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...

1. Combine like terms on the left side of =

4x-2x =2x

New left side is 2x + 1

2. Combine like terms on right side of =

5 - 7 = -2

New right side is x - 2

3. New problem is 2x + 1 = x - 2

4. Now get all x terms on left side of =

and all constants on right side of = by adding the opposites of the corresponding terms as you move them across the =

2x - x = -2 - 1

5. Combine the like terms on each side of the =

Left side: 2x - x = x and Right side: -2 -1 = -3

So x = -3

6. Now check your solution by substituting your answer back in to the original problem.

4(-3) - 2(-3) + 1 = 5 + (-3) -7

7. Now do the arithmatic and hopefully the left and right sides are equal.

-12 + 6 +1 = 5 -3 -7

-5 = -5

Since this is a true statement I know my answer of x = -3 is the correct solution to this problem.

4x-2x =2x

New left side is 2x + 1

2. Combine like terms on right side of =

5 - 7 = -2

New right side is x - 2

3. New problem is 2x + 1 = x - 2

4. Now get all x terms on left side of =

and all constants on right side of = by adding the opposites of the corresponding terms as you move them across the =

2x - x = -2 - 1

5. Combine the like terms on each side of the =

Left side: 2x - x = x and Right side: -2 -1 = -3

So x = -3

6. Now check your solution by substituting your answer back in to the original problem.

4(-3) - 2(-3) + 1 = 5 + (-3) -7

7. Now do the arithmatic and hopefully the left and right sides are equal.

-12 + 6 +1 = 5 -3 -7

-5 = -5

Since this is a true statement I know my answer of x = -3 is the correct solution to this problem.

May 17, 2012 | 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...

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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