Example 1: Simplify the expression
2(4a  5b)  (8 + b) + b + (2b + 4)  5a
Solution to Example1

given
2(4a  5b)  (8 + b) + b + (2b + 4)  5a

multiply factors
8a  10b  8  b + b 2b + 4  5a

group like terms
 13a  12b  4
Matched Exercise 1 Simplify the expression
2(a  8b)  (5  b) + b + (6b  9)  a
solution
Example 2: Solve the equation
2(3x  5)  (8  x) = 2(2x + 4) + 12
Solution to Example 2

given
2(3x  5)  (8  x) = 2(2x + 4) + 12

multiply factors
6x 10  8 + x = 4x  8 + 12

group like terms
5x  18 = 4x + 4

add 18 to both sides
5x 18 + 18 = 4x + 4 + 18

group like terms
5x = 4x + 22

add 4x to both sides
5x + 4x = 4x + 22 +4x

group like terms
x = 22

multiply both sides by 1
x = 22

Check the solution
left side:2(3*(22)  5)  (8  (22)) = 92
right side:2(2(22) +4) + 12 = 92

Conclusion
x = 22 is the solution to the given equation
Matched Exercise 2: Solve the equation
2(x  5)  (6 + x) = 3(2x + 4) + 12
solution
/* scriptreplace75dd8afeaeaadb46709c0be645677fc6 */
Example 3: If x > 2, simplify the expression
2 x + 2   3x  (2  x) +  6  9 
Solution to Example 3

To simplify the given expression, we need to simplify the terms with absolute value using definition of absolute value.
if x > = 0 ,  x  = x
if x < 0 ,  x  = x

According to the definition of the absolute value above,
x >  2 (given above) is equivalent to x + 2 > 0
if x + 2 > 0 then  x + 2  = x + 2

the above definition gives
 6  9  =   3  = 3

the whole expression given above can now
be written as
2(x + 2)  3x  (2  x) + 3

multiply factor
2x + 4 3x + 2 + x + 3

group like terms
9
Matched Exercise 3: If x
> 3, simplify the expression
2 x  3  + 6x  (2  3x) +  9  20 
solution
Example 4: Find the slope and the yintercept of the line given by the equation
2y  3x = 10
Solution to Example 4

We first write the equation in slope
intercept form y = mx +b. Put terms in x and constant terms on the right side
2y = 3x + 10

Divide both sides by 2
y = (3/2)x + 5

Now that the equation is in slope intercept form, we identify the slope as the coefficient of x and is equal to 3/2 and the y intercept as (0 , 5).
Matched Exercise 4: Find
the slope and the yintercept of the line given by the equation
3y  6x = 7
solution
Example 5: Find the equation of the line
passing through the points (2 , 3) and (4 , 1).
Solution to Example 5

We first calculate the slope m
m = (1  3) / (4  2) = 1

We now use the pointslope form of a line to find the equation of the line
y  y_{1} = m(x  x_{1}) , where m is the slope and (x_{1},y_{1}) is any of the two points given above.

Substitute m by its value 1 and x_{1} and y_{1} by 2 and 3 respectively, we obtain the equation of the line.
y  3 = 1(x  2)

in slope intercept form the equation is written as
y = x + 5
Matched Exercise 5: Find
the equation of the line passing through the points (0 , 3) and (1 , 1).
solution
More links and references to pages with algebra problems, tutorials and self tests.