Algebra Tutorials!
Rational Expressions
Graphs of Rational Functions
Solve Two-Step Equations
Multiply, Dividing; Exponents; Square Roots; and Solving Equations
Solving a Quadratic Equation
Systems of Linear Equations Introduction
Equations and Inequalities
Solving 2nd Degree Equations
Review Solving Quadratic Equations
System of Equations
Solving Equations & Inequalities
Linear Equations Functions Zeros, and Applications
Rational Expressions and Functions
Linear equations in two variables
Lesson Plan for Comparing and Ordering Rational Numbers
Solving Equations
Radicals and Rational Exponents
Solving Linear Equations
Systems of Linear Equations
Solving Exponential and Logarithmic Equations
Solving Systems of Linear Equations
Solving Quadratic Equations
Quadratic and Rational Inequalit
Applications of Systems of Linear Equations in Two Variables
Systems of Linear Equations
Test Description for RATIONAL EX
Exponential and Logarithmic Equations
Systems of Linear Equations: Cramer's Rule
Introduction to Systems of Linear Equations
Literal Equations & Formula
Equations and Inequalities with Absolute Value
Rational Expressions
Steepest Descent for Solving Linear Equations
The Quadratic Equation
Linear equations in two variables
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Solving Equations & Inequalities

Solutions to Examples

(a) Solve for x: 4x+5 = 0

subtract −5 from both sides
multi. both sides by 1/4
combine and done!

Solving an equation consists of a sequence of legal steps: adding
the same quantity to both sides (see equation (1)); multiplying
both sides by the same quantity (see equation (2)) were used
Presentation of Answer:

(b) Solve for x: 1/2x − 4 = 6.

add 4 to both sides
multi. both sides by 2
combine and done!

Note this equation initially was not of the form ax + b = 0.
There was a nonzero constant on right-hand side of the equation.
Obviously, the equation (1/2)x−4 = 6 is equivalent to (1/2)x−10 = 0;
consequently, no big deal was made of it.
Presentation of Solution:

(c) Solve for x: 7 − 3x = 2.

multiply both sides by −1
add 7 to both sides
multiply both sides by 1/3
combine and done!

Comments: Here the coefficient of x was negative. I opted to
multiplyb y −1 to ‘change the sign’ so that I would feel more
comfortable solving.
Presentation of Answer:

7.2. Solutions:
(a) Solve for x: (x − 1)2 = x2.

sub. x2 both sides: see (1)
diff. of squares!
multipy by −1: see (2)
add 1 to both sides: see (1)
divide by 2: see (3)

Comments: Note the recommended way of handling an equation
with squares on both sides. Rather than taking the square root
of both sides, take everything to one side of the equation and
factor it as a difference of squares: this is a much nicer method.
The other option is to expand all binomials and combine.
Presentation of Answer:

(b) Solve for x:

given equation
multiply both sides by x + 1
expand r.h.s.
sub. 2x2 from both sides: (1)
two steps in one!
divide both sides by 2

Presentation of Answer:

(c) Solve for x:

given equation
multiply both sides by 2x + 1
expand r.h.s.
add −8x − 1 to both sides
divide both sides by 11

Presentation of Answer:

(d) Solve for x:

the given equation
multiply both sides by 3x + 1
transpose equation: a = b <=> b = a
add −1 to both sides
divide both sides by 3

Presentation of Answer:

7.3. Solutions:
(a) Solve x2 − 5x+6 = 0.

factor l.h.s.


From (S-1) and the Zero-Product Principle, we deduce that

Presentation of Solution:

Comments: Note that this equation has two distinct solutions.
An alternate method of presenting the solutions to an
equation is to present the solution set: the set of all solutions.
Here we could have written,

(b) Solve x2 + 4x+4 = 0.


given equation
factor: perfect square!

Zero-Product Principle

Presentation of Answer:

Comments: The left-hand side is a quadratic polynomial that,
it turned out, was a perfect square. Note that this equation has
only one solution.

(c) Solve 6x2 − x − 2 = 0.
Solution: Let’s take the polynomial aside and factor it using the
factoring techniques explained in Lesson 6.

Factor: 6x2 −x−2 = 0. Begin bytry ing a factorization of the
6x2 − x − 2 = (3x + r1)(2x + r2),
where r1 and r2 are chosen so that r1r2 = −2 and the “crossproduct”
is correct. After some trial and error we get r1 = −2
and r2 = 1; thus,
6x2 − x − 2 = (3x − 2)(2x + 1). (S-2)
Solve: Now let’s take the factorization (S-2) and return to our
original equation:
(3x − 2)(2x+1) = 0

We can now obtain the solutions by applying the Zero-Product
Principle: (3x − 2)(2x + 1) = 0 implies either . . .

We deduce the solution set is
Presentation of Solution:

7.4. Solution: This will be short. All the work has been alreadydone.

complete the square
equation of the form (7)
divide by 2
take square root

Presentation of Solution:

7.5. Solution: Just follow the algorithm.

Step 2 & 3
Step 4: add/sub 1/2 coeff. of x
perfect square!
divide by 3
take square root both sides
solve for x and done


Presentation of Solution:

7.6. Solutions: We simply apply the Quadratic Formula!
(a) Solve for x: x2 − 5x + 6 = 0. This is the case where a = 1,
b = −5, and c = 6:

Presentation of Answers:

(b) Solve for x: x2+4x+4 = 0. This is the case where a = 1, b = 4,
and c = 4.

Presentation of Solution:

Comments: This is the case of the discriminant of zero. Therefore
we have onlyone solution.

(c) Solve for x: 6x2 − x − 2 = 0. Here a = 6, b = −1, and c = −2.

Presentation of Solutions:

(d) Solve for x: 3x2 − 3x + 1 = 0. This is the case of a = 3, b = −3
and c = 1. Before starting the standard calculations, we might
check the discriminant:
b2 − 4ac = (−3)2 − 4(3)(1) = 9 − 12 = −3 < 0.

We’ve saved ourselves some work. This equation has no solutions.

7.7. Solve 1 + 4y3 ≥ 0. We simply use the tools:

add −1 both sides: see (10)
multiply 1/4 both sides: see (11)
by (13)
apply properties of radicals: see (1)

The answer can be presented in three ways.
1. Inequalities. Solution:
2. Interval Notation. Solution:
3. Set Notation. Solution:

7.8. Solutions:
(a) Solve for x: 5x + 7 < 0.

add −7 to both sides (10)
multiply both sides by 1/5: see (11)

Presentation of Answer. Interval Notation:

(b) Solve for x: 3 − 9x ≥ 4.

add −3 to both sides
mul. by −1/9, inequality reversed! See (12)

Presentation of Answer. Set Notation:

(c) Solve for x: 3x5 + 4 ≥ 9.

add −4 to both sides
multiply by 1/3
take 5th root both sides: See (1)

Presentation of Answer. Inequalities.

(d) Solve for x: 3x2 + 4 ≤ 3.

add −4 to both sides
divide by 3

Now it is apparent that there are no solutions to this equation.
For any x, x2 ≥ 0 so x2 cannot be less than a negative number.
Presentation of Solution. Set Notation. The solution set is
the empty set.

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