STEVE'S MATH PAGE

 

algebra ii problems BY: STEVE DAVIS
solving linear equations

1. 4x - 6 = x+9
       -x         -x
 
 2. 3x - 6 = 9
            - 6   -6

 3. 3x = 9
       3x/3 = 9/3
                x=3

 1. First you subtract x from both sides and you come up with 3x -6 = 9.

2. Now you subtract 6 from each side and you come up with 3x = 9.

3.Next you divide each side by 3 and you get the answer, x = 3. 

 
 
 
 
finding equations 4 a line.  using slope intercept form.
write the equation for a line that has a slope of 3 and goes through the point (2, 9).

     y = mx + b
1. y = 3x + b
 
 

2. 9= 3(2) + b
     9= 6 + b
     9= 6 + 3
 

3. y= 3x + 3
 
 
 
 
 
 
 

1. the variable m represents slope, so if the slope is 3 than 
m = 3.    y= 3x + b
 

2. the point was given to you so now you have to plug the point (2, 9) into the equation and solve for b.

3. after you solve for b then you plug it into the original equation and you get y= 3x + 3
 

 
 
 
 
Solve for b1
A=(b1+b2)h
         2

1. 2A=(b1+b2)h
     h         h

2. 2A=b1+b2
     h
   -b2          -b2

3. 2a- 2=b1
        H
 
 
 

1. First you multiply both sides by 2 to get rid of the fraction.

2. Now you subtract b2 from both sides so you can get b1 by itself.
 

3. Now you have the answer to the problem.
 
 
 
 
 
Things 2 remember when solving equations
1.add, subtract, multiply, or divide a " thing"

3. substitute/simplify

 2. distribute

4. square root,  square, cube, etc

 visit Union Grove High School
 go to Period 5
 see Mrs Felz

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