Question about SoftMath Algebrator - Algebra Homework Solver (689076614429)

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

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!

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...

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

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...

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

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

Feb 13, 2013 | SoftMath Algebrator - Algebra Homework...

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).

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...

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)

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

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

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...

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.

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...

__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____so add them together__

__(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...

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

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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