4x-(2x-3)=0 - SoftMath Algebrator - Algebra Homework Solver (689076614429)

If all sides are equal then 2x + 5 = 4x - 3
2x = 8
x = 4

so the perimeter is 3 times either expression above
3 * (8 + 5) = 39
.

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

I assume the 10x2 means 10 squared. (I will show it as 10x**2)
(22x+11)(10x**2+5x) factors into"
11(2x+1)*5(2x**2+1x)
=55(2x+1)(2x**2+1x)
=(55x)(2x+1)(2x+1)
Multiplied out, it becomes 55x(4x**2+4x+1)
= 220x**3 + 220x**2 + 55x

x equals 1

First multiply the numbers outside the parentheses by what is inside the parentheses, which results in:

6 - 2x + 3x - 3 = 4x

then combine what you can:

6 - 3 = 3,
-2x + 3x = 1x

which leaves:

3 + x = 4x

Subtract the x on the left from the 4x on the right and you get:

3 = 3x

Divide by 3 to get the x alone

3 / 3 = 1

x = 1

We can write this polynomial as:

• (x-(-2))*(x-1)*(x-3)=
• (x+2)(x-1)(x-3)=
• (x+2)[x*(x-3)-1*(x-3)]=
  
• (x+2)*(x^2-3x-x+3)=
• (x+2)(x^2-4x+3)=
• x*(x^2-4x+3)+2*(x^2-4x+3)=
• x^3-4x^2+3x+2x^2-8x+6=
• x^3-2x^2-5x+6

x^3-2x^2-5x+6 is polynomial with roots -2, 1, 3.

You can see this polynomial in following picture:



Notice that it intersects x axis for x=-2, 1 and 3 (because these are roots of polynomial).

Hii,you can use the following solution x^3 - 4x^2-12x=0 x(x^2-4x-12)=0 replacing -4x from -6x+2x x(x^2-6x+2x-12)=0
Now Factoring x(x(x-6) +2(x-6))=0 x(x-6)(x+2)=0 Solution is x=0, x-6=0,x+2=0 1) x=0 2) x-6=0 =>x=6 3) x+2=0 =>x=-2

Hi,

Let's say your 3 numbers are a, b and c. Because they are consecutive and odd, you can write them like this:

a = 2x-1, b = 2x+1, c = 2x+3

Now, your problem looks like this:

2(2x-1)(2x+1) = (2x+1)(2x+3) + 7
2(4x^2-1) = 4x^2+6x+2x+3 + 7
8x^2-2 = 4x^2+8x+10
4x^2-8x-12 = 0
4(x^2-2x-3) = 0
x^2-2x-3 = 0

The solutions for this quadratic equation ( see more about this kind of equations here: http://en.wikipedia.org/wiki/Quadratic_equation ) are x=3 or x=-1. So, you've got two sets of solutions:

[1] a = 5, b = 7, c = 9
[2] a = -3, b = -1, c = 1


Take care,
Alex

6x+6=4x+12
Since 4x contains the variable to solve for, move it to the left-hand side of the equation by subtracting 4x from both sides.
6x+6-4x=12
Since 6x and -4x are like terms, add -4x to 6x to get 2x.
2x+6=12
Since 6 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 6 from both sides.
2x=-6+12
Add 12 to -6 to get 6.
2x=6
Divide each term in the equation by 2.
(2x)/(2)=(6)/(2)
Simplify the left-hand side of the equation by canceling the common factors.
x=(6)/(2)
Simplify the right-hand side of the equation by simplifying each term.
x=3

Good Luck

sec^4X- sec^2X = 1/cot^4X + 1/cot^2X
RHS
1/cot^4X + 1/cot^2X
=1/(Cos^4X/Sin^4X) + 1/(Cos^2X/Sin^2X)
=Sin^4X/Cos^4X + Sin^2X/Cos^2X
=Sin^4X/Cos^4X + Cos^2X.Sin^2X/Cos^4X
=Sin^2X/Cos^4(Sin^2X + Cos^2X)
=Sin^2X/Cos^4X
=(1-Cos^2X)/Cos^4X
=1/Cos^4X - Cos^2X/Cos^4X
=1/Cos^4X - 1/Cos^2X
=Sec^4X - Sec^2X
=LHS

Paul 3 years younger to his friend peter. In seven years, the product of their ages would be 5 more than the product of their ages 5 years ago. How old are they now?