Linear inequalities and systems of linear inequalities on the SAT Test

SAT Subscore: Heart of Algebra

Studying linear inequalities and systems of linear inequalities

Note that linear inequalities topic includes linear inequalities and systems of linear inequalities.

On the SAT test linear inequalities are part of heart of algebra subscore that includes 4 fundamental topics that appear in many SAT questions. The 4 topics are:

Linear functions– Start studying heart of algebra subscore with linear functions topic.

Linear equations and systems of linear equations– Continue to these 2 topics after you study linear functions topic.

linear inequalities– The current topic, it is the last topic of heart of algebra subscore.

Linear inequalities topic is divided into sections from easy to difficult (the list of the sections appears on the left menu). Each section includes detailed explanations of the required material with examples followed by a variety of self-practice questions with solutions.

Linear inequalities and systems of linear inequalities- summary

A linear inequality is linear equation in which the equal sign is replaced by one of the symbols of inequality. The solution of a linear equation is a range of values, rather than one specific value.

A system of linear inequalities is like a system of linear equations, the difference is in the sign. We use the system of inequalities when we a have a word problem with number of constrains instead of one (we write an inequality for each constrain).

Creating and solving a linear inequality requires creating a linear inequality from a word problem and solving it. Note that solving a linear inequality may require changing the direction of the inequality sign.

Finding values that don’t satisfy an inequality– In these questions we need to find values that are not a solution for a linear inequality.

Graphic presentation of a linear inequality requires identifying a graphic presentation of an inequality. The answers to the inequality are represented by the area below or above the line of its equation.

Creating a system of linear inequalities- In these questions we need to create a system of linear inequalities from a word problem.

Finding possible solutions for a system of linear inequalities– In these questions we need to decide which values satisfy a given system of linear inequalities. Sometimes we will need to translate a word problem into a system of inequalities as explained before.

Solving a system of linear inequalities requires turning a system of linear inequalities to one inequality that can be solved easily. This can be achieved using additional data or by turning one of the inequalities to an equation. We may also need to translate a word problem to a system of inequalities.

Graphic presentation of a system of linear inequalities– In these questions we need to identify a graphic presentation of a system of inequalities given its formulas or to write the formulas of the system that correspond to a given graphic presentation.

Creating and solving a linear inequality

In these questions you need to create a linear inequality from a word problem and solve it.

The skills required:
Translating a word problem into an inequality.
Defining x variable.
Solving inequality with one variable.

Statements representing the inequalities in the context of the question

First of all, you need to recognize that the question is about inequality, look for the statements representing the inequalities in the context of the question:
Statements for greater than sign ≥: at least, more than, no less than, greater than, minimum value.
Statements for less than sign ≤: less than, at most, no more than, smaller than, maximum value, the greatest value.

Rules for solving a linear inequality

The direction of the inequality sign:
When we divide by a negative number, we must change the inequality sign direction. This happens when the x variable sign is negative, therefore we need to divide by a negative number to isolate x.

Range of solutions instead of one:
In contrast to a linear equation where we can find a single solution there is a range of solutions for a linear inequality. If the solution is x>5 the value of the solution can be 6, 7,8……

Rounding the answer:
If the answer is a decimal value and it needs to be a whole number, you need to round it.
Round up example: x≥4.5, when x represents the minimum number of students in the class. The answer is x≥5.
Round down example: x≤38.6, when x represents the maximum number of students in the bus. The answer is x≤38.

Solving a linear inequality without changing the direction of the inequality sign

Solving linear inequalities is similar to solving of linear equations except for treating the inequality sign. When we divide by a negative number, we must change the inequality sign direction.

Consider this example:

In the previous working day, the machine produced 152 products. Every hour the machine produces 10 more products. Twenty percent of the products are disqualified (turning out to be defective) after a quality control check. The minimum quantity of products for delivery is 170. If the current working day starts at 9 am, at what time will it be possible to prepare the shipment?

Since the question is about a minimum quantity, we will need to construct an inequality in order to reach a solution. The quantity should be at least 170 so we will write an expression that should be greater than 170 (instead of equal to 170)

Step 1: X definition: We know the initial quantity of the products and their rate of production. The question is how soon will we reach 170 products, therefore the variable x should be the time (we define x as the number of hours).

Step 2: Writing the expressions in each side of the inequality: The expression on one side of the inequality will be the products in stock from yesterday plus the products that the machine produces today and on the other side of the inequality we will write 170.
152+10*(1-0.2)*x>=170

Step 3: Solving inequality:
152+10*0.8x≥170
152+8x≥170
8x≥18
x≥2+_²⁄8
x≥2+_¹⁄4
x≥2_¹⁄4

Since the working day starts at 9 o’clock and it takes at least 2.25 hours to reach 170 products, the hour when it will be possible to prepare the delivery will be 11 o’clock plus 0.25 hours. That is 60 minutes double 0.25 (quarter) = 15 minutes. The answer is that the time when it’s possible to prepare the shipment is after 11:15.

Step 4: checking the answer:
In order to test the answer we will solve the left side of the inequality: 152+10*0.8*2.25= 152+8*2.25= 152+16+8*0.25=152+16+2=170

Solving a linear inequality that requires a change in the direction of the inequality sign

Pay attention that when we divide by a negative number, we must change the inequality sign direction!

Consider this example:

Solve the inequality -4x+10<18

Step 1: Solving the inequality:
-4x+10<18
-4x<8-18-10
-4x<8
Now we need to divide both sides by a negative number -4 so we will change the sign direction:
-4x/-4>8/-4
x>-2

Step 2: Checking the answer
We can place any number that is bigger than -2 and see if the inequality holds. Let’s take x=0.
-4*0+10<18
10<18 is a true statement

Note that if we don’t change the direction of the sign the answer x<-2 is incorrect.
For x=-3 for example we will get
-4*-3+10<18
22<18 is a false statement.

Finding values that don't satisfy an inequality

In these questions you need to find values that are not a solution for a linear inequality.

The skills required:
Solving inequality with one variable.
Identifying values that are not included in the solution of the inequality.

If we are asked to find values that are not a solution for the inequality, we first need to solve the inequality. All the values that are not included in the solution are values that don’t satisfy the inequality.

Consider this example:

Which values are not a solution to the inequality -4x+10<18?

We solved this inequality in the previous example so we know that the solution is x>-2.
Therefore the values that are not a solution are x≤-2.

Checking the solution:
Let’s plug -2 and a number that is smaller than -2 like -3 into the inequality -4x+10<18

For x=-2:
-4*-2+10<18
8+10<18
18<18 is a false statement

For x=-3:
-4*-3+10<18
12+10<18
22<18 is a false statement

Graphic presentation of a linear inequality

In these questions you need to identify a graphic presentation of an inequality. The answers to the inequality are represented by the area below or above the line of its equation.

The skills required:
Writing an inequality in a slope intercept form.
Graphing the line of the inequality in the xy plane.
Marking the area that corresponds to the inequality sign.

Steps for graphing an inequality

Step 1: Write the inequality in a slope intercept form y>mx+b or y<mx+b.
Step 2: Draw the graph of the inequality as a straight line (as if the sign is =).
Step 3: Shade the range of inequality by its sign:
If y<mx+b shade the area below the graph.
If y>mx+b shade the area above the graph.

Consider this example:

Graph the inequality y+4x>14.

Step 1: Writing the inequality in a slope intercept form y=mx+b:
y+4x>14
y>-4x+14

Step 2: Drawing the equality graph: the graph is represented by the line y=-4x+14 in orange.

Step 3: Shading the area above the graph:
For every x, instead of marking the point where y is equal to -4x+14, we need to mark all the points for which y is bigger than -4x+14. That will result in a yellow area above the graph.

Note that the region y<-4x+14 is the white area below the graph.

Creating a system of linear inequalities SAT questions

In these questions you need to create a system of linear inequalities from a word problem.

The skills required:
Defining variables.
Finding the statements representing the inequalities in the context of the question.
Writing two inequalities with two variables.

Statements representing the inequalities in the context of the question

Consider this example:

On the first day the tourists rode the bus at a speed of x km per hour. On the second day they walked at a speed of y km per hour. In total the tourists traveled z km and walked k km. The tour guide wants to cover a total distance of at most 100 km and the time she wants to walk and ride the bus is at most 15 hours. Write a system of inequalities that represents her plan.

Step 1: writing the first inequality: The tour guide wants to cover a total distance of at most 100 km and we know that the tourists traveled z km and walked k km.
Therefore the first inequality is simple: z+x≤100.

Step 2: writing the second inequality: The time that the guide wants to walk and ride the bus is at most 15 hours so we need to write an equation from riding time plus walking time.
Riding time equals to the riding distance z km divided by riding speed x km.
Walking time equals to the walking distance k km divided by walking speed y km.
Therefore the second equation will be z/x+k/y≤15

Finding possible solutions for a system of linear inequalities SAT questions

In these questions you need to decide which values satisfy a given system of linear inequalities. Sometimes you will need to translate a word problem into a system of inequalities as explained before.

The skills required:
Finding the values that meet the constrains of both inequalities by plugging possible solutions into each inequality.

Consider this example:

Daniel needs to send gifts to ten customers. Each customer should receive at least one gift. The price of a book is $7 and the price of a chocolate box of is $5. The shipping cost is $ 1 per gift. Daniel’s budget is $80. Offer 3 options for gift combinations.

Step 1: Variables definition– Define x as the books amount and y as the chocolate boxes amount.

Step 2: Writing the inequalities:
x+y≥10
(7+1)*x+(5+1)*y≤80

Step 3: Writing possible solutions: Since there is a budget limit in this question, we should try to choose first the smallest x and y. We will start with a x=1 and choose the smallest value of y so that the first condition x + Y> 10 will hold in equality y=9. We will plug these x and y values in the second condition and check whether the second condition is met.
1+9≥10
8*1+6*9=62, 62≤80
We see that the conditions are met, therefore x=1, y=9 is a possible solution for the inequalities system.
To get more answers we can continue to increase x and y values as shown in the table below. Once the second condition is not met (the budget gets above 80) we continue to examine options where x=2.

Solving a system of linear inequalities SAT questions

In these questions you need to turn a system of linear inequalities to one inequality that can be solved easily. This can be achieved using additional data or by turning one of the inequalities to an equation.
You may also need to translate a word problem to a system of inequalities.

The skills required:
Plugging a given value into an inequality.
Finding x or y maximum or minimum value.
Solving inequality with one variable

Solving a system of linear inequalities using additional data

If we receive x or y value we have only 1 variable so we can find a solution. Since the inequalities remain, we may get a number of solutions.

Steps for solving the system using an additional data:
Step 1: Plugging the additional data into each one of the inequalities.
Step 2: Finding the solution by combining the answers from two inequalities.

Consider this example:

If Daniel buys 5 books, what is the possible number of chocolate boxes that he can buy?

Step 1: Plugging the additional data into each one of the inequalities:

We need to substitute x=5 into the first inequality and solve it:
x+y≥10
5+y≥10
y≥5

We can substitute x=5 into the second inequality and solve it:
8x+6y≤80
8*5+6y≤80
6y≤40
y≤40/6
y≤6 4/6
y≤6 2/3
The answer is y<6 since y can’t be a fraction.

Step 2: Combining the answers from two inequalities:
The answer is x=5, 5≤y≤6 meaning we have 2 answers (5,5) or (5,6).

Step 3: Checking the answers:
We can check the answer (5,5): 5+5≥10, 8*5+6*5=70 ≤80.
We can check the answer (5,6): 5+6≥10, 8*5+6*6=76 ≤80.
We can check a bigger answer (5,7) to see that it doesn’t satisfy the second condition: 5+7≥10, 8*5+6*7=82 ≥80.

Finding maximum or minimum value of a variable

There are questions in which you can be asked about x or y maximum or minimum value.

Consider this example:

Daniel needs to send gifts to ten customers. Each customer should receive at least one gift. The price of a book is $ 7 and the price of a chocolate box of is $ 5. The shipping cost is $ 1 per gift. Daniel’s budget is $ 80.

What is the maximum number of books that Daniel can buy?

In this case we need to turn one of the inequalities to an equality and isolate y by writing it in a form of y= . Then we need to plug the value of y into the inequality and solve it for x that we are asked about.

Note that we are asked about the value of x so we need to leave the x variable and isolate y variable.

Also note that we are also limited by the budget therefore we need to solve the second inequality with the x variable and turn the first inequality x+y≥10 to y=10-x.

Step 1: Turning the first inequality to an equation and finding y=:
x+y≥10
y=10-x

Step 2: Substituting 10-x for y in the second inequality:
8x+6y≤80
8x+6(10-x)≤80
8x+60-6x≤80
2x≤20
x≤10

The maximum number of books is 10.

Step 3: Checking the answer:
Plugging x=10 in the second inequality that was limited will give us 80+6y≤80. For y=0 this inequality exists as 80≤80.
Plugging x=11 which is not in the answer range will give us 88+6y≤80. Even for y=0 this inequality doesn’t exist because 88≤80 is false statement.
As we see the answer is x≤10.

Graphic presentation of a system of linear inequalities SAT questions

In these questions you need to identify a graphic presentation of a system of inequalities given its formulas or to write the formulas of the system that correspond to a given graphic presentation.
The answer to the system of the inequalities is represented by the area below or above the lines of both equations.

The skills required:
Writing an inequality in a slope intercept form.
Graphing the line of the inequality in the xy plane.
Finding areas that correspond to the inequality signs of each equation and both equations.
Finding the formulas from the graphic presentation of an inequality system.

Graphic presentation of a system of linear inequalities that has a solution

Drawing the system of the inequalities will make it easy to see their possible solutions.

Steps for graphing a system of inequalities:
Step 1: Write each inequality in a slope intercept form y<mx+b or y>mx+b
Step 2: Draw the graphs of the inequalities as straight lines (as if the signs are =).
Step 3: Shade the ranges of the inequalities by their signs:
If y<mx+b shade the area below the graph.
If y>mx+b shade the area above the graph.
Step 4: The solution is only the area that was shaded twice, because both of the conditions must be met.

Consider this example:

Graph the solution of the system of the linear inequalities
8<4x+y
6>-2x+y

Step 1: Writing each inequality in a slope intercept form:
8<4x+y
y>-4x+8

6>-2x+y
y<2x+6

Step 2: Drawing the graphs of the inequalities as straight lines: The graphs are in blue and orange.

Step 3: Shading the ranges of the inequalities by their signs:
For y>-4x+8 we shade the area above the graph, this area is marked in yellow.
For y<2x+6 we shade the area below the graph, this area is marked in grew.

Step 4: Finding the solution: The solution is only the area that was shaded twice, because both of the conditions must be met. The area is marked in dark grew.

Graphic presentation of a system of linear inequalities with no solution

If there is no area that is shaded twice then the system of the inequalities that has no solution.

Consider this example:

Graph the solution of the system of the linear inequalities
y>-_²⁄3x+_⁵⁄3
y<-_²⁄3x+_⁷⁄9

Step 2: Drawing the graphs of the inequalities as straight lines: The graphs are in blue and orange.

Step 3: Shading the ranges of the inequalities by their signs:
For y>-_²⁄3x+_⁵⁄3 we shade the area above the graph, this area is marked in gray.
For y<-_²⁄3x+_⁷⁄9 we shade the area below the graph, this area is marked in yellow.

Step 4: Finding the solution: The solution is only the area that was shaded twice, because both of the conditions must be met. The is no such area, therefore there is no solution for the inequalities system.

Note that you can also understand why this system of equations has no solution by testing the equations. Plug x=1 in the equations, according to the first equation y should be greater than 1 and according to the second equation y should be less than 1/9. Those cannot exist at the same time.

Writing inequalities represented by a given graphs of linear equations system

In these questions the graphs of the inequality system are given and you need to find the formulas.

Consider this example:

Write the inequalities represented by the green area below for the system of the linear equations
y=3x-3
y=-2x+5

The equation y=3x-3 is represented by the blue graph. We can see that because of its positive slope (m=3) and its y intercept (for x=0 y=-3). Therefore, we know that the shaded area is above the graph, this means that the inequality should be y>3x-3.

The equation y=-2x+5 is represented by the orange graph. We can see that because of its negative slope (m=-2) and its y intercept (for x=0 y=5). Therefore, we know that the shaded area is under the graph, this means that the inequality should be y<-2x+5.

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