Guide :

neg. two subtract neg. six ?

-2subtract -6 = ?

Research, Knowledge and Information :


Subtracting Positive and Negative Numbers - Wyzant


Subtracting Positive and Negative Numbers. ... Subtracting a negative number from a negative number ... sign followed by a negative sign, turn the two signs into ...
Read More At : www.wyzant.com...

Search › adding and subtracting negative numbers | Quizlet


Study sets matching "adding and subtracting negative numbers" Study sets. Classes
Read More At : quizlet.com...

MathSteps: Grade 6: Negative Numbers: What Is It?


The rules for operations with negative numbers may ... and when adding two negative numbers the sum is negative. ... subtracting with negative numbers on ...
Read More At : eduplace.com...

What is negative 6 subtract negative 2 - Answers.com


What is negative 6 subtract negative 2? SAVE CANCEL. already exists ... You pretend that the second number is a positive number and then you add the two numbers ...
Read More At : www.answers.com...

Adding and Subtracting Positive and Negative Numbers


How to Add and Subtract Positive and Negative Numbers Numbers Can be Positive or Negative: ... Two unlike signs become a negative sign. Example: What is 5+(−2) ?
Read More At : www.mathsisfun.com...

Subtracting a Negative Number from a Positive Number - YouTube


Apr 16, 2011 · ... Learn how to subtract a negative number from a positive ... rewrite the problem making the two negative ... Adding and Subtracting Negative ...

A negative minus a negative equal - Answers.com


Two things to note here: Any number subtracted from itself equals zero.Subtracting a negative number is the same ... A negative minus a negative turns the latter ...
Read More At : www.answers.com...

Positive and Negative Numbers Flashcards | Quizlet


Positive and Negative Numbers. STUDY. PLAY. Add two positive numbers. 5+5=10 (pos+pos= pos) ... When subtracting positive numbers there's two ways to do it.
Read More At : quizlet.com...

Adding and Subtracting Negative Fractions - Khan Academy


Practice adding and subtracting positive and negative fractions. Learn for free about math, ... Khan Academy is a nonprofit with the mission of providing a free, ...
Read More At : www.khanacademy.org...

Subtracting Positive and Negative Integers


An interactive math lesson about subtraction of two-digit negative or positive integers. ... Subtracting Negative and Positive Integers. To subtract integers, ...
Read More At : aaamath.com...

Suggested Questions And Answer :


neg. two subtract neg. six ?

-2 -(-6)=6-2=4 ...................
Read More: ...

3x + 6y – 6z = 9 2x – 5y + 4z = 6 -x +16y + 14z = -3 what is the answer

3x + 6y – 6z = 9 2x – 5y + 4z = 6 -x +16y + 14z = -3 what is the answer how do you solve it and what are the answers ? 1)  3x + 6y – 6z = 9 2)  2x – 5y + 4z = 6 3)  -x +16y + 14z = -3 The first objective is to eliminate z so we can solve for x and y. It's a multi-step process, so follow along. Multiply equation one by 4. 4 * (3x + 6y – 6z) = 9 * 4 4)  12x + 24y - 24z = 36 Multiply equation 2 by 6. 6 * (2x – 5y + 4z) = 6 * 6 5)  12x - 30y + 24z = 36 Now, we have two equations with a "24z" term. Add the equations and the z drops out. Add equation five to equation four.    12x + 24y - 24z = 36 +(12x - 30y + 24z = 36) ----------------------------------    24x -  6y           = 72 6)  24x - 6y = 72 The same process applies to equations two and three. Multiply equation two by 7 this time. 7 * (2x – 5y + 4z) = 6 * 7 7)  14x - 35y + 28z = 42 Multiply equation three by 2. 2 * (-x + 16y + 14z) = -3 * 2 8)  -2x + 32y + 28z = -6 Subtract equation eight from equation seven.   14x - 35y + 28z = 42 -(-2x + 32y + 28z = -6) ---------------------------------   16x - 67y          = 48 9)  16x - 67y = 48 Looking at equations six and nine, it would be simpler to eliminate the x. The multipliers are smaller. Multiply equation six by 2. 2 * (24x - 6y) = 72 * 2 10)  48x - 12y = 144 Multiply equation nine by 3. 3 * (16x - 67y) = 48 * 3 11)  48x - 201y = 144 Subtract equation eleven from equation 10.   48x -   12y = 144 -(48x - 201y = 144) ---------------------------          189y  =   0 189y = 0 y = 0  <<<<<<<<<<<<<<<<<<<<<<<<<<<<<< Substitute that value into equations six and nine to solve for x and verify. Six: 24x - 6y = 72 24x - 6(0) = 72 24x - 0 = 72 24x = 72 x = 3  <<<<<<<<<<<<<<<<<<<<<<<<<<<<<< Nine: 16x - 67y = 48 16x - 67(0) = 48 16x - 0 = 48 16x = 48 x = 3    same answer for x To solve for z,substitute the x and y values into the three original equations. One: 3x + 6y – 6z = 9 3(3) + 6(0) – 6z = 9 9 + 0 - 6z = 9 9 - 6z = 9 -6z = 9 - 9 -6z = 0 z = 0  <<<<<<<<<<<<<<<<<<<<<<<<<<<<<< Two: 2x – 5y + 4z = 6 2(3) – 5(0) + 4z = 6 6 - 0 + 4z = 6 6 + 4z = 6 4z = 6 - 6 4z = 0 z = 0     same answer Three: -x +16y + 14z = -3 -(3) +16(0) + 14z = -3 -3 + 0 + 14z = -3 -3 + 14z = -3 14z = -3 + 3 14z = 0 z = 0     once again, same answer x = 3, y = 0, z = 0
Read More: ...

6x-6y-4z=-10 -5x+4y-z=-12 2x+3y-2z=9

6x-6y-4z=-10 -5x+4y-z=-12 2x+3y-2z=9 1)  6x - 6y - 4z = -10 2)  -5x + 4y - z = -12 3)  2x + 3y - 2z = 9 Multiply equation two by 4. 4 * (-5x + 4y - z) = -12 * 4 4)  -20x + 16y - 4z = -48 Subtract equation four from equation one.       6x -    6y - 4z = -10 -(-20x + 16y - 4z = -48) -----------------------------    26x - 22y      = 38 5)  26x - 22y = 38 Multiply equation two by 2. 2 * (-5x + 4y - z) = -12 * 2 6)  -10x + 8y - 2z = -24 Subtract equation six from equation three.       2x + 3y - 2z =    9 -(-10x + 8y - 2z = -24) ----------------------------    12x - 5y      = 33 7)  12x - 5y = 33 Multiply equation five by 5. 5 * (26x - 22y) = 38 * 5 8)  130x - 110y = 190 Multiply equation seven by 22. 22 * (12x - 5y) = 33 * 22 9)  264x - 110y = 726 Subtract equation eight from equation nine.   264x - 110y = 726) -(130x - 110y = 190) --------------------------   134x        = 536 134x = 536 x = 4  <<<<<<<<<<<<<<<<<<< Plug the value of x into equation seven. 12x - 5y = 33 12(4) - 5y = 33 48 - 5y = 33 -5y = -15 y = 3  <<<<<<<<<<<<<<<<<<< Plug the values for x and y into equation two. -5x + 4y - z = -12 -5(4) + 4(3) - z = -12 -20 + 12 - z = -12 -z = -12 + 20 - 12 -z = -4 z = 4  <<<<<<<<<<<<<<<<<<< x = 4, y = 3, z = 4 Check the answers by plugging all three values into the original equations. One: 6x - 6y - 4z = -10 6(4) - 6(3) - 4(4) = -10 24 - 18 - 16 = -10 6 - 16 = -10 -10 = -10 Two: -5x + 4y - z = -12 -5(4) + 4(3) - 4 = -12 -20 + 12 - 4 = -12 -8 - 4 = -12 -12 = -12 Three: 2x + 3y - 2z = 9 2(4) + 3(3) - 2(4) = 9 8 + 9 - 8 = 9 9 = 9 All values are correct.
Read More: ...

Solve {3x-2y+2z=30, -x+3y-4z=-33, 2x-4y+3z=42}

Solve {3x-2y+2z=30, -x+3y-4z=-33, 2x-4y+3z=42} Please just solve the set provided above!!!! This will be a bit more involved than the systems with two unknowns, but the process is the same. The plan of attack is to use equations one and two to eliminate z. That will leave an equation with x and y. Then, use equations one and three to eliminate z again, leaving another equation with x and y. Those two equations will be used to eliminate x, leaving us with the value of y. I'll number equations I intend to use later so you can refer back to them. That's enough discussion for now. 1)  3x-2y+2z=30 2)  -x+3y-4z=-33 3)  2x-4y+3z=42 Equation one; multiply by 2 so the z term has 4 as the coefficient. 3x - 2y + 2z = 30 2 * (3x - 2y + 2z) = 30 * 2 4)  6x - 4y + 4z = 60 Add equation two to equation four:   6x - 4y + 4z =  60 +(-x + 3y - 4z = -33) ----------------------   5x - y       = 27 5)  5x - y = 27 Multiply equation one by 3. Watch the coefficient of z. 3 * (3x - 2y + 2z) = 30 * 3 6)  9x - 6y + 6z = 90 Multiply equation three by 2. Again, watch the coefficient of z. 2 * (2x - 4y + 3z) = 42 * 2 7)  4x - 8y + 6z = 84 Subtract equation seven from equation six.   9x - 6y + 6z = 90 -(4x - 8y + 6z = 84) ----------------------   5x + 2y      =  6 8)  5x + 2y = 6 Subtract equation eight from equation five. Both equations have 5 as the coefficient of x. We eliminate x this way.   5x -  y = 27 -(5x + 2y = 6) ---------------       -3y = 21 -3y = 21 y = -7  <<<<<<<<<<<<<<<<<<< At this point, I am confident that I followed the correct procedures to arrive at the value for y. Use that value to determine the value of x. ~~~~~~~~~~~~~~~ Plug y into equation five to find x. 5x - y = 27 5x - (-7) = 27 5x + 7 = 27 5x = 27 - 7 5x = 20 x = 4  <<<<<<<<<<<<<<<<<<< Plug y into equation eight, too. 5x + 2y = 6 5x + 2(-7) = 6 5x - 14 = 6 5x = 6 + 14 5x = 20 x = 4    same value for x, confidence high Proceed, solving for the value of z. ~~~~~~~~~~~~~~~ Plug both x and y into equation one. We will solve for z. Equation one: 3x - 2y + 2z = 30 3(4) - 2(-7) + 2z = 30 12 + 14 + 2z = 30 26 + 2z = 30 2z = 30 - 26 2x = 4 z = 2  <<<<<<<<<<<<<<<<<<< Continue using the original equations to check the values. Equation two: -x + 3y - 4z = -33 -(4) + 3(-7) - 4z = -33 -4 - 21 - 4z = -33 -25 - 4z = -33 -4z = -33 + 25 -4z = -8 z = 2   same value for z, looking good Equation three: 2x - 4y + 3z = 42 2(4) - 4(-7) + 3z = 42 8 + 28 + 3z = 42 36 + 3z = 42 3z = 42 - 36 3z = 6 z = 2  satisfied with the results We have performed several checks along the way, thus proving all three of the values. x = 4, y = -7 and z = 2
Read More: ...

solve three way equation with z. 1:3x-2y+2z=30 2:-x+3y-4z=-33 3:2x-4y+3z=42

Solve {3x-2y+2z=30, -x+3y-4z=-33, 2x-4y+3z=42} Please just solve the set provided above!!!! 1)  3x-2y+2z=30 2)  -x+3y-4z=-33 3)  2x-4y+3z=42 Equation one; multiply by 2 so the z term has 4 as the coefficient. 3x - 2y + 2z = 30 2 * (3x - 2y + 2z) = 30 * 2 4)  6x - 4y + 4z = 60 Add equation two to equation four:   6x - 4y + 4z =  60 +(-x + 3y - 4z = -33) ----------------------   5x - y       = 27 5)  5x - y = 27 Multiply equation one by 3. Watch the coefficient of z. 3 * (3x - 2y + 2z) = 30 * 3 6)  9x - 6y + 6z = 90 Multiply equation three by 2. Again, watch the coefficient of z. 2 * (2x - 4y + 3z) = 42 * 2 7)  4x - 8y + 6z = 84 Subtract equation seven from equation six.   9x - 6y + 6z = 90 -(4x - 8y + 6z = 84) ----------------------   5x + 2y      =  6 8)  5x + 2y = 6 Subtract equation eight from equation five. Both equations have 5 as the coefficient of x. We eliminate x this way.   5x -  y = 27 -(5x + 2y = 6) ---------------       -3y = 21 -3y = 21 y = -7  <<<<<<<<<<<<<<<<<<< ~~~~~~~~~~~~~~~ Plug y into equation five to find x. 5x - y = 27 5x - (-7) = 27 5x + 7 = 27 5x = 27 - 7 5x = 20 x = 4  <<<<<<<<<<<<<<<<<<< Plug y into equation eight, too. 5x + 2y = 6 5x + 2(-7) = 6 5x - 14 = 6 5x = 6 + 14 5x = 20 x = 4    same value for x Proceed, solving for the value of z. ~~~~~~~~~~~~~~~ Plug both x and y into equation one. We will solve for z. Equation one: 3x - 2y + 2z = 30 3(4) - 2(-7) + 2z = 30 12 + 14 + 2z = 30 26 + 2z = 30 2z = 30 - 26 2x = 4 z = 2  <<<<<<<<<<<<<<<<<<< Continue using the original equations to check the values. Equation two: -x + 3y - 4z = -33 -(4) + 3(-7) - 4z = -33 -4 - 21 - 4z = -33 -25 - 4z = -33 -4z = -33 + 25 -4z = -8 z = 2   same value for z Equation three: 2x - 4y + 3z = 42 2(4) - 4(-7) + 3z = 42 8 + 28 + 3z = 42 36 + 3z = 42 3z = 42 - 36 3z = 6 z = 2  satisfied with the results   x = 4, y = -7 and z = 2
Read More: ...

A number increased by twenty-six is forty-two.

1 3 6
Read More: ...

Solve the system using the elimination method.

Solve the system using the elimination method. Solve the system using the elimination method.  -8x    - 4y     + 2z    =     -26 -2x    + 4y    + 2z    =     10 -6x    - 8y    - 5z    =     -41 It appears that the three equations are: 1)  -8x - 4y + 2z = -26 2)  -2x + 4y + 2z = 10 3)  -6x - 8y - 5z = -41 Subtract equation two from equation one, eliminating the z.   -8x - 4y + 2z = -26 -(-2x + 4y + 2z =  10) ---------------------------   -6x   -  8y        = -36 4)  -6x - 8y = -36 Multiply equation two by 5. 5 * (-2x + 4y + 2z) = 10 * 5 5)  -10x + 20y + 10z = 50 Multiply equation three by 2. 2 * (-6x - 8y - 5z) = -41 * 2 6) -12x - 16y - 10z = -82 Add equation six to equation five, again eliminating the z.    -10x + 20y + 10z = 50 +(-12x - 16y - 10z = -82) --------------------------------    -22x  + 4y           = -32 7)  -22x + 4y = -32 Multiply equation seven by 2. 2 * (-22x + 4y) = -32 * 2 8)  -44x + 8y = -64 Add equation eight to equation four, eliminating the y.      -6x -  8y = -36 +(-44x + 8y = -64) -----------------------    -50x         = -100 -50x = -100 x = 2  <<<<<<<<<<<<<<<<<<<< Substitute the value of x into equation four. -6x - 8y = -36 -6(2) - 8y = -36 -12 - 8y = -36 -8y = -36 + 12 -8y = -24 y = 3  <<<<<<<<<<<<<<<<<<<< Substitute both x and y into equation one. -8x - 4y + 2z = -26 -8(2) - 4(3) + 2z = -26 -16 - 12 + 2z = -26 -28 + 2z = -26 2z = -26 + 28 2z = 2 z = 1  <<<<<<<<<<<<<<<<<<<< Substitute all three values into equation two. -2x + 4y + 2z = 10 -2(2) + 4(3) + 2(1) = 10 -4 + 12 + 2 = 10 -4 + 14 = 10 10 = 10 Substitute all three values into equation three. -6x - 8y - 5z = -41 -6(2) - 8(3) - 5(1) = -41 -12 - 24 - 5 = -41 -36 - 5 = -41 -41 = -41 Everything checks. x = 2, y = 3, z = 1
Read More: ...

Write the expression. Then, check all that apply.

1 3 4 5
Read More: ...

what is x when x over two plus six equals ten

written out, the problem looks like this: (x/2) + 6 = 10 first, subtract 6 from both sides (x/2) = 4 now multiply 2 on both sides so x = 8 in this equation
Read More: ...

i need to find and equation that equals 16 and 17

Can you use 11, eleven, as two of the one's? .1 repeating is 1/9 and when you divide 1 by it you get 9. 1 / .1(repeating) x (1+1) -1/1=17 1/.1(repeating) x(1+1) -1-1=16
Read More: ...

Tips for a great answer:

- Provide details, support with references or personal experience .
- If you need clarification, ask it in the comment box .
- It's 100% free, no registration required.
next Question || Previos Question
  • Start your question with What, Why, How, When, etc. and end with a "?"
  • Be clear and specific
  • Use proper spelling and grammar
all rights reserved to the respective owners || www.math-problems-solved.com || Terms of Use || Contact || Privacy Policy
Load time: 0.1095 seconds