Abbey's Helpful
Algebra Page



| x
= 2x-6
5 |
|
| 5(x=2x-6)
x=10x-30
5 |
First--multiply the entire equation by 5 to cancel out the 5 in x
and you get x=10x-30
5 |
| x=10x-30
x+30=10x
+30 +30 |
add 30 to both sides to cancel out the -30 on the right side and you get x+30=10x |
| x+30=10x
30=9x
-x -x |
subtract x from each side to remove the x from the left side and you get 30=9x |
| 30=9x
30=x
9 9 9 |
divide each side by 9 to cancel out the 9x from the right side and
you get 30=x
9 |
| 30=x
10=x
9 3 |
reduce the fraction 30 to 10
9 3 and you get x=10 3 |
| x=10
3 |
|
Now lets try another type of problem...
A LITERAL EQUATION
| E=pA(T4-To4) | Here is the original problem. You will solve for A. |
| E = pA(T4-To4)
E =pA
(T4-To4) (T4-To4) (T4-To4) |
First you should divide the everything by (T4-To4)
and you get
E =pA
(T4-To4) |
| E = pA
E=p
(T4-To4) p p(T4-To4) p |
Next you divide both sides by p to get A by itself but since you have
already have a denominator in the left side you can just add the p on to
that. You get
E=p
p(T4-To4) |
|
E=p
p(T4-To4) |
This is your answer!! |
How would you like to learn how to get an equation for a line???
| Draw and equation for a line that crosses through (3,4) and has a slope
of -2
3 |
Here is our 3rd Problem!! |
| y=mx+b
y=-2 x +B
3 |
You know that the equation for a line is y=mx+b where m=slope and b=y-intercept.
You know that the slope is -2 so you can put that in 3 in place of the m in the equation. |
| y=-2 x +B
4=-2 (3) +B
3 3 |
Next you can use the equation that you have so far and plug in the point that you know is on the line in place of the y and the x. Then you should be able to find the y-intercept (B). |
| 4=-2 (3) +B
4=-2+B
3 |
To solve you should first multiply the -2 by 3
3 and you get -2. |
| 4=-2+B
6=B
+2 +2 |
Add 2 to both sides to cancel out the -2 on the right side and you get 6=B |
| 6=B
y=-2 x +6
3 |
Plug in the 6 in place of the B (y-intercept) and you now have your equation!!! |
| y=-2 x +6
3 |
This is the equation you were looking for!!!!!! |