Simplify 2x 8

Simplify 2x 8 DEFAULT

Simplifying Expressions versus Solving Equations

Purplemath

Related to "simplification" problems are some "solving" problems. Just as some expressions contain parentheticals and even nested grouping symbols, so also do some equations. The simplification process is the same, whether we're working with expressions (and thus only simplifying) or equations (so we're also solving). But, because the "equals" sign can cause confusion, especially if the instructor has been a bit careless, I'm going to work some "solving" exercises and, as we go, highlight where there might be pitfalls. (In other words, this page is an expansion of my comment on the previous page.)

Let's look at a typical linear equation that contains some parentheticals.

  • Solve 3 + 2[4x – (4 + 3x)] = –1

This equation has some nested grouping symbols on the left-hand side. As usual, I'll simplify from the inside out. I'll start by inserting the "understood" 1 in front of that innermost set of parentheses:

3 + 2[4x – (4 + 3x)] = –1

3 + 2[4x – 1(4 + 3x)] = –1

3 + 2[4x – 1(4) – 1(3x)] = –1

3 + 2[4x – 4 – 3x] = –1

3 + 2[1x – 4] = –1

3 + 2[1x] + 2[–4] = –1

3 + 2x – 8 = –1

2x + 3 – 8 = –1

2x – 5 = –1

2x – 5 + 5 = –1 + 5

2x = 4

x = 2

It is not required that you write out this many steps; once you get comfortable with the process, you'll probably do a lot of this in your head. But until you reach that comfort zone, you should write things out at least this clearly and completely.

Always remember, by the way, that you can check your answers in "solving" problems by plugging the numerical answer back in to the original equation. In this case, the variable is only in terms on the left-hand side (LHS) of the equation; my "check" (that is, my evaluation at the solution value) looks like this:

LHS: 3 + 2[4x – (4 + 3x)]:

3 + 2[4(2) – (4 + 3(2))]

3 + 2[8 – (4 + 6)]

3 + 2[8 – (10)]

3 + 2[–2]

3 – 4

–1

Since this is what I was supposed to get for the right-hand side (that is, I've shown that the LHS is equal to the RHS), my solution value was correct.


Note the difference between the exercise at the top of this page and all the preceeding exercises. This was a "solving" problem, whereas the previous ones had been "simplifying" problems. That is, the exercise above started with an equation (that is, it started something with an "equals" sign in it) and I was supposed to find a solution (that is, I was supposed to find something of the form "(some variable) equals (some number)"). On the other hand, for the previous exercises I had started with an expression (that is, I had started with something containing no "equals" sign) and had ended up with a different version of the same expression (in particular, I ended up with something that still contained no "equals" sign).

Why am I making a big deal about this? Because (warning!) many students, for whatever reason, conflate "equations" and "expressions". That is, given an expression to simplify, the student will somehow turn the problem into an equation to solve. This error will look something like the following:

  • Simplify 6 – 4(2x + 3) + 10x

incorrect "solution": 6 - 4(2x + 3) + 10x becomes 6 - 8x - 12 + 10x, becomes 2x - 6, which magically [as shown by a red arrow] becomes 2x = 6, which magically solves as x = 3

As you can see, the student somehow (at the red arrow) converted the expression "2x – 6" into the equation "2x = 6", and then somehow "solved" the expression. But the original problem didn't have an "equals" sign in it, so there was nothing to solve!

Be careful that you don't do this. And, to help yourself be careful, strive to use "equals" signs only when you're working with equations. If what they gave you didn't start with an "equals" sign in it, then your answer probably shouldn't have one, either.

(Yes, you've probably seen at least one instructor who would "simplify" by saying "this expression equals this expression, which equals this expression, which equals...", and so forth, putting "equals" signs in between each step of the simplification process. I'm fairly certain that it is exactly this process which later leads to students trying to solve expressions. Those instructors knew what they meant; to make sure that you know what you mean, reserve the "equals" signs for equations.)


Let's look at some more examples:

  • Simplify 6 – 10[x + (2x – 3)] + 12x

This is a "simplification" problem, since (1) there is no "equals" sign in the original problem and (2) they told me to simplify. I will simplify the expression from the inside out, and will end up with an equivalent, but simpler, expression as my answer. (In particular, I will not end up with a value for the variable.)

6 – 10[x + (2x – 3)] + 12x

6 – 10[x + 2x – 3] + 12x

6 – 10[3x – 3] + 12x

6 – 10[3x] – 10[–3] + 12x

6 – 30x + 30 + 12x

–30x + 12x + 6 + 30

–18x + 36

I can't simplify any further, so this is my answer.


What if they gave me something that was almost the same as the "simplify the expression" exercise above, but they'd turned it into an equation? Then I'd have something with an "equals" sign in it, and they'd be telling me to "solve", like this:

  • Solve 6 – 10[x + (2x – 3)] = –12x

This is almost like the previous exercise, but it's an equation (that is, it's something with an "equals" in it) and I'm told to solve. I'll still need to simplify the nested parentheticals on the left-hand side (LHS), but my steps after the simplification will be different.

6 – 10[x + (2x – 3)] = –12x

6 – 10[x + 2x – 3] = –12x

6 – 10[3x – 3] = –12x

6 – 10[3x] – 10[–3] = –12x

6 – 30x + 30 = –12x

36 – 30x = –12x

I've done as much simplifying as I can do. Now I turn to the "solving" part:

36 – 30x = –12x

6 – 5x = –2x

6 = 5x – 2x

6 = 3x

2 = x

They gave me something with an "equals" in it and told me to solve; I've ended up with something with an "equals" in it, and it's the solution value they wanted. So my answer is:


By the way, if you take the expression in the previous exercise and set it equal to zero (thus creating an equation which can be solved), you'll get a solution of x = 2.

This is not a mere coincidence. Later, when you study about the real roots of polynomials and how they relate to factorizations, you'll learn more about this. For now, just make sure that you're keeping straight the differences between simplifying expressions and solving equations.


  • Solve 2(x + 3) = 4 – (2 – x)

This is a "solving" problem, since (1) there is an "equals" sign in the original problem and (2) they told me to solve. In order to find the solution, I will work on both sides of the equation, simplifying and rearranging things as I go, and will eventually end up with a solution of the form "(the variable) equals (some number)" as my answer.

Note: Even though this is an equation, even though I'm solving, and even though they gave me something that started out with an "equals" sign in it, I will not be putting an "equals" sign between my lines of working! Each line stands alone, and my working looks like this:

2(x + 3) = 4 – (2 – x)

2(x) + 2(3) = 4 – 1(2) – 1(–x)

2x + 6 = 4 – 2 + 1x

2x + 6 = 2 + x

2xx + 6 = 2 + xx

x + 6 = 2

x + 6 – 6 = 2 – 6

x = –4

I'm down to a solution value for the variable x, so I'm done.


  • Simplify 2(x + 3) – 4 + (2 – x)

The parentheses aren't nested, so this is an easier exercise. My simplification process is:

2(x + 3) – 4 + (2 – x)

2x + 6 – 4 + 2 – x

2xx + 6 + 2 – 4

x + 4

I can't go any further, so I'm done.


Let me stress once again: To do "simplification" problems successfully, you need to take the time to write out each step; you must work from the inside out, being especially careful with the "minus" signs when they're right in front of grouping symbols; don't forget the Order of Operations (you'll be needing this for the rest of your life); and don't make the mistake of confusing "simplifying" an expression with "solving" an equation.


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Answer Math Problems

All grade answer math problems are solved here from different topics. A large number of objective answer math problems are solved here to help the students quickly to test their knowledge and skills on different topics.

1. Simplify: (4x2 - 2x) - (-5x2 - 8x).

Solution:


(4x2- 2x) - (-5x2- 8x)

= 4x2- 2x + 5x2+ 8x.

= 4x2+ 5x2- 2x + 8x.

= 9x2+ 6x.

= 3x(3x + 2).

Answer: 3x(3x + 2)



2. Simplify: (-7x2 - 3x) - (4x2 - x)

Solution:


(-7x2- 3x) - (4x2- x)

= -7x2- 3x - 4x2+ x

= -7x2- 4x2- 3x + x

= -11x2– 2x

= -x(11x + 2)

Answer: -x(11x + 2)



3. Simplify: (-5x2 - 8x + 7) - (5x2 - 4)

Solution:


(-5x2- 8x + 7) - (5x2- 4)

= -5x2- 8x + 7 - 5x2+ 4

= -5x2- 5x2- 8x + 7 + 4

= -10x2– 8x + 11

Answer: -10x2 – 8x + 11



4. Simplify: (4x2 + 5x) + (x2 - 5x + 1)

Solution:


(4x2+ 5x) + (x2- 5x + 1)

= 4x2+ 5x + x2- 5x + 1

= 4x2+ x2+ 5x - 5x + 1

= 5x2+ 0 + 1

= 5x2+ 1

Answer: 5x2 + 1



5. Simplify: (-6x + 3) - (-x2 + 5x)

Solution:


(-6x + 3) - (-x2+ 5x)

= -6x + 3 + x2- 5x

= -6x - 5x + x2+ 3

= x2– 11x + 3

Answer: x2 – 11x + 3



6. Simplify: (-9x - 5) - (-9x2 + x - 5)

Solution:


(-9x - 5) - (-9x2+ x - 5)

= -9x - 5 + 9x2- x + 5

= 9x2- 9x – x – 5 + 5

= 9x2– 10x

= x(9x – 10)

Answer: x(9x – 10)

7. Simplify: (2x - 4) - (6x + 6)

Solution:


(2x - 4) - (6x + 6)

= 2x - 4 - 6x – 6

= 2x – 6x – 4 – 6

= -4x – 10

= -2(2x + 5)

Answer: -2(2x + 5)


8. Simplify:(8+4x)/4x

Solution:


(8+4x)/4x

= 4(2 + x)/4x

[Cancel 4 from the numerator and denominator]

= (2 + x)/x

Answer: (2 + x)/x


9. Decide whether the lines are parallel, perpendicular or neither.

x + 4y = 7 and 4x – y = 3

Solution:


x + 4y = 7

4y = -x + 7

y = (-1/4) x + 7

Slope of the equation x + 4y = 7 is -1/4.

Again, 4x – y = 3

y = 4x – 3

Slope of the equation 4x – y = 3 is 4.

Since multiplying both the slope of the equation = -1/4 × 4 = -1

Therefore, the given two equations are perpendicular to each other.

Answer: Perpendicular


10. Simplify: 3x – 5 + 23x – 9

Solution:


3x – 5 + 23x – 9

= 3x + 23x – 5 – 9

= 26x – 14

= 2(13x – 7)

Answer: 2(13x – 7)

11. Use the discriminant to determine the number of real roots the equation has. 3x2 – 5x + 1 =0

(a) One real root (a double root),

(b) Two distinct real roots,

(c) Three real roots,

(d) None (two imaginary roots)

Solution:


Discriminant = bx2– 4ac

Compare the above equation 3x2– 5x + 1 =0 with ax2+ bx + c = 0

We get, a = 3, b = -5, c = 1

Put the value of a, b and c;

Discriminant = bx2– 4ac

Discriminant = (-5)2- 4 × 3 × 1

= 25 – 12

= 13 [13 > 0]

Therefore, discriminant is positive.

So the given equation has two distinct real roots.

Answer: (b)



12. Use the discriminant to determine the number of real roots the equation has. 7x2 + 3x + 1 =0.

(a) One real root (a double root),

(b) Two distinct real roots,

(c) Three real roots,

(d) None (two imaginary roots)

Solution:


Discriminant = b2– 4ac

Compare the above equation 7x2+ 3x + 1 =0 with ax2+ bx + c = 0

We get, a = 7, b = 3, c = 1

Put the value of a, b and c;

Discriminant = b2– 4ac

Discriminant = (3)2- 4 × 7 × 1

= 9 – 28

= -19 [13 < 0]

Therefore, discriminant is negative.

So the given equation has none (two imaginary roots).

Answer: (d)


13. Fill in the blank:

(a) The point with coordinates (0,0) is called ........of a rectangular coordinate system.

(b) To find the x-intercept of a line, we let....equal 0 and solve for ......; to find y-intercept, we let ......equal 0 and solve for.......

Solution:


(a) The point with coordinates (0,0) is called origin of a rectangular coordinate system.

(b) To find the x-intercept of a line, we let y equal 0 and solve for x ; to find y-intercept, we let x equal 0 and solve for y .

14. Name the quadrant, if any, in which each point is located.

(a) (1, 6)

(b) (-4, -2)

Solution:


(a) (1, 6) --------- I Quadrant.

(b) (-4, -2) --------- III Quadrant.


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