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Posted on Jun 17, 2008

Perform the indicated operation.

Subtract 8x-y+5z,from the sum of 2x+3y-6z and 5x-3y.

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Calculator can't do it, its Transposition, you don't say what it equals....
I don't believe the brackets are needed, I've put them in to seperate the parts as you have written them.

((2x+3y-6z)+(5x-3y)) - (8x-y+5z) =
(7x-6z) - (8x-y+5z) =
-x-y-11z =

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Im trying to solve the equation 5x=10-3x

5x (-3x) = 2x -3x - -3x = 0
The equation then becomes 2x=10 2x/2 = x and 10/2 =5
then x=5. does it work 5x5=10+3*5 YES
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Find two numbers such that the sum of the first and three times the second is 5 and the sum of the second and 2 times the first is 8

Let the two numbers be X and Y:
X + 3Y = 5
2X + Y = 8

Rewrite the second equation to Y = 8 - 2X and substitute in first equation:
X + 3(8-2X) = 5
X + 24 - 6X = 5
-5X + 24 = 5
-5X = 5 - 24
-5X = -19
X = 19/5 = 3.8
Y = 8 - 2(3.8) = 8 - 7.6 = 0.4

Check: 3.8 + 3(0.4) = 3.8 + 1.2 = 5
2(3.8) + .4 = 7.6 + 0.4 = 8
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What ia the sulution of 3y-6x=-3

To find the solution, first find the value of y for each equation.
Then substitue one equation into the other so that you only the x variable left.
Then just solve for x.
Once you have a value for x, then you can easily solve for y.


So for the first equation:

3y - 6x = -3
3y = 6x - 3

y = 2x - 1


Now for the second equation:

2y + 8x = 10
2y = -8x + 10

y = -4x + 5

Since both equations equal y, they also equal each other, therefore:

2x - 1 = -4x + 5

Now just solve for x:

2x + 4x = 5 + 1
6x = 6

x=1

Now substitute x=1 into either original equation:

y = 2x - 1
y = 2 (1) - 1
y = 2 - 1

y = 1


Therefore the solution is x=1 and y=1



Good luck, I hope that helps.


Joe.
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1answer

2x^5+4x^4-3x^2+7,7x^4-3x^3+x^2-9

5x -2(x-2) = 6 -2(-3x-3)
5x -2x +4 = 6 +6x +6 (use the distributive property)
3x + 4 = 6x + 12 (collect terms)
4 = 3x + 12 (subtract 3x from both sides to get all x on one side)
-8 = 3x (subtract 12 from both sides to get only x on one side)
-8/3 = x (divide both sides by 3 to get x by itself)

Check
5(-8/3) -2((-8/3)-2) = 6 - 2(-3(-8/3)-3)
-40/3 -2(-8/3-6/3) = 6 - 2(8-3)
-40/3 -2(-14/3) = 6 -2(5)
-40/3 + 28/3 = 6 - 10
-12/3 = -4 (yes)
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4x-y-2z=-7 -4x+8y+5z=46 8x-3y+z=13

Type solve(exp1 and exp2 and exp3, {x,y,z})
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How do I solve x and y in the following equations? 5x+3y=6 2x-4y=5 Thank you

You can solve it with following method.

5x+3y=6 2x-4y=5
So 5x=6-3y so 2[(6-3y)/5]-4y=5
So x=(6-3y)/5 so 12-6y-20y=25
so -26y=25-12
so -26y=13
so y= -(1/2)
2x-4y=5
so 2x=5+4y
so 2x=5+4(-1/2)
so 2x=(10-4)/2
so 2x=6/4
so x =3/2

The value of x=3/2 and value of y= -1/2

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Thanks for using FixYa.
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Algebra 1

B. 2((2-Y)/3)-Y=3
2(2-Y)-Y=9
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-3Y=5
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Y=-1.67

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