Linear Equations
Literal Equations
Simplifying Expressions & Solving Equations
Two Equations containing Two Variables
Solving Linear Equations
Plane Curves Parametric Equation
Linear Equations and Matrices
Trigonometric Identities and Conditional Equations
Solving Quadratic Equation
Solving Systems of Linear Equations by Graphing
Exponential and Logarithmic Equations
Quadratic Equations
Homework problems on homogeneous linear equations
Solving Quadratic Equations
Functions, Equations, and Inequalities
Solving Multiple-Step Equations
Test Description for Quadratic Equations and Functions
Solving Exponential Equations
Linear Equations
Linear Equations and Inequalities
Literal Equations
Quadratic Equations
Linear Equations in Linear Algebra
Investigating Liner Equations Using Graphing Calculator
represent slope in a linear equation
Linear Equations as Models
Solving Quadratic Equations by Factoring
Solving Equations with Rational Expressions
Solving Linear Equations
Solve Quadratic Equations by Completing the Square
Solving a Quadratic Equation

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Linear Equations

3.1 Linear equations

We encounter many situations in which we have to solve linear equations (that is,
equations in which the exponent of the unknown is 1). The skill to solve such equations
is indispensable. So, we now review the method of solving such equations.

Consider the problem: What is the number x which when multiplied by 8 gives 125?

We know at once that the number x must be greater than 10 since 10 times 8 is 80 but less
than 20 since 20 times 8 is 160. We can narrow down our estimates and obtain a
reasonable answer in a few minutes. However, we want a more efficient method of
finding the answer. To do so, we formulate the problem into an equation:

x(8) = 125 ___ or ___

Since 8 times x is to be 125, we reason that x must be the number that is obtained by
dividing 125 by 8. That is,

So, the number is ___ or ___ . The number ___ or ___ is called the solution of
the equation

To solve more complicated problems, we make use of the following properties of

(i) If equals are added to equals, the results are equal.
(ii) If equals are multiplied by equals, the results are equal.

Since subtraction of a number is defined to be the addition of its negative, the statement,
"If equals are subtracted from equals, the results are equal." is included in Statement (i).
Similarly, since division by a number is defined to be the multiplication by its reciprocal,
the property of equality about division is included in Statement (ii).

We can formally solve the simple equation given above as follows:

Of course, we do not write in such great detail, but it is important to know the principles
being used.

Example 1: Find the solution of the equation

Solution: We will write out the solution in great detail. Once you get used to, you
can omit many of the steps.

The idea is to bring the equation to the form ax = b by making use of the
properties of equality. To get rid of ___ from the left-hand side, we add
37 to both sides of the equation:


To get rid of ___ from the right-hand side, we add ___ to both sides of
this equation:


and so

Example 2: Find the solution of the equation 5(3t+ 2) = 2(5t+ 7)+ 8 .

Solution: Here the first problem is to simplify both sides of the equation:

Therefore, the solution of the equation is 2.4.

Example 3: Find the solution of the equation

Solution: Here again the first step is to simplify the left-hand side of the equation.
Combine the two terms into one fraction:

The solution is 3.55.
Since the procedure involved many steps, we should check the solution by
computing the left-hand side of the original equation for u = 3.55.

Exercises 3.1

1. Find the solution of each of the following equations: (You should check your
solution at least for some of the problems.)

3.2 Literal equations

We often have to solve an equation for one of the variables in terms of the other
variables. What we are doing in such cases is essentially postponing the substitution of
the values of the variables to the last. Consider the following problem for example.

Example 1: A man wants to buy a new car priced at $18,000, financing it at 5.75%
per annum and paying back the loan in monthly installments for 5 years.
What is the monthly payments?

We could have substituted the values into the installment purchase formula

and obtained the equation

We can solve this equation as follows:

So, the monthly payment is $345.90.

Or we can substitute the values in the monthly payment formula

and get

which is exactly the same as the value obtained above.

Let us review how we obtained the monthly payment formula from the installment
purchase formula.

We let

Then, we have

We have to solve this equation for M. Adding to both sides of the equation, we get

Multiplying both sides of the equation by C and then dividing both sides of the equation
by B, we obtain

Substituting back, we get the monthly payment formula:

Exercises 3.2

Solve each of the equations for the indicated variable: