how do i solve: (x-13) -(x-5) -2x = 0?

Answer 1

First get rid of the parentheses (reversing the sign in front of the 5, because negative times negative is positive):

x - 13 - x + 5 - 2x = 0

Then add the exes and the numbers

-2x -8 = 0

Now add eight to both sides

-2x = 8

Divide by -2

x = -4

Answer 2

you should really be doing your own homework. Maybe you should try looking into your book. Maybe this will help.

Quadratic Equations and Graphs Worksheet
Carl V. Lutzer
Topics:

Using the Quadratic Formula
Solving quadratic equations graphically
Finding the vertex of parabolas
Roots of quadratic functions
The discriminant

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"
Finish the following sentence: The quadratic formula tells us that the solutions of ax2+bx+c= 0 will be x=...

Solve x2+x-13 = 0 algebraically.

Solve x2-3x-2 = 1 graphically.

Solve 2x2-3x-2 = x2+x-13...

...graphically.
...algebraically.

Suppose f(x) = (x-3)2 - 4.

Find the x-intercept(s).

Find the y-intercept(s).

Use what you know of translations of graphs to determine the location of the vertex of the parabola described by y=f(x).

Plot the x and y intercepts, and the vertex which you found above. From that information, sketch a graph of the function f.

Algebraically determine the x and y intercepts, and the vertex of the parabolas described by the following equations. Also, without graphing these parabolas, determine whether they are "cupped up," or "cupped down."

y = 2x2 - 3.
y = (x+1)2 + 1.
y = -2(x-3)2 + 1.

Find real numbers a,b,c will so that ax2 + bx + c will have no real roots. (Hint: Use what you know of the discrimiant.)

Find real numbers a,b,c will so that ax2+bx+c has only one root. (Hint: Use what you know of the discrimiant.)

Suppose f(x)=ax2 + bx + c, where a>0 and c<0. How many roots does f have?

For each of the following, find a function f of the form f(x) = ax2+bx+c that satisfies the given condition or explain why such a function does not exist.

f(1)=0, f(9)=0, and f(2)=1.

f(1)=0 and the vertex of the graph of y=f(x) is at (5,4).

f(1)=2 and the vertex of the graph of y=f(x) is at (0,-4).

f(1)=0, f(9)=0, and the vertex of the graph of y=f(x) is at (4,5).

f(1)=0, f(4)=5, and f(9)=0.

This question concerns the equation



Determine the number of real solutions to this equation algebraically (hint: there is a common factor in every term.)
Determine the number of real solutions to this equation graphically.

Solve the following equation (hint: get rid of the radical by...)



Solve the following equation (hint: clear the denominators by...)

"

Answer 3

(x-13)-(x-5) -2x = 0

=> x-13 - x +5 = 2x
=> x-13+5=3x
=> -8=2x
=> x=-4

Answer 4

x-13-x+5-2x=0
-13+5-2x=0
-8-2x=0
-2x=8
x= -4

Answer 5

Write the equation out as a list of separate terms to make it clearer to see (multiply the second parenthetical term by -1):

x - 13 - x + 5 - 2x = 0

Now add all the numbers together and all the x's together:

-x -8 = 0

Now move the x to the opposite side by adding -x to both sides:

-8 = x

Answer 6

(x-13)-(x-5)-2x=0
-8 - 2x=0
2x = 0-8
divide 2x = 8 by 2

answer is x=4

then solution is
change all x = 4
(4-13)-(4-5)-2(4)=0
(-9)-(-1)-(-8)=0
0=0

Answer 7

-2x-8=0

U may take x=-4

It will be correct

Answer 8

You know what?
It really doesn't matter.

Go play outside.
Do something productive.

Answer 9

Will you remember in a test?

Answer 10

I hit a blank! dont I feel stupid!