Lesson SOLVING EQUATIONS WHERE THE VALUE ON ONE SIDE OF THE EQUATION IS EQUAL TO THE VALUE ON THE OTHER SIDE OF THE EQUATION

an equation has 2 sides to it and a symbol in the middle that is telling you the relationship between the value on the left side of the equation and the value on the right side of the equation.

NOTE
while showing you the symbols involved in equations where the value on one side of the equation does not equal the value on the other side of the equation, this lesson will address only those equations where the value on the left side of the equation equals the value on the right side of the equation.

TERMS USED IN THIS LESSON

EXPRESSION whatever is on either side of an equation is called an expression.
example:
x^2 + 2x + 7 = 3
x^2 + 2x + 7 is the expression on the left side of the equation.
3 is the expression on the right side of the equation.

EQUATION describes the relationship between the expression on the left side of the
equation and the expression on the right side of the equation.
example:
x^2 + 2x + 7 = 3 expresses equality
x^2 + 2x + 7 > 3 expresses inequality

EQUATION SYMBOLS

the symbols used to evaluate expressions in equations are:

symbol     meaning

=          the value on the left side of the equation is equal to: 
           the value on the right side of the equation.

>          the value on the left side of the equation is greater than:
           the value on the right side of the equation.


>=         the value on the left side of the equation is greater than or equal to:
           the value on the right side of the equation.


<          the value on the left side of the equation is less than:
           the value on the right side of the equation.


<=         the value on the left side of the equation is less than or equal to:
           the value on the right side of the equation.


<>         the value on the left side of the equation is not equal to:
           the value on the right side of the equation.

note that if the value on the left side of the equation is not equal to the value on the right side of the equation, then the value on the left side of the equation will either be less than the value on the right side of the equation (<) or it will be greater than the value on the right side of the equation (>). this is the reason for the symbol expressing inequality to be shown as (<>).
some examples:

x = 5            (x is equal to 5)
x > 4            (x is greater than 4)
x >= 5           (x is greater than or equal to 5)
x < 5            (x is less than 5)
x <= 4           (x is less than or equal to 4)
x <> 4           (x is not equal to 4)

variables are letters that are used to represent values that can vary (not always the same).
that's why they are called variables.

for example:
y can represent the number 5 at one time and the number 10 at another time.
x can represent the number 1 at one time and the number 2 at another time.

DEPENDENT AND INDEPENDENT VARIABLES

if the value of one variable doesn't depend on the value of another variable, then it is called the independent variable.

if the value of one variable depends on the value of another variable, then it is called the dependent variable.

the dependent variable and the independent variable are tied to each other through the operations indicated by the equation.

an example:

the values of 1 and 2 are assigned to the value of x.
x is the independent variable because it's value doesn't depend on the value of any other variable.

an equation is created that determines the relationship between x and y.
for example, the equation is:
y = 5*x

this equation states that the value of y will always be 5 times the value of x.

when x = 1, the value of y will be equal to 5 * 1 = 5
when x = 2, the value of y will be equal to 5 * 2 = 10

the value of x was assigned and does not depend on the value of any other variable. this makes it the independent variable.

the value of y was determined by the value of x after the operations indicated in the equation were performed. this make it the dependent variable.

note that independence and dependence itself is variable.

if my equation is y = 5*x, then x is the independent variable and y is the dependent variable because it depends on the value of x.

if i turn the equation around and solve for x, then the equation becomes x = y/5.
now y is the independent variable and x is the dependent variable because the value of x is now dependent on the value of y.

the independent variable is assigned a value.
the dependent variable is tied to the independent variable by an equation that defines the procedure required to generate the value of y based on the value of x.

in the equation of y = 5*x, the procedures is to multiply the value of x by 5 to get the value of y.
in the equation of x = y/5, the procedure is to divide the value of y by 5 to get the value of x.


EQUATION RULES

equations define the relationship between the value on the left side of the equation and the value on the right side of the equation.

in order for the equation to be true, that relationship needs to be preserved.
the relationship we are talking about here is an equality relationship.
the value on the left side of the equation is equal to the value on the right side of the equation.
inequality relationships are left to another lesson.

when you work with equality equations, there is one basic rule that has to be followed:

WHATEVER YOU DO TO ONE SIDE OF THE EQUATION, YOU HAVE TO DO TO THE OTHER SIDE OF THE EQUATION IN ORDER TO PRESERVE THE RELATIONSHIP.

here's some examples.

original equation:
x = y

revised equations:

x + c = y + c
x - c = y - c
c*x = c*y
x/c = y/c
x^c = y^c
root(c,x) = root(c,y)
log(c,x) = log(c,y)

examples for the revised equations where x = 3 and y = 3 and c = 2

x = y becomes 3 = 3
x + c = y + c becomes 3 + 2 = 3 + 2 becomes 5 = 5
x - c = y - c becomes 3 - 2 = 3 - 2 becomes 1 = 1
c*x = c*y becomes 2 * 3 = 2 * 3 becomes 6 = 6
x/c = y/c becomes 3/2 = 3/2
x^c = y^c becomes 3^2 = 3^2 becomes 9 = 9
root(c,x) = root(c,y) becomes root(2,3) = root (2,3) becomes sqrt(3) = sqrt(3)

note that the expressions on each side of the revised equation are not the same as the expressions on each side of the original equation, but the equality is preserved.

3 = 3
5 = 5
1 = 1
6 = 6
3/2 = 3/2
9 = 9
sqrt(3) = sqrt(3)

the whole idea behind solving equality equations is to preserve the equality.
preserving the equality allows us to solve for the value of specific variables.

SOLVING PROBLEMS INVOLVING EQUATIONS

the main goal in solving equation is to isolate the variable you are solving for to one side of the equation while the remaining variables and constants are on the other side of the equation.

for example:

if the original equation is x + y + z = 32, and you want to solve for x, then you have to isolate x to one side of the equation and move or keep 32 and y and z on the other side of the equation.

you can isolate x to the left side of the equation or to the right side of the equation.
it really doesn't matter.
most of the times, however, you will want to isolate x to the left side of the equation.

keep in mind that you can switch sides in an equality equation and the equality will still be preserved.

if the equation is x = y, then the equation y = x means the same thing.
whether x is on the left or the right doesn't matter.
if you see an equation that says x^2 + 2x = y, then you can flip it around so that it says y = x^2 + 2x and it will mean the same thing.

back to our original equation being x + y + z = 32
we want to solve for x
we will want to keep x on the left side of the equation in this case.
this means that we want x on the left side of the equation and we want y and z and 32 on the right side of the equation.

REVERSE OPERATIONS ON VARIABLES OR CONSTANTS

we move variables or constants from one side of an equation to the other side of an equation by performing reverse operations.

if the variable or constant is an addition, we apply a subtraction.
if the variable or constant is a subtraction, we apply an addition.
if the variable or constant is a multiplication, we apply a division.
if the variable or constant is a division, we apply a multiplication.

back to our example:
our equation is x + y + z = 32
we subtract y and z from both sides of the equation to get:
x + y + z - y - z = 32 - y - z
y and -y cancel out on the left side of the equation.
z and -z cancel out on the left side of the equation.
we are left with:
x = 32 - y - z
x has been isolated to the left side of the equation.
y and z have been moved to the right side of the equation by performing reverse operations.
they were being added on the left side of the equation so we subtracted them from both sides of the equation.
this had the effect of canceling them out on the left side of the equation because any value subtracted from itself will be equal to 0, and 0 is not shown.
y - y = 0
z - z = 0
expression on the left side of the equation becomes x + 0 + 0 which becomes x.

now we'll solve some simple problems so you can see how the rule works in practice.

EXAMPLE 1

3x + 2y = 3
solve for y
subtract 3x from both sides of the equation to get:
3x + 2y - 3x = 3 - 3x
reorder the terms to get:
3x - 3x + 2y = 3 - 3x
combine like terms to get:
0 + 2y = 3 - 3x
simplify to get:
2y = 3 - 3x
divide both sides of the equation by 2 to get:
2y / 2 = (3 - 3x) / 2
simplify to get:
y = (3 - 3x) / 2

EXAMPLE 2:

5x = 10
solve for x
divide both sides of the equation by 5 to get:
5x/5 = 10/5
simplify to get:
x = 2

EXAMPLE 3:

sqrt(x) = 6
square both sides of the equation to get:
x = 6^2
simplify to get:
x = 36
this example did not move any variable or constant from one side of the equation to the other, but it did take the square root of both sides of the equation in order to get the value of x.

EXAMPLE 4:

x/5 = 7/3
multiply both sides of the equation by 5 to get:
x/5 * 5 = (7/3) * 5
simplify to get:
x = (7/3) * 5
this is equivalent to:
x = (7*5) / 3 which becomes:
x = 35 / 3

EXAMPLE 5:

7/3 = 5/x
multiply both sides of this equation by x to get:
(7/3) * x = (5/x) * x
this is equivalent to:
7x/3 = (5x)/x
the x in the numerator and denominator on the right side of the equation cancels out and you are left with:
7x/3 = 5
multiply both sides of this equation by 3 to get:
7x/3 * 3 = 5 * 3 which becomes:
7x = 15
divide both sides of this equation by 7 to get:
7x / 7 = 15 / 7 which becomes:
x = 15/7 because the 7 in the numerator and denominator on the left side of the equation cancel out.

in all of these equations, you confirm that the answer is correct by replacing the value of x in the original equation with the value of x that you solved for to determine if the original equation is still true. it it is still true, then the variable that you solved for has been solved for successfully.

let's do that for all of these examples.

CONFIRMING THE VALUE OF X IN THE ORIGINAL EQUATIONS IS TRUE.

CONFIRMING EXAMPLE 1:

our original equation is:
3x + 2y = 3
our final equation is:
y = (3 - 3x) / 2
we had been asked to solve for the value of y and we did.
we did not know the value of x, so there was no real way of checking to see if we did it correctly.
we assign a value to x arbitrarily.
any value of x will do since the equation must be true for all values of x.
assign the value of 15 to x
we get:
x = 15
y = (3 - 3x) / 2 becomes y = (3 - 3*15) / 2 which becomes (3 - 45) / 2 which becomes -42 / 2 which becomes -21.
the value of y is equal to -21 when the value of x is equal to 15.
now we go back to the original equation and replace x with 15 and y with -21.
the original equation is:
3x + 2y = 3
after replacement, the original equation becomes:
(3 * 15) + (2 * -21) = 3
performing the indicated operations gets:
45 - 42 = 3
simplifying gets:
3 = 3
since the original equation is still true, we solved for the value of y successfully.

CONFIRMING EXAMPLE 2:

the original equation is:
5x = 10
the final equation is:
x = 2
replace x with 2 in the original equation to get:
5*2 = 10
simplify to get:
10 = 10
since the original equation is still true, we solved for the value of x successfully.

CONFIRMING EXAMPLE 3:

the original equation is:
sqrt(x) = 6
the final equation is:
x = 36
replace x with 36 in the original equation to get:
sqrt(36) = 6
use your calculator to find the square root of 36 to get:
6 = 6
since the original equation is still true, we solved for the value of x successfully.

CONFIRMING EXAMPLE 4:

original equation is:
x/5 = 7/3
final equation is:
x = 35/3
replace x with 35/3 in the original equation to get:
(35/3) / 5 = 7/3
(35/3) / 5 is the same as (35/3) * (1/5) which is equivalent to (35)/(3*5)
simplify to get:
35/15 = 7/3
divide both numerator and denominator of the fraction on the left side of the equation by 5 to get:
7/3 = 7/3
since the original equation is still true, we solved for the value of x successfully.

CONFIRMING EXAMPLE 5:

original equation is:
7/3 = 5/x
final equation is:
x = 15 / 7
replace x with 15/7 in the original equation to get:
7/3 = 5/(15/7)
5 / (15/7) is the same thing as 5 * (7/15) which is equivalent to (5*7)/15 which is equivalent to 35/15.
our equation becomes:
7/3 = 35/15
divide the numerator and the denominator on the right side of the equation by 5 to get:
7/3 = 7/3
since the original equation is still true, we solved for the value of x successfully.


SOME MORE EXAMPLES:

EXAMPLE 6:

this one's a little more complicated, but it shows you that the same rule applies regardless of the equation.

the equation of a circle is (x-a)^2 + (y-b)^2 = r^2
a is the x value of the coordinate of the center of the circle.
b is the y value of the coordinate of the center of the circle.
r is the radius of the circle.
a coordinate in 2 dimensions is expressed as a pair of values in parentheses, the standard form being (x,y).

the equation of our circle is:
(x-5)^2 + (y-3)^2 = 36
the coordinates of the center of this circle will be (5,3)
5 is the x value of the coordinate.
3 is the y value of the coordinate.
the radius of this circle will be 6.

we want to graph the equation of this circle.
in order to do that we need to solve for y.
in order to do that we need to get y on the left side of the equation all by itself and we need to get the rest of the variables and constants on the right side of the equation.

we start with:
(x-5)^2 + (y-3)^2 = 36
we subtract (x-5)^2 from both sides of the equation to get:
(y-3)^2 = 36 - (x-5)^2
we take the square root of both sides of the equation to get:
y-3 = +/- sqrt(36 - (x-5)^2)
we add 3 to both sides of this equation to get:
y = 3 +/- sqrt(36 - (x-5)^2)
this translates to:
y = 3 + sqrt(36 - (x-5)^2)
or
y = 3 - sqrt(36 - (x-5)^2)
we have solved for y and are now ready to graph this equation.
(x-5)^2 + (y-3)^2 = 36 is graphed as 2 separate equations.
they are:
y = 3 + sqrt(36 - (x-5)^2) and y = 3 - sqrt(36 - (x-5)^2)
the graph of these equations is shown below:

in this particular case, the graph of our equations shows that we have successful transformed the equation of:
(x-5)^2 + (y-3)^2 = 36 into the equations of:
y = 3 + sqrt(36 - (x-5)^2) and y = 3 - sqrt(36 - (x-5)^2)
by solving for y.
the center of our circle is (5,3).
the radius of our circle is equal to 6.
this is the expected values for the center of the circle and the radius of the circle so the graph confirms we solved for y successfully and are able to accurately present the original equation on it.