May I please have assistance on inequalities: *solution by calculation*. Thank you for the lessons, they are helpful.

Solving inequalities is a lot like solving normal equations. You still shift around numbers and variables, trying to isolate the variable on one side. The one difference is that whenever you multiply or divide by a negative number, you have to switch the sides of the inequality. Think about it. When you multiply or divide both sides by a normal number, they get larger or smaller by a set amount, and the side that was greater before is still greater now. However, if you multiply by a negative number, the greater side is now more negative than the lesser side will be, which makes the bigger side less than the smaller side. To fix this, we switch the inequality sign. Besides that detail, the process is basically the same as solving an equation.

I was working through this problem for PRACTICE: DETERMINE THE CONDITION FOR NO SOLUTION, but I don't understand why you can claim that 3ax=6x and ignore the -11 and +6. Can you please explain further? "So we have 3ax-11=6x+6. We know that if 3ax=6x, we can remove the x-terms from the equation altogether, leaving us with the impossible equation -11=6 and no solution for the equation."

The goal for this problem is to end up with a value for a that means that the equation has no solution. What does it mean to have no solution? On a graph, you can think of this as both lines being parallel to each other and never intersecting. Algebraically, it means that there is some way to remove the variable completely from both sides of the equation so that you end up with a false statement, like 2 = 3. How do we get there? Well, we have to find a way to do one operation and remove both variables from both sides. This means that the variables should be the same on both sides, so we can subtract/add to cancel them both at once. For the problem, this means that 3ax = 6x, and a = 2. We can ignore the -11 = 6 because that part is added to x and not a coefficient, and that -11 = 6 is what'll be left when we subtract 6x from both sides, leaving us with the false statement we need to say the equation has no solution. Hope this helps!

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SAT

Course: SAT > Unit 6

Lesson 3: Heart of Algebra: lessons by skill

Solving linear equations and linear inequalities | Lesson

Google Classroom

What are linear equations and inequalities, and how frequently do they appear on the test?

Linear equations and inequalities are composed of

and

Linear equations use the equal sign (equals).

[Example]

Linear inequalities use inequality signs (is greater than, is less than, is greater than or equal to, and is less than or equal to).

[Example]

In this lesson, we'll learn to:

Solve linear equations
Solve linear inequalities
Recognize the conditions under which a linear equation has one solution, no solution, and infinitely many solutions

On your official SAT, you'll likely see 2 to 4 questions that test your ability to solve linear equations and inequalities. Additionally, you'll need to know how to solve linear equations and inequalities as a part of solving word problems.

Note: If you're taking the SAT, then chances are you have a good understanding of how to solve linear equations and inequalities. However, since they appear so frequently on the test, it's important to solve them in their various forms with consistency. We recommend that you write out your steps (instead of doing everything in your head) to avoid careless errors, and we will do the same in our examples!

You can learn anything. Let's do this!

How do I solve linear equations?

Reasoning with linear equations

Khan Academy video wrapper

Reasoning with linear equationsSee video transcript

Types of linear equations

The goal of solving a linear equation is to find the value of a variable; we isolate the variable step by step until only the variable is on one side of the equation and only a constant is on the other.

When solving linear equations, the most important thing to remember is that the equation will remain equivalent to the original equation only if we always treat both sides equally: whenever we do something to one side, we must do the exact same thing to the other side.

Linear equations in one variable

Most of these questions on the SAT contain only one variable.

Example: If 2, x, plus, 1, equals, 5, what is the value of x ?

[Show me!]

We may be asked to combine like terms and distribute coefficients when solving.

When combining like terms, recall that:

a, x, plus minus, b, x, equals, left parenthesis, a, plus minus, b, right parenthesis, x

Example: If 2, x, minus, 4, equals, 5, minus, x, what is the value of x ?

[Show me!]

When distributing coefficients, recall that:

a, left parenthesis, b, x, plus minus, c, right parenthesis, equals, a, b, x, plus minus, a, c

Example: If 2, left parenthesis, x, plus, 1, right parenthesis, equals, 5, what is the value of x ?

[Show me!]

Fractions and negative numbers

The presence of fractions and negative numbers can make linear equations more difficult to solve.

When solving a linear equation with fraction coefficients or constants:

If the equation has only a fraction coefficient, consider leaving the fraction until the last step in isolating x.

Example: If start fraction, 1, divided by, 2, end fraction, x, plus, 3, equals, 5, what is the value of x ?

[Show me!]

If the equation has both fraction coefficients and fraction constants, consider getting rid of the fractions in the first step.

Example: What is the solution to the equation start fraction, 1, divided by, 2, end fraction, x, plus, start fraction, 1, divided by, 3, end fraction, equals, start fraction, 1, divided by, 5, end fraction ?

[Show me!]

When working with negative numbers, remember that:

start text, n, e, g, a, t, i, v, e, end text, dot, start text, n, e, g, a, t, i, v, e, end text, equals, start text, p, o, s, i, t, i, v, e, end text
start text, p, o, s, i, t, i, v, e, end text, dot, start text, n, e, g, a, t, i, v, e, end text, equals, start text, n, e, g, a, t, i, v, e, end text

Example: If minus, 2, left parenthesis, x, minus, 5, right parenthesis, equals, 1, what is the value of x ?

[Show me!]

Linear equations in two variables

Sometimes, we're given an equation in two variables and we're told the value of one of the variables. Plug the value of the known variable into the equation and solve.

Example: If 2, x, plus, 5, y, equals, 1 and x, equals, 3, what is the value of y ?

[Show me!]

Using linear equations to evaluate expressions

Sometimes, we'll be given a linear equation in one variable and be asked to evaluate a different expression containing the variable. We can approach this type of question in two ways:

Solve the linear equation, then plug the value of the variable into the expression to evaluate it.
Find the relationship between the equation and the expression, then evaluate the expression without solving for the variable.

Knowing the second approach is not required, though it may save you valuable time on test day.

Example: If 2, x, plus, 1, equals, 5, what is the value of 8, x, plus, 4 ?

[Show me!]

Absolute value equations

Note: Absolute value equations appear very rarely on the SAT.

The absolute value of a number is equal to the number's distance from 0 on the number line, which means the absolute value of a nonzero number is always positive. For example:

The absolute value of 2, or vertical bar, 2, vertical bar, is 2.
The absolute value of minus, 2, or vertical bar, minus, 2, vertical bar, is also 2.

Practically, this means every absolute value equation can be split into two linear equations. For example, if vertical bar, 2, x, plus, 1, vertical bar, equals, 5:

The absolute value equation is true if 2, x, plus, 1, equals, 5.
The absolute value equation is also true if 2, x, plus, 1, equals, minus, 5 since vertical bar, minus, 5, vertical bar, equals, 5.

When solving absolute value equations, rewrite the equation as two linear equations, then solve each linear equation. Both solutions are solutions to the absolute value equation.

Example: What are the solutions to the equation vertical bar, 2, x, minus, 1, vertical bar, equals, 5 ?

[Show me!]

Try it!

Try: identify the steps to solving a linear equation

7, minus, 3, x, equals, 28

To solve the equation above, we can first

both sides of the equation to isolate the x-term.

Next, we can

both sides of the equation by minus, 3.

What is the value of x ?

How do I solve linear inequalities?

Multi-step inequalities

Khan Academy video wrapper

Inequalities with variables on both sidesSee video transcript

Types of linear inequalities

The steps for solving linear inequalities are similar to those for solving linear equations. For inequalities, we have to pay attention to the direction of the inequality signs.

Linear inequalities that do not require reversing the inequality sign

When the coefficient of x is positive, the inequality sign maintains its direction when we divide by the coefficient to isolate x.

Example: What values of x satisfy the inequality 2, x, plus, 1, is greater than, 5 ?

[Show me!]

Linear inequalities that require reversing the inequality sign

When the coefficient of x is negative, we must reverse the direction of the inequality sign when we divide by the coefficient to isolate x.

Example: What values of x satisfy the inequality minus, 2, x, plus, 1, is greater than, 5 ?

[Show me!]

Try it!

try: identify the steps to solving a linear inequality

minus, 2, x, is greater than, 9, plus, 4, x

To solve the inequality above, we can first

both sides of the equation.

Next, we can

both sides of the equation by minus, 6 and

x is

minus, start fraction, 3, divided by, 2, end fraction.

How do I alter the number of solutions for linear equations?

Note: Questions about the number of solutions for linear equations do not appear on every test.

Creating an equation with no solutions

Khan Academy video wrapper

Creating an equation with no solutionsSee video transcript

Creating an equation with infinitely many solutions

Khan Academy video wrapper

Creating an equation with infinitely many solutionsSee video transcript

How many solutions can a linear equation have?

Most linear equations on the SAT have exactly one solution. Linear equations with no solutions or infinitely many solutions must be engineered by specifying the values of constants.

For a linear equation in one variable:

If the equation can be rewritten in the form x, equals, a, where a is a constant, then that equation has one solution.
If the variable can be eliminated from the equation, and what remains is the equation a, equals, b, where a and b are different constants, then the equation has no solution. (No value of x can make 1 equal to 2!)
If the equation can be rewritten in the form x, equals, x, then the equation has infinitely many solutions. (No matter what the value of x is, it will always equal itself!)

Let's look at some examples!

2, x, minus, 1, equals, a, x, minus, 2

If a, equals, 2 in the equation above, what value of x satisfies the equation?

[Show me!]

2, x, minus, 4, equals, a, left parenthesis, x, minus, 2, right parenthesis

If a, equals, 2 is the equation above, what value of x satisfies the equation?

[Show me!]

Try it!

try: change the number of solutions for a linear equation

5, x, plus, 3, equals, a, x, plus, b

In the equation above, a and b are constants.

The equation has a single solution if a

and b is a real number.

The equation has infinitely many solutions if a is

and b is

Your turn!

Practice: evaluate a linear expression

If 6, x, plus, 10, equals, 24, what is the value of 3, x, plus, 5 ?

Practice: solve a linear equation in one variable

2, left parenthesis, 3, x, plus, 1, right parenthesis, minus, left parenthesis, 9, minus, 2, x, right parenthesis, equals, 25

What value of x satisfies the equation above?

Practice: find a value that does not satisfy an inequality

Which of the following numbers is NOT a solution to the inequality 7, x, minus, 3, is less than or equal to, 3, x, plus, 11 ?

practice: determine the condition for no solution

3, a, x, minus, 11, equals, 4, left parenthesis, x, plus, 2, right parenthesis, plus, 2, left parenthesis, x, minus, 1, right parenthesis

In the equation above, a is a constant. If no value of x satisfies the equation, what is the value of a ?

Things to remember

For the absolute value equation vertical bar, a, x, plus, b, vertical bar, equals, c, rewrite the equation as the following linear equations and solve them.

a, x, plus, b, equals, c
a, x, plus, b, equals, minus, c

Both solutions are solutions to the absolute value equation.

When solving linear inequalities:

If the coefficient of x is positive, the inequality sign maintains its direction when we divide by the coefficient to isolate x.
If the coefficient of x is negative, we must reverse the direction of the inequality sign when we divide by the coefficient to isolate x.

When determining the number of solutions for a linear equation:

If the equation can be rewritten in the form x, equals, a, where a is a constant, then that equation has one solution.
If the variable can be eliminated from the equation, and what remains is the equation a, equals, b, where a and b are different constants, then the equation has no solution.
If the equation can be rewritten in the form x, equals, x, then the equation has infinitely many solutions. (No matter what the value of x is, it will always equal itself!)

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omereliwa51
Posted 3 years ago. Direct link to omereliwa51's post “thank you very much khan ...”
thank you very much khan academy this is so helpful.Can you please do some lessons for the additional topics in math
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SheIs
Posted 2 years ago. Direct link to SheIs's post “May I please have assista...”
May I please have assistance on inequalities: solution by calculation. Thank you for the lessons, they are helpful.
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Answer
- Hecretary Bird
  Posted 2 years ago. Direct link to Hecretary Bird's post “Solving inequalities is a...”
  Solving inequalities is a lot like solving normal equations. You still shift around numbers and variables, trying to isolate the variable on one side. The one difference is that whenever you multiply or divide by a negative number, you have to switch the sides of the inequality. Think about it. When you multiply or divide both sides by a normal number, they get larger or smaller by a set amount, and the side that was greater before is still greater now. However, if you multiply by a negative number, the greater side is now more negative than the lesser side will be, which makes the bigger side less than the smaller side. To fix this, we switch the inequality sign. Besides that detail, the process is basically the same as solving an equation.
  Button navigates to signup page
  (10 votes)
Kenzzzzzzzzzzie
Posted 2 years ago. Direct link to Kenzzzzzzzzzzie's post “I was working through thi...”
I was working through this problem for PRACTICE: DETERMINE THE CONDITION FOR NO SOLUTION, but I don't understand why you can claim that 3ax=6x and ignore the -11 and +6. Can you please explain further?

"So we have 3ax-11=6x+6. We know that if 3ax=6x, we can remove the x-terms from the equation altogether, leaving us with the impossible equation -11=6 and no solution for the equation."
Button navigates to signup pageComment on Kenzzzzzzzzzzie's post “I was working through thi...”
(3 votes)
Answer
- Hecretary Bird
  Posted 2 years ago. Direct link to Hecretary Bird's post “The goal for this problem...”
  The goal for this problem is to end up with a value for a that means that the equation has no solution. What does it mean to have no solution? On a graph, you can think of this as both lines being parallel to each other and never intersecting. Algebraically, it means that there is some way to remove the variable completely from both sides of the equation so that you end up with a false statement, like 2 = 3.
  
  How do we get there? Well, we have to find a way to do one operation and remove both variables from both sides. This means that the variables should be the same on both sides, so we can subtract/add to cancel them both at once.
  For the problem, this means that 3ax = 6x, and a = 2. We can ignore the -11 = 6 because that part is added to x and not a coefficient, and that -11 = 6 is what'll be left when we subtract 6x from both sides, leaving us with the false statement we need to say the equation has no solution. Hope this helps!
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ayesha31
Posted a year ago. Direct link to ayesha31's post “Thank you so much Sir. I ...”
Thank you so much Sir. I keep Remember You in my prayers.
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- clark.curtis
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  Were all stars to disappear or die,
  I should learn to look at an empty sky
  And feel its total dark sublime,
  Though this might take me a little time.
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Question
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Bryan100
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how many honëybünzz in lëh glëëka spëëk?
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Aryan
Posted 2 years ago. Direct link to Aryan's post “i have a doubt,'A sound t...”
i have a doubt,'A sound technician analyzes the audio feedback by placing a microphone at certain distances from a speaker. If the microphone is connected to the speaker, then the microphone senses 60 decibels (dB) at a distance of 0 meters (m) from the speaker with the decibel level decreasing by half of itself for every additional meter from the speaker. If the microphone is not connected to the speaker, then the microphone senses 30 dB at a distance of 0 m from the speaker with the decibel level decreasing by 8 for every additional meter from the speaker. Three meters from the speaker, what is the difference between the decibel level when it is connected to the speaker versus when it is not connected to the speaker?
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- Hecretary Bird
  Posted 2 years ago. Direct link to Hecretary Bird's post “So we have two different ...”
  So we have two different scenarios here, and want to compare them. The easiest thing to do is to calculate the decibel level for each situation (whether or not the mic is connected) and then find the difference after. If it's connected to the speaker, the decibel level decreases by half. This means we have to write an exponential function to model this. 60 is our initial value, .5 is our rate, and the distance will be the exponent (which tells you how much the rate compounds). You should get this:
  dB = 60(.5)^m
  If the mic isn't connected, it's more straightforward. We have a straight line equation here as it's decreasing by a fixed amount, with a slope of -8 and y-intercept of 30:
  dB = -8*m + 30
  
  Now that we have two equations, we just have to calculate the values for 3 meters, and then subtract one from the other to get a difference:
  dB_1 = 60(.5)^3
  dB_1 = 7.5
  dB_2 = -8(3) + 30
  dB_2 = 6
  
  dB_1 - dB_2 = 1.5 decibels.
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