Download Solving Linear Equations in One Variable

Solving Linear Equations
with One Variable
To solve a linear equation in one variable:
1. Simplify both sides of the equation.
2. Use the addition and subtraction properties to get all variable terms
on the left-hand side and all constant terms on the right-hand side.
3. Simplify both sides of the equation.
4. Divide both sides of the equation by the coefficient of the variable.
Example: Solve x + 1 = 3(x  5).
x + 1 = 3(x  5)
Original equation
x + 1 = 3x  15
Simplify right-hand side.
x = 3x  16
Subtract 1 from both sides.
Subtract 3x from both sides.
 2x =  16
Divide both sides by 2.
x=8
The solution is 8.
Check the solution: (8) + 1 = 3((8)  5)  9 = 3(3) True
2
Equations with fractions can be simplified by multiplying both
sides by a common denominator.
1
2 1
x

 ( x  4). The lowest common denominator
Example: Solve
of all fractions in the equation is 6.
2
3 3
1
2
1
6  x    6  ( x  4)  Multiply everything on each side by 6.
3
2
3

3x + 4 = 2x + 8
Simplify.
3x = 2x + 4
Subtract 4.
Subtract 2x.
x=4
1
2 1
(4)   ((4 )  4) Check.
2
3 3
2 1
2   (8)
3 3
8 8
True

3 3
3
Alice has a coin purse containing $5.40 in 10 cents coins and
20 cents coins. There are 30 coins all together. How many 10
cents coins are in the coin purse?
Let the number of 10 cents coins in the coin purse = d.
Then the number of 20 cents coins = 30  d.
10d + 20(30  d) = 540
Linear equation
10d + 600  20d = 540
Simplify left-hand side.
10d  20d =  60
10d =  60
d=6
Subtract 480.
Simplify right-hand side.
Divide by 10.
There are 6, 10 cents coins in Alice’s coin purse.
4
The sum of three consecutive integers is 54. What are the
three integers?
Three consecutive integers can be represented as
n, n + 1, n + 2.
n + (n + 1) + (n + 2) = 54
3n + 3 = 54
3n = 51
n = 17
Linear equation
Simplify left-hand side.
Subtract 3.
Divide by 3.
The three consecutive integers are 17, 18, and 19.
17 + 18 + 19 = 54.
Check.
5
x and y -intercepts
● The x-intercept is the point where a line crosses
the x-axis.
The general form of the x-intercept is (x, 0).
The y-coordinate will always be zero.
● The y-intercept is the point where a line crosses
the y-axis.
The general form of the y-intercept is (0, y).
The x-coordinate will always be zero.
To summarize….
●To find the x-intercept, plug in 0
for y.
●To find the y-intercept, plug in 0
for x.
Find the x and y- intercepts
of x = 4y – 5
● x-intercept:
● Plug in y = 0
x = 4y - 5
x = 4(0) - 5
x=0-5
x = -5
● (-5, 0) is the
x-intercept
● y-intercept:
● Plug in x = 0
x = 4y - 5
0 = 4y - 5
5 = 4y
5
=y
4
5
)
4
● (0, is the
y-intercept
Example
Question: Draw the graph of 2x + y = 4
Solution
x=0
y=0
2(0) + y = 4
2x + 0 = 4
y=4
2x = 4
x=2
1st Co-ordinate = (0,4)
2nd Co-ordinate = (2,0)
So the graph will look like this.
y
8
2x + y = 4
7
6
5
4
3
2
1
–7 –6
–5 –4 –3
–2 –1
-1
-2
-3
-4
-5
-6
1
2
3 4
5
6
7 8
x
Simultaneous Equations
What are they?
• Simply 2 equations
– With 2 unknowns
– Usually x and y
• To SOLVE the equations means we find
values of x and y that
– Satisfy BOTH equations [work in]
– At same time [simultaneously]
Elimination Method
A
B
2x – y = 1
+
3x + y = 9
5x
= 10
=2
x
2x2–y=1
4–y=1
y=3
We have the same
number of y’s in each
If we ADD the equations, the y’s disappear
Divide both sides by 5
Substitute x = 2 in equation A
Answer
x = 2, y = 3
What if NOT same number of x’s or y’s?
A
B
A
3x + y = 10
5x + 2y = 17
If we multiply A by 2 we
get 2y in each
6x + 2y = 20
B 5x + 2y = 17
x
=3
In B 5 x 3 + 2y = 17
15 + 2y = 17
y=1
Answer
x = 3, y = 1
Equations
y
Linear Simultaneous Equations
Substitution Example
2x - y = -1
x + 2y = 7
1.
2.
Make ‘y’ the subject in 1.
y = 2x + 1
15
1 + 2y = 7
2y = 6
1a.
Substitute 1a into equation 2.
x + 2(2x + 1) = 7 5x + 2 = 7
5x = 5
Sub x = 1 into 2.
x=1
y=3
Solution (1, 3)
Slope of a Line
• Slope of a line:
y2  y1 rise
m

x2  x1 run
 x1, y1 
rise
 x2 , y2 
run
Slope-Intercept Form
The equation
y = mx+b
is called the slope-intercept form of an
equation of a line.
The letter m represents the slope and b
represents the y-intercept.
Point-Slope Form
The point-slope form of the equation of a line is
y  y1  m( x  x1 )
where m is the slope and (x1, y1) is a given point.
It is derived from the definition of the slope of a line:
Cross-multiply and
y2  y1
m
substitute the more
x2  x1
general x for x2
Example
Find the equation of the line through the points (–5, 7) and (4, 16).
Example
Find the equation of the line through the points (–5, 7) and (4, 16).
Solution:
16  7 9
m
 1
4  (5) 9
Now use the point-slope form with m = 1 and (x1, x2) = (4, 16).
(We could just as well have used (–5, 7)).
y  16  1( x  4)
y  x  4  16  x  12
Distance between two
points
The formula
Why?
• The formula, derived from the Pythagoras
theorem states that square a and square b,
add them together, and square root them,
you get the length of c, hypotenuse.
Example
Given the coordinates of two points we can use the formula
( x2  x1 ) 2 + ( y2  y1 ) 2
to directly find the distance between them. For example:
What is the distance between the points A(5, –1) and B(–4, 5)?
A(5, –1)
x1 y1
B(–4, 5)
x2 y2
(4  5)2 +(5  1)2 = ( 9)2 + 62
= 81+36
= 117
= 3 13
The Mid-Point of a Line
In general, the coordinates of the mid-point of the line segment
joining (x1, y1) and (x2, y2) are given by:
 x1 + x2 y1 + y2 
,


2
2


y
(x2, y2)
(x1, y1)
0
 x1 + x2 y1 + y2 
,


2 
 2
x
x1 + x2
2
is the mean of the xcoordinates.
y1 + y2
2
is the mean of the ycoordinates.
Example:
END
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