Question about SoftMath Algebrator - Algebra Homework Solver (689076614429)

# Math algebra like 2x=4 or 6x=12 and what number do i devide both sides by to solver the equation

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Divide the first equation by 2, the second by 6 (i.e. the number that comes before the x)

Posted on Oct 05, 2008

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Posted on Jan 02, 2017

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## Related Questions:

### What is the graph linear equation for y=2*+1

Now is the equation supposed to be y=2x+1? If so,

The graph of y = 2x+1 is a straight line
When x increases, y increases twice as fast, hence 2x
When x is 0, y is already 1. Hence +1 is also needed
So: y = 2x + 1
Here are some example values:
xy = 2x + 1
-1y = 2 × (-1) + 1 = -1
0y = 2 × 0 + 1 = 1
1y = 2 × 1 + 1 = 3
2y = 2 × 2 + 1 = 5
Check for yourself that those points are part of the line above!

Jan 16, 2015 | SoftMath Algebrator - Algebra Homework...

### How do I solve (3x-2y)2(3xy-3)

First, we will find y in terms of x. We will use the first equation to determine this.
4x+2y=2
We can subtract 4x from both sides:
2y=2-4x
And then divide both sides of the equation by two:
y=1-2x
Since we now have y in terms of x, we can substitute this into our second equation.
-3x-y=-3
-3x-(1-2x)=-3
Then, we can distribute the minus sign
-3x-1+2x=-3
-x-1=-3
Next, we can add 1 to both sides of the equation.

-x=-2
Finally, we divide both sides by negative one to isolate x.
x=2
Now that we have x's value, we can find y's value.
The first thing that we determined is:
y=1-2x
We can substitute in the value of x to this equation.
y=1-2x

y=1-4
y=-3
Therefore, we now have the values of both variables.
x=2
y=-3

Jan 13, 2015 | SoftMath Algebrator - Algebra Homework...

### 5I+5(I-31)=55 how do i solve this

Start by using the distributive property to expand the term 5(I-31)=5*I-5*31=5*I-155.
Rewrite the equation as 5*I+5*I-155=55
Combine the terms with I, this yields 10*I-155=55
Add 155 to both sides of the equation 10*I-155+155=55+155
you now get 10*I=210
Divide both sides of the equation by 10
(10*I)/10=210/10
Cancel the 10's on the left and divide 210 by 10 to get 21
Thus I=21

Basically you use the priority order of operations BEDMAS backwards

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### Find the radius of the equation x^2+y^2-2x-2y=14

Try casting the equation of the circle into the form (x-h)^2+(y-k)^2=R^2. Here (h,k) would be the center and R the radius.
To do that complete the two squares.
x^2-2x=x^2-2x+1-1=(x^2-2x+1)-1=(x-1)^2 -1.
Do the same for the y terms.
y^2-2y=(y-1)^2 -1.
Substitute these two expressions in your original equation.
(x-1)^2+(y-1)^2 -1-1=14 or (x-1)^2+(y-1)^2=14+2=16
The radius is square root of 16 or 4, and the center is C(1,1).

Sep 20, 2011 | SoftMath Algebrator - Algebra Homework...

### Evaluating math equations

http://www.brothersoft.com/free-universal-algebra-equation-solver-376429.html

This one is pretty good.
You can also simply type some equations into google and it will solve them. (sometimes)

Jun 07, 2011 | Computers & Internet

### 2x^2-19x+22

I'm assumeing the problem is 2x^2 - 19x+22 = 0, solve for x.

I don't see any easy to factor this so I'll either need to complete the square or use the quadratic equation.
Using the quadratic equation, where A = 2, B = -19, and C = 22

x = [-(-19) +/- squareroot( 19^2 - 4*2*22)] / (2*2) =
(19 +/- squareroot(185))/4 =
8.150367 or 1.349632

Feb 01, 2011 | SoftMath Algebrator - Algebra Homework...

### Y=7-x x-y=-3

Substitute (7 - X) in for Y in the 2nd equation and you get:

X- (7 - X) = -3 => 2X - 7 = -3 => 2X = 4

Solve this for X and you get X =2.

Plug X = 2 back into the first equation and you get Y = 7 - 2 = 5.

Therefore, the answer is X = 2 and Y = 5

I hope this helps you out.

Jan 27, 2011 | SoftMath Algebrator - Algebra Homework...

### X+y=10 x-y=2 use elimination

• The 1st equation contains a ‘+y’, while the 2nd contains a ‘-y’ term. they will cancel if added together, so we will add the equations to eliminate ‘y’.
• x-y=2
• x+y=10
• (x-y)+(x+y)=12
• 2x = 12
• X= 12/2 = 6
• subsitutue for
• 6+y=10
• y=10-6=4

• Answer X = 6 and Y = 4

Apr 11, 2010 | SoftMath Algebrator - Algebra Homework...

### The perimeter of a rectangular field is 628 meters. The length of the field exceeds its width by 6 meters. What is the demensions.

Let x = the short side. Then the long side is the short side plus 6 meters
now we have 2 sides that are x meters long, add those together gives us 2x for the length
of both short sides

The length of the long side is x + 6 for one side and x + 6 for the other side. Added together we have x + 6 + x + 6 = 2x + 12

Adding the 4 sides together we get 2x ( the 2 short sides) + 2x + 12 (the long sides) we get 4x + 12 = the perimeter or 628. 4x + 12 = 628. Subtract 12 from both sides of the equation leaves 4x = 616. divide both sides by 4 leaves x = 154 So the width is 154 meters and the length is 160 meters.

Hope this helps

Good luck Loringh

Oct 06, 2008 | SoftMath Algebrator - Algebra Homework...

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