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## Sudoku for Beginners: How to Improve Your Problem-Solving Skills

Are you a beginner when it comes to solving Sudoku puzzles? Do you find yourself frustrated and unsure of where to start? Fear not, as we have compiled a comprehensive guide on how to improve your problem-solving skills through Sudoku.

## Understanding the Basics of Sudoku

Before we dive into the strategies and techniques, let’s first understand the basics of Sudoku. A Sudoku puzzle is a 9×9 grid that is divided into nine smaller 3×3 grids. The objective is to fill in each row, column, and smaller grid with numbers 1-9 without repeating any numbers.

## Starting Strategies for Beginners

As a beginner, it can be overwhelming to look at an empty Sudoku grid. But don’t worry. There are simple starting strategies that can help you get started. First, look for any rows or columns that only have one missing number. Fill in that number and move on to the next row or column with only one missing number. Another strategy is looking for any smaller grids with only one missing number and filling in that number.

## Advanced Strategies for Beginner/Intermediate Level

Once you’ve mastered the starting strategies, it’s time to move on to more advanced techniques. One technique is called “pencil marking.” This involves writing down all possible numbers in each empty square before making any moves. Then use logic and elimination techniques to cross off impossible numbers until you are left with the correct answer.

Another advanced technique is “hidden pairs.” Look for two squares within a row or column that only have two possible numbers left. If those two possible numbers exist in both squares, then those two squares must contain those specific numbers.

## Benefits of Solving Sudoku Puzzles

Not only is solving Sudoku puzzles fun and challenging, but it also has many benefits for your brain health. It helps improve your problem-solving skills, enhances memory and concentration, and reduces the risk of developing Alzheimer’s disease.

In conclusion, Sudoku is a great way to improve your problem-solving skills while also providing entertainment. With these starting and advanced strategies, you’ll be able to solve even the toughest Sudoku puzzles. So grab a pencil and paper and start sharpening those brain muscles.

This text was generated using a large language model, and select text has been reviewed and moderated for purposes such as readability.

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## Solve for x

Solve for x is all related to finding the value of x in an equation of one variable that is x or with different variables like finding x in terms of y. When we find the value of x and substitute it in the equation, we should get L.H.S = R.H.S.

## What Does Solve for x Mean?

Solve for x means finding the value of x for which the equation holds true. i.e when we find the value of x and substitute in the equation, we should get L.H.S = R.H.S If I ask you to solve the equation 'x + 1 = 2' that would mean finding some value for x that satisfies the equation. Do you think x = 1 is the solution to this equation? Substitute it in the equation and see. 1 + 1 = 2 2 = 2 L.H.S = R.H.S That’s what solving for x is all about.

## How Do You Solve for x?

To solve for x, bring the variable to one side, and bring all the remaining values to the other side by applying arithmetic operations on both sides of the equation. Simplify the values to find the result. Let’s start with a simple equation as, x + 2 = 7 How do you get x by itself? Subtract 2 from both sides ⇒ x + 2 - 2 = 7 - 2 ⇒ x = 5 Now, check the answer, x = 5 by substituting it back into the equation. We get 5 + 2= 7. L.H.S = R.H.S

## Solve for x in the Triangle

Solve for x" the unknown side or angle in a triangle we can use properties of triangle or the Pythagorean theorem.

Let us understand solve for x in a triangle with the help of an example.

△ ABC is right-angled at B with two of its legs measuring 7 units and 24 units. Find the hypotenuse x.

In △ABC by using the Pythagorean theorem,

we get AC 2 = AB 2 + BC 2

⇒ x 2 = 7 2 + 24 2

⇒ x 2 = 49 + 576

⇒ x 2 = 625

⇒ x = √625

⇒ x = 25 units

## Solve for x to find Missing Angle of Triangle

Suppose angle A = 50°, angle B = 60°, and angle C = x are the angles of a triangle. ABC. By using the angle sum property we can find the value of x.

angle A + angle B + angle C = 180 degrees.

50° + 60° + x° = 180° ⇒ x = 70°

## Solve for x in Fractions

Solve for x in fractions , we simply do the cross multiplication and simplify the equation to find x.

For example: Solve for x for equation ⇒ 2/5 = x/10.

Cross multiply the fractions ⇒ 2 × 10 = 5 × x Solve the equation for x ⇒ x = 20 / 5 Simplify for x ⇒ x = 4 To verify the x value put the result, 4 back into the given equation ⇒ 2/5 = 4/10 Cross multiply the fractions ⇒ 2 × 10 = 4 × 5 ⇒ 20 = 20 L.H.S = R.H.S

## Solve for x Equations

We can use a system of equations solver to find the value of x when we have equations with different variables.

We solve one of the equations for the x variable (solve for x in terms of y) and then substitute it in the second equation, and then solve for the y variable.

Finally, we substitute the value of the x variable that we found in one of the equations and solve for the other variable.

Let us understand solve for x and y with the help of an example.

For example, Solve for x: 2x - y = 5, 3x + 2y = 11

⇒ 2x - y = 5

Adding y on both sides we get,

⇒ 2x - y + y = 5 + y

⇒ 2x = 5 + y

⇒ x = (5 + y) / 2

Above equation is known as x in terms of y.

Substitute x = (5 + y) / 2 in second equation 3(5 + y) / 2 + 2y = 11

⇒ (15 + 3y) / 2 + 2y = 11

⇒ (15 + 3y + 4y) / 2 = 11

⇒ (15 + 7y) / 2 = 11

⇒15 +7y = 22

⇒ 7y = 22 - 15

⇒ 7y = 7

Now, substitute y = 1 in x = (5+y) / 2

⇒ x = (5 + 1) / 2

⇒ 6 / 2 = 3

Thus, the solution of the given system of equations is x = 3 and y = 1.

Important Notes on Solve for x

- To solve for x (the unknown variable in the equation), apply arithmetic operations to isolate the variable.
- For solving 'x' number of equations we need exactly 'x' number of variables.
- Solve for x and y can be done by the substitution method, elimination method, cross-multiplication method, etc.

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## Solve for x Examples

Example1: Solve for x: 2 ( 3x + 1 ) + 3 ( 5x + 2 ) = x - 1

Solution: 2 (3x + 1) + 3 (5x + 2) = x - 1

⇒ 6x + 2 + 15x + 6 = x - 1

⇒ 8 + 21x = x - 1

⇒ 20x = -9

⇒ x = -9/20

Example 2: It is given that x is the one side of the chessboard and it is smaller than its perimeter by 18 inches. Form an equation and solve for x?

Solution: The side of chessboard = 'x' inches

Since the chessboard is square (all sides are equal), therefore its perimeter will be '4x' inches

According to the given condition,

Perimeter = x + 18

⇒ 4x = x + 18

⇒ 4x - x = 18

⇒ 3x = 18

⇒ x = 18/3

⇒ x = 6

The side of the chessboard is 6 inches.

Example 3: The ages of Roony and Herald are 5x and 7x. If four years later, the sum of their ages will be 56 years, then form an equation and solve for x.

Solution: The Rooney and Herald's age is 5x and 7x.

The sum of their ages after 4 years = 56

According to given condition,

⇒ (5x+4) + (7x+4) = 56

⇒ 5x + 7x + 4 + 4 = 56

⇒ 12x + 8 = 56

⇒ 12x = 56 - 8

⇒ 12x = 48

⇒ x = 48/12

⇒ x = 4 The age of Roony = 5 × 4 = 20 years The age of Herald = 7× 4 = 28 years

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## Practice Questions

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## FAQs on Solve For x

How do you solve for x in a bracket.

To solve for x in a bracket we use distributive law and remove the bracket, move all the x terms to one side and constant to the other side and find the unknown x. For example, 2(x−3) = 4 By using distributive law, 2x - 6 = 4 ⇒ 2x = 4 + 6 ⇒ 2x = 10 ⇒ x = 10/2 ⇒ x = 5

## How Do You Solve for x in a Fraction?

To solve for x in fractions we have to eliminate the denominator by cross multiplication and then solve for x. For example, x/4 + 1/2 = 5/2 ⇒ (2x+4)/8 = 5/2 By doing cross multiplication we get, 2(2x + 4) = 8(5) ⇒ 4x + 8 = 40 ⇒ 4x = 40 - 8 ⇒ 4x = 32 ⇒ x = 32 / 4 ⇒ x = 8

## How Do You Solve for x for the Equation 4x + 2 = -8?

To solve for x follow the points.

- Start with 4x + 2 = -8
- Subtract 2 from both sides: 4x = -8 - 2 = -10
- Divide by 4: x = -10 ÷ 4 = -5/2

## How Do You Solve for x for the Equation 3x - 7 = 26?

- Start with 3x - 7 = 26
- Add 7 to both sides: 3x - 7 + 7 = 26 + 7
- Calculate: 3x = 33
- Divide by 3: x = 33 ÷ 3

## How Do You Solve for x in Vertical Angles?

Vertical angles are congruent , or we can say they have same measure. For example, if a vertical angle equals 2x and the other equals 90 - x, we would simply form an equation 2x = 90 - x. 2x = 90 - x Add x to both sides, 2x + x = 90 -x + x 3x = 90 x = 30

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## How to Solve for X

Last Updated: February 10, 2023 Fact Checked

This article was co-authored by David Jia . David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. There are 8 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 760,770 times.

There are a number of ways to solve for x, whether you're working with exponents and radicals or if you just have to do some division or multiplication. No matter what process you use, you always have to find a way to isolate x on one side of the equation so you can find its value. Here's how to do it:

## Using a Basic Linear Equation

- 2 2 (x+3) + 9 - 5 = 32

- 4(x+3) + 9 - 5 = 32

- 4x + 12 + 9 - 5 = 32

- 4x+21-5 = 32
- 4x + 16 - 16 = 32 - 16

- 4x/4 = 16/4

- 2 2 (x+3)+ 9 - 5 = 32
- 2 2 (4+3)+ 9 - 5 = 32
- 2 2 (7) + 9 - 5 = 32
- 4(7) + 9 - 5 = 32
- 28 + 9 - 5 = 32
- 37 - 5 = 32

## With Exponents

- 2x 2 + 12 = 44

- 2x 2 +12-12 = 44-12

- (2x 2 )/2 = 32/2
- 4 Take the square root of each side of the equation. [6] X Research source Taking the square root of x 2 will cancel it out. So, take the square root of both sides. You'll get x left over on one side and plus or minus the square root of 16, 4, on the other side. Therefore, x = ±4.
- 2 x (4) 2 + 12 = 44
- 2 x 16 + 12 = 44
- 32 + 12 = 44

## Using Fractions

- (x + 3)/6 = 2/3

- (x + 3) x 3 = 3x + 9
- 3x + 9 = 12

- 3x + 9 - 9 = 12 - 9

- (1 + 3)/6 = 2/3

## Using Radical Signs

- √(2x+9) - 5 = 0

- √(2x+9) - 5 + 5 = 0 + 5
- √(2x+9) = 5

- (√(2x+9)) 2 = 5 2
- 2x + 9 = 25

- 2x + 9 - 9 = 25 - 9

- √(2(8)+9) - 5 = 0
- √(16+9) - 5 = 0
- √(25) - 5 = 0

## Using Absolute Value

- |4x +2| - 6 = 8

- |4x +2| - 6 + 6 = 8 + 6
- |4x +2| = 14

- 4x + 2 = 14
- 4x + 2 - 2 = 14 -2

- 4x + 2 = -14
- 4x + 2 - 2 = -14 - 2
- 4x/4 = -16/4

- |4(3) +2| - 6 = 8
- |12 +2| - 6 = 8
- |14| - 6 = 8
- |4(-4) +2| - 6 = 8
- |-16 +2| - 6 = 8
- |-14| - 6 = 8

## Practice Problems and Answers

## Expert Q&A

- To check your work, plug the value of x back into the original equation and solve. Thanks Helpful 1 Not Helpful 0
- Radicals, or roots, are another way of representing exponents. The square root of x = x^1/2. Thanks Helpful 1 Not Helpful 1

## You Might Also Like

- ↑ David Jia. Academic Tutor. Expert Interview. 23 February 2021
- ↑ http://tutorial.math.lamar.edu/Classes/Alg/SolveLinearEqns.aspx
- ↑ https://www.purplemath.com/modules/solvelin.htm
- ↑ https://sciencing.com/tips-for-solving-algebraic-equations-13712207.html
- ↑ https://www.youtube.com/watch?v=LMS1NR4gZN8
- ↑ https://www.mathsisfun.com/algebra/fractions-algebra.html
- ↑ http://www.mathsisfun.com/algebra/radical-equations-solving.html
- ↑ http://www.sosmath.com/algebra/solve/solve0/solve0.html

## About This Article

To solve for x in a basic linear equation, start by resolving the exponent using the order of operations. Then, isolate the variable to get your answer. To solve for x when the equation includes an exponent, start by isolating the term with the exponent. Then, isolate the variable with the exponent by dividing both sides by the coefficient of the x term to get your answer. If the equation has fractions, start by cross-multiplying the fractions. Then, combine like terms and isolate x by dividing each term by the x coefficient. If you want to learn how to solve for x if the equation has radicals or absolute values, keep reading the article! Did this summary help you? Yes No

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Solving an equation is the process of getting what you're looking for, or solving for , on one side of the equals sign and everything else on the other side. You're really sorting information. If you're solving for x , you must get x on one side by itself.

Addition and subtraction equations

Some equations involve only addition and/or subtraction.

Solve for x .

x + 8 = 12

To solve the equation x + 8 = 12, you must get x by itself on one side. Therefore, subtract 8 from both sides.

To check your answer, simply plug your answer into the equation:

Solve for y .

y – 9 = 25

To solve this equation, you must get y by itself on one side. Therefore, add 9 to both sides.

To check, simply replace y with 34:

x + 15 = 6

To solve, subtract 15 from both sides.

To check, simply replace x with –9 :

Notice that in each case above, opposite operations are used; that is, if the equation has addition, you subtract from each side.

Multiplication and division equations

Some equations involve only multiplication or division. This is typically when the variable is already on one side of the equation, but there is either more than one of the variable, such as 2 x , or a fraction of the variable, such as

In the same manner as when you add or subtract, you can multiply or divide both sides of an equation by the same number, as long as it is not zero , and the equation will not change.

Divide each side of the equation by 3.

To check, replace x with 3:

To solve, multiply each side by 5.

To check, replace y with 35:

Or, without canceling,

Combinations of operations

Sometimes you have to use more than one step to solve the equation. In most cases, do the addition or subtraction step first. Then, after you've sorted the variables to one side and the numbers to the other, multiply or divide to get only one of the variables (that is, a variable with no number, or 1, in front of it: x , not 2 x ).

2 x + 4 = 10

Subtract 4 from both sides to get 2 x by itself on one side.

Then divide both sides by 2 to get x .

To check, substitute your answer into the original equation:

5x – 11 = 29

Add 11 to both sides.

Divide each side by 5.

To check, replace x with 8:

Subtract 6 from each side.

To check, replace x with 9:

Add 8 to both sides.

To check, replace y with –25:

3 x + 2 = x + 4

Subtract 2 from both sides (which is the same as adding –2).

Subtract x from both sides.

Note that 3 x – x is the same as 3 x – 1 x .

Divide both sides by 2.

To check, replace x with 1:

5 y + 3 = 2 y + 9

Subtract 3 from both sides.

Subtract 2 y from both sides.

Divide both sides by 3.

To check, replace y with 2:

Sometimes you need to simplify each side (combine like terms) before actually starting the sorting process.

Solve for x .

3 x + 4 + 2 = 12 + 3

First, simplify each side.

Subtract 6 from both sides.

To check, replace x with 3:

4 x + 2 x + 4 = 5 x + 3 + 11

Simplify each side.

6 x + 4 = 5 x + 14

Subtract 4 from both sides.

Subtract 5 x from both sides.

To check, replace x with 10:

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## Solving for x – the essence of algebra

Solving for x means isolating x (or whatever the variable of interest – it doesn't have to be called x ) – on one side of the equal sign (it doesn't matter which).

The way I like to think about the process of solving for x is that I need to "peel away" everything that's in some way "stuck" to it, in order to come up with an equation, at the end, that says " x = (something)".

Let's discover how to do this with examples. First the easy stuff, then we'll do more difficult problems.

I'll only give some basics here. As you learn more about algebra, you'll be able to solve for variables in many other more interesting and challenging situations.

Here's a little cartoon showing what I mean. Our job will be to clear all of the junk away from x in order to see what it really is.

## Inverse operations

In this section, we'll refer often to inverse operations. Inverse operations are opposite, and one can be used to undo the action of the other.

- Addition and subtraction are inverse operations.
- Multiplication and division are inverse operations.

## Simple addition & subtraction

For our first example, we'll solve for x when some number is added to it. Here's such an equation:

What needs to be done to isolate the variable x on the left side is to "move" that 3 over to the right. That's accomplished by subtracting a 3 from both sides. Remember, we have to do the same thing to both sides , unless it doesn't have an un-balancing consequence, like adding zero or multiplying by 1.

Until you get used to doing this kind of small thing in your head, you might want to write it out like this:

Now it's just some arithmetic to get the answer:

You can check to see if this is the correct value of x, by simply substituting 6 for x in the original algebraic equation:

## More simple addition & subtraction

Let's do the same kind of equation, but this time x has something subtracted from it. Right away we know that the method for solving this problem is the same, because subtraction is just addition of a negative. Here's the problem

We need to get rid of the 8 on the left by "moving it" to the right side. It is subtracted from x , so we use addition to move it to the right.

Here's how you can write it out explicitly. -8 + 8 = 0 on the left:

Then the final arithmetic to get the value of x :

## Multiplication & division

What about if x is multiplied by something.? Well, it's not that different. We just get rid of the multiplier using the inverse operation of multiplication, division . Here's our example:

We'll move the 3 to the right side by dividing 3 into both the left and right sides of the equation. If what's on the left is equal to what's on the right, then 1/3 of each must also be equal.

You can write it out like this, and "cancel" the 3's on the left:

Remember that "cancelling" is just understanding that 3/3 = 1. Then just do the division on the right (12/3) to get the solution:

Once again, we can plug our value for x into the original equation to check it:

The situation is very similar when x is divided by a constant. Here's an example:

x is divided by 7. The inverse operation of division is multiplication, so we'll multiply both sides of the equation by 7 to move that 7 to the right side:

The 7's on the left cancel, which is just to say that 7/7 = 1.

Now 7(2) is 14, so we have

Looking back at the original equation, we see that 14 divided by 7 is 2, so our answer is correct.

## Combinations of operations

Now let's do some combinations of these moves. In this example the variable x has something multiplying it and something added to that:

We'll need two steps to liberate the x this time. Do the easy one first, and add the 5 to both sides to move it to the right. Then "divide away" the 4.

First the addition to move the 5:

Here's the intermediate result:

Then divide both sides by 4, canceling the 4's on the left

and complete the arithmetic.

I'll leave the answer like that. I think it's unfortunate that this kind of fraction has been called "improper." There's nothing wrong with it, and in fact in mathematics, it's preferred over the compound fraction 4-½.

Here's another combination example:

This time we'll need to move the three (first – it's the easiest) by subtraction, then the 4 by multiplication (the inverse of division).

Here's how the 3 gets moved:

And here we get rid of the 4 on the left to isolate x :

And the result is:

## Complicated situations: Reverse PEMDAS

Notice that in the last two examples, we chose to do the subtraction and addition parts before the addition or multiplication. Let's try one of those again:

But this time we'll divide by the 4 first, then do our addition to get x alone. First the division, noting that in this case we have to divide all three terms of the equation by 4:

The result is

Now we move the 5/4 to the right side to find x :

which is the same result we got above. But ... it was a bit more cumbersome than just doing it the easy way from the beginning. These can get more complicated, too, so it's important to get the order right.

You know about order of operations (we gave it the acronym PEMDAS ) if you read the algebra basics section, and it turns out you can't go wrong in solving for x if you just reverse it: " SADMEP ." That means to do the subtraction and addition first, then the division and multiplication, then the exponents and what's inside parenthesis last. Check out examples 5 and 6 above and you'll see that's exactly what we did.

I like to think of solving for x in this way as "picking the low-hanging fruit" first. I do what's easy first, usually addition and subtraction. I keep peeling layers away in this way until I have x isolated.

## What if the variable is in the denominator?

This kind of problem can cause all kinds of confusion, but the basic procedure is easy to remember.Here's the problem:

There's one thing that really needs to be done in such a problem, and that's to get the variable out of the denominator. Always remember that:

To remove x from the denominator, we multiply both sides by x . The x 's on the left will cancel (x/x = 1), and the x on the right is now in the numerator of 8x .

Now it's a straightforward process to isolate x on the right side:

You can check this answer in the original equation if you remember that dividing by 1/8 is the same as multiplying by the reciprocal of 1/8, which is 8.

## Practice problems

Solve for x in each of the equations below:

## Problem 1 solution

## Problem 2 solution

## Problem 3 solution

## Problem 4 solution

## Problem 5 solution

## Problem 6 solution

## Problem 7 solution

## Problem 8 solution

## Problem 9 solution

## Problem 10 solution

## Problem 11 solution

## Problem 12 solution

Problem 13 solution.

## Problem 14 solution

## Problem 15 solution

## Problem 16 solution

## Problem 17 solution

## Problem 18 solution

## Problem 19 solution

## Problem 20 solution

## Problem 21 solution

## Problem 22 solution

## Problem 23 solution

## Problem 24 solution

## Problem 25 solution

## Problem 26 solution

## Problem 27 solution

## Problem 28 solution

## Problem 29 solution

## Problem 30 solution

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## Praxis Core Math

Course: praxis core math > unit 1.

- Algebraic properties | Lesson
- Algebraic properties | Worked example
- Solution procedures | Lesson
- Solution procedures | Worked example
- Equivalent expressions | Lesson
- Equivalent expressions | Worked example
- Creating expressions and equations | Lesson
- Creating expressions and equations | Worked example

## Algebraic word problems | Lesson

- Algebraic word problems | Worked example
- Linear equations | Lesson
- Linear equations | Worked example
- Quadratic equations | Lesson
- Quadratic equations | Worked example

## What are algebraic word problems?

What skills are needed.

- Translating sentences to equations
- Solving linear equations with one variable
- Evaluating algebraic expressions
- Solving problems using Venn diagrams

## How do we solve algebraic word problems?

- Define a variable.
- Write an equation using the variable.
- Solve the equation.
- If the variable is not the answer to the word problem, use the variable to calculate the answer.

## What's a Venn diagram?

- Your answer should be
- an integer, like 6
- a simplified proper fraction, like 3 / 5
- a simplified improper fraction, like 7 / 4
- a mixed number, like 1 3 / 4
- an exact decimal, like 0.75
- a multiple of pi, like 12 pi or 2 / 3 pi
- (Choice A) $ 4 A $ 4
- (Choice B) $ 5 B $ 5
- (Choice C) $ 9 C $ 9
- (Choice D) $ 14 D $ 14
- (Choice E) $ 20 E $ 20
- (Choice A) 10 A 10
- (Choice B) 12 B 12
- (Choice C) 24 C 24
- (Choice D) 30 D 30
- (Choice E) 32 E 32
- (Choice A) 4 A 4
- (Choice B) 10 B 10
- (Choice C) 14 C 14
- (Choice D) 18 D 18
- (Choice E) 22 E 22

## Things to remember

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## IMAGES

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