Download Solving Linear Equations

Solving Linear Equations:
In solving an equation, we are trying to find the value of the variable(s)
that would make the quation a true equation.
E.g. x + 3 = 5
In this equation, if x = 2, then the equation is true. The basic idea in
solving a linear equation is to isolate the variable. I.e. we need the
variable on one side of the equation and the rest of the numbers in the
other side.
Central Idea in solving equations
For any equation, whatever we do to one side of the equation, if we do
the same action to the other side of the equation, the two sides will
remain equal. For example, if we add −3 to one side of the equation,
as long as we add −3 to the other side of the equation, the two sides
are still equal. Using this idea, we can perform operations on both
sides of the equation to try to isolate the variable.
E.g. 2x + 4 = 10
1
To solve this equation, we need to isolate the 2x term. We add −4 to
both sides of the equation to eliminate the 4 on the right side. (You
may also think of this as subtracting 4 from both sides)
(1) 2x + 4 = 10
(2) 2x + 4 + (−4) = 10 + (−4)
(3) 2x = 6
After adding −4 on both sides of the equation, the left side is simplified
to just 2x, and we get equation (3). Equations (1) and (3) are said
to be equivalent equations because they have the same solution(s).
The idea of solving an equation is to using operations to simplify an
originally complicated equation to an equivalent equation whose solution is easy to determine.
2
To solve equation (3), we need to isolate x by getting rid of the factor of
2. Remember that 2x means 2 times x, to get rid of the 2, we multiply
both sides of the equation by 1/2 (you may think of it as dividing by
2), we have:
1
1
(4) ( )2x = ( )6
2
2
(5) x = 3
The solution is now clear to us, x must be equal to 3 for the equation
to be true.
We can plug the value of 3 into the original equation to check:
2(3) + 4 = 6 + 4 = 10
The left and right hand side are both equal to 10, therefore 3 is a
proper solution.
3
In solving any equation, you can always do whatever operation you
need to the equation to try to isolate the variable, just keep in mind
that whatever you do to one side of the equation, you must do the
same operation to the other side of the equation.
E.g. 4x + 3 = 2x − 9
4x + 3 + (−3) = 2x − 9 + (−3) (add −3 (subtract 3) on both sides of
the equation)
4x = 2x − 12 (collect like terms)
4x + (−2x) = 2x − 12 + (−2x) (add −2x (subtract 2x) on both sides
of the quation because we only want x on one side of the quation)
2x = −12 (collect like terms)
1
1
1
(2x) = (−12) (multiply both sides by (divide by 2))
2
2
2
x = −6
4
Sometimes the distributive property needs to be applied to get ride of
parenthesis involved in an equation:
5(x + 2) − 5 = 2(x − 1) + x
5x + 10 − 5 = 2x − 2 + x (distributive property)
5x + 5 = 3x − 2 (simplify by collecting like terms)
5x + 5 + (−5) = 3x − 2 + (−5) (adding −5 to both sides)
5x = 3x − 7 (simplify)
5x − 3x = 3x − 7 − 3x (subtract 3x from both sides)
2x = −7 (simplify by collecting like terms)
1
1
(2x) = (−7) (multiply both sides by 1/2)
2
2
7
x=−
2
5
In solving an equation that involve fractions, multiply both sides of
the equation by the least common multiple of the denominators of the
fractions to get rid of the fraction first. Remember to distribute the
factor to each term of the expression.
E.g.
9
2
33
y+ y=
5
5
5
The LCM of the denominators is 5, therefore we multiply 5 to both
sides of the equation, we get:
2
33
9
5( y + y) = 5( )
5
5
5
9
2
33
5( y) + 5( y) = 5( ) (distribute 5 to each of the terms in both sides)
5
5
5
9y + 2y = 33 (simplify and get rid of the fractions)
11y = 33 (collect like terms)
y = 3 (divide by 11)
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E.g.
4
1
3
x− = x+2
3
5 15
The LCM of the denominators is 15, so we multiply this number to
both sides of the equations:
4
1
3
15( x − ) = 15( x + 2)
3
5
15
4
1
3
15( x) − 15( ) = 15( x) + 15(2) (distribute 15 to each term on each
3
5
15
side)
5(4)x − 3(1) = 3x + 30 (simplify the fractions)
20x − 3 = 3x + 30
20x = 3x + 33 (add 3 to both sides)
17x = 33 (subtract 3x from both sides)
33
x=
(divide by 17)
17
It is important that you do not selectively multiply the LCM to just the
term you need. If you multiply the LCM to one side of the equation,
it must be distributed to every term of the expression.
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A formula is an equation that represents a relationship between two or
more quantities. The quantities involved are represented by variables.
Sometimes it is useful to try to change a formula from one form the
the other.
E.g.
C = 2πr
This is the formula that expresses the relationship between the circumference of a circle (C) and its radius (r). What if we want to isolate
r? We can solve the equation for r just like we did with numbers:
To isolate r, we divide both sides by 2 and π, we get:
1
1
( )C = ( )2πr
2π
2π
C
=r
2π
C
r=
2π
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E.g.
(b1 + b2 )h
2
This is the formula that expresses the area of a trapazoid (A) in terms
of its two bases (b1 , b2 ) and height (h). If we want to isolate b1 , we can
do:
(b1 + b2 )h
) (multiply both sides by 2)
(2)A = (2)(
2
2A = (b1 + b2 )h (simplify)
1
1
2A = ((b1 + b2 )h) (divide by h)
h
h
2A
= b1 + b2 (simplify)
h
2A
− b2 = b1 (subtract b2 from both sides)
h
2A
b1 =
− b2 (subtract b2 from both sides)
h
A=
9
We may use the techniques of equation solving to solve some application problems. The most difficult task in any of these problems is
to set up the correct equation. Some guidelines in trying to solve the
problem:
Read the problem carefully, ask yourself, what (quantity) does the problem ask for?
Use a variable to represent the quantity you are looking for. When
choosing the variable name, try use meaningful names instead of generic
letters like x or y. For example, if the problem askes for the age of a
person, try using a as the variable for the unknown. If the problem
askes for the amount of time, try using t.
Always lable your variable. In your solution, there should always
be descriptions as to what the variable represent. It should never be
just an equation with variables. For example, in your solution, write
something like: ”Let t = the amt of time to reach city A”.
Check your answer for sensibility. If the question askes for the age of
a person and you get 208, or if the question askes for the number of
coins in a pocket, and you get a decimal number, something is wrong.
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E.g.
The sum of the page numbers on the facing pages of a book is 373,
what are the page numbers?
Ans: What does the question askes for? We randomly turn a book to
a page, the sum of the two pages facing us, is 373, the question askes
for what is each of the number of the page. Let p = the number of
the smaller of the two pages. Then what is the number of the larger
of the two pages? The two pages are next to each other, so the larger
of the two pages must be one more than the smaller one, so the large
one is p + 1, the sum of them is p + (p + 1), which is 373. This is the
equation we can set up:
p + (p + 1) = 373
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Solving inequalities:
To solve an inequality like x + 3 < 10 means we want to find the
values of x that would make the inequality true. Unlike an equation,
however, there are usually more (a lot more) values that could solve
an inequality. For the above inequality, x could be 4, could be 6, could
be 0, or any value that is less than 7. To express the solution to an
inequality, we usually use a graph or an interval:
(−∞, 7)
The −∞ symbol means any number to the left of 7 are all acceptable
values.
The technique of solving an inequality is the same as solving an equation, the only difference is that, when multiplying or dividing by a
negative number, the inequality sign much be switched.
E.g. 3x + 7 ≥ 19
3x ≥ 12 (subtract 7 from both sides)
x ≥ 4 (divide both sides by 3)
[4, ∞)
Note that we always use open parenthesis for the infinity symbol, because infinity is not a number, it is just a symbol to express an idea.
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E.g. −4x − 8 ≤ 2x + 16
−4x ≤ 2x + 24 (add 8 to both sides of inequality)
−6x ≤ 24 (subtract 2x from both sides)
1
1
− (−6x) ≥ − (24) (divide both sides by −6, notice that the direction
6
6
of the inequality symbol is switched)
x ≥ −4
[−4, ∞)
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Compound Inequality
x + 3 < 7 or 2x − 4 > 10
If we have two inequalities connected by an or conjunctive, we need to
solve each inequality individually, and any values that are solution to
one or the other, will be part of the solution.
x < 4 or x > 7
(−∞, 4) ∪ (7, ∞)
The union ∪ symbol means we want both of the two intervals together.
Any number that is in the first interval (numbers that are less than
4) or any number in the second interval (numbers greater than 7) are
solutions.
If two inequalities are connected by and conjunctive, solutions must be
values that solve both inequalities.
E.g.
3x − 5 < 16 and 2x + 6 > −2
3x < 21 and 2x > −8
x < 7 and x > −4
We need numbers that are less than 7 and at the same time greater
than −4, the solution set written in interval notation is:
(−4, 7)
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Absolute Value Equations and Inequalities
|x| = 3
How many solutions does this equation have? If x = 3, this solves
the equation, if x = −3, this still solves the equation. In general, the
equation:
|x| = c, where c is a non-negative real number, has two solutions,
namely, x = c or x = −c.
15
We can generalize this concept:
|3x + 5| = 22
The 3x + 5 is the expression inside the absolutel value, if it is equal
to 22 or −22, the absolute value of that will be 22, so we have two
equivalent equations:
3x + 5 = 22 or 3x + 5 = −22
Solving these two equations give us:
3x = 17 or 3x = −27
17
x=
or x = −9
3
16
E.g.
|x| < 5
What kind of values must x be to solve this: If x is larger than 5, this
will not work. If x is too negative, say if x < −5. If x = −6, then
|x| = 6, and this still fails, so we need:
x < 5 and −5 < x
What about:
|x| > 3
If x > 3, this works. If x < −3, say x = −4, then |x| = 4, and 4 is still
greater than 3, so the solution will be:
x > 3 or x < −3
In general, if c > 0, then the inequality
|x| < c has the solution: x < c and −c < x
|x| > c has solution: x > c or x < −c
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