Question

Please help i really need an A and i'm horrible at math ):

1. solve the following equation algebraically, show your work
6 = x+2 over 3

2.solve the following equation algebraically, show your work
13 + w over 7 = - 18

3. solve the following equation algebraically, show your work
17 = -13 - 8x

Example: 1 over 5 would mean a fraction I just don't want anyone confused

Answer 1

You'll get better at math if you do your own work. We'll be glad to check your answers for you.

Fractions are simply written like this: 1/4 and 2/5.

Answer 2

Also the way you typed it without brackets ...
is it:
6 = x + 2/3
or
6 = (x+2)/3 ???

Answer 3

Reiny, It's x + 2/3

Answer 4

If you can't solve any of the problems, you need a one-on -one tutor.
1. 6 = x + 2/3.
6 - 2/3 = x,
6 = 18/3,
18/3 - 2/3 = x,
16/3 = x,
5 1/3 = X.
You should try solving the other two and practice working similar problems.
Remember, if you cannot do the homework; you cannot pass the test.

Answer 5

I did all the other questions, I was just having trouble with this one

Answer 6

Solve the following equation. Show all your work.

x
x
−
2
+
x
−
1
x
+
1
=
−
1

Answer 7

First, we need to find a common denominator for all the fractions.

The common denominator is (x+1)(x-2)(x+1)= (x+1)^2(x-2)

Now we can multiply every term by this common denominator to get rid of the fractions.

(x+1)(x-2) - (x+1)(x+1) + x(x-2)(x+1) = -1(x+1)^2(x-2)

Expand each term and simplify:

x^2 - x - 2 - (x^2 + 2x + 1) + x^3 - x^2 + x - 2x^2 - 2x = -1(x^2 + 2x + 1)(x-2)

x^3 - 5x^2 - 3x - 2 = -1(x+1)^2(x-2)

Distribute the negative on the right side:

x^3 - 5x^2 - 3x - 2 = -(x+1)^2(x-2)

Expand the right side:

x^3 - 5x^2 - 3x - 2 = -x^3 - 4x^2 - 4x + 2

Combine like terms:

2x^3 - x^2 - 7x - 4 = 0

Now we can try to factor this expression.

First, we can see that 2 is a possible root since it divides evenly into the constant term (-4). We can use synthetic division to test this:

2 | 2 -1 -7 -4
- 2 2 -10
--------------
2 1 -5 -14

The remainder is non-zero, which means 2 is not a root.

Next, we can try to use the rational root theorem to find possible rational roots. The rational root theorem states that any rational root is of the form p/q, where p divides evenly into the constant term (-4) and q divides evenly into the leading coefficient (2). The possible rational roots are then:

±1/2, ±1, ±2, ±4

We can test these using synthetic division or direct substitution to see if they are roots.

Trying x = -1:

(-1)^3 - 5(-1)^2 - 3(-1) - 2 = -1 + 5 - 3 - 2 = -1

So, x = -1 is a root. Using synthetic division, we get:

-1 | 2 -1 -7 -4
- -1 2 5
---------------
2 -2 -5 1

This gives us the quadratic factor 2x^2 - 2x - 5. We can use the quadratic formula to solve for the remaining roots:

x = [2 ± sqrt(4 + 40)]/4
x = [1 ± sqrt(11)]/2

Therefore, the solutions to the equation are:

x = -1, [1 + sqrt(11)]/2, [1 - sqrt(11)]/2