draw a solid line for the boundary line because it is part of the graph.
Step 3 Find the correct region to shade.Substitute (0, 0) into the inequality. Because 0 + 0 ≤ 12 is true, shade the region that contains (0, 0).
(0, 3)
(4, 0)
Holt Algebra 2
2-5 Linear Inequalities in Two Variables
Graph 3x – 4y > 12 using intercepts.
(4, 0)
(0, –3)
Holt Algebra 2
2-5 Linear Inequalities in Two Variables
Many applications of inequalities in two variables use only nonnegative values for the variables. Graph only the part of the plane that includes realistic solutions.
A school carnival charges $4.50 for adults and $3.00 for children. The school needs to make at least $135 to cover expenses.
A. Using x as the adult ticket price and y as the child ticket price, write and graph an inequality for the amount the school makes on ticket sales.
B. If 25 child tickets are sold, how many adult tickets must be sold to cover expenses?
Holt Algebra 2
2-5 Linear Inequalities in Two Variables
Let x represent the number of adult tickets and y represent the number of child tickets that must be sold. Write an inequality to represent the situation.
An inequality that models the problem is 4.5x + 3y ≥ 135.
135y3.00+x4.50
total.is at least
number of child tickets
timeschild price
plusnumber of
adult tickets
timesAdult price
• •
Holt Algebra 2
2-5 Linear Inequalities in Two Variables
Find the intercepts of the boundary line.
Graph the boundary line through (0, 45) and (30, 0) as a solid line. Shade the region above the line that is in the first quadrant, as ticket sales cannot be negative.
4.5(0) + 3y = 135 4.5x + 3(0) = 135
y = 45 x = 30
Holt Algebra 2
2-5 Linear Inequalities in Two Variables
If 25 child tickets are sold,
Substitute 25 for y in 4.5x + 3y ≥ 135.
Multiply 3 by 25.
A whole number of tickets must be sold.
At least 14 adult tickets must be sold.
4.5x + 3(25) ≥ 135
4.5x + 75 ≥ 135
4.5x ≥ 60, so x ≥ 13.3_
Holt Algebra 2
2-5 Linear Inequalities in Two Variables
Solve for y. Graph the solution.
ADVANCED LEVEL
Solve 2(3x – 4y) > 24 for y. Graph the solution.