For example, let’s look at what happens when you manipulate the equation 3x
+ 2 = 5, with x = 1.
- Subtract 2 from both sides:
3x + 2 - 2 = 5 - 2
3x + 0 = 3
3 (1) = 3
3=3
- Multiply both sides by 2:
2 ( 3x + 2 ) = 2 ( 5)
6x + 4 =10
6 (1 ) + 4 = 10
10 = 10
- Add 4 to both sides:
3x + 2 + 4 = 5 + 4
3x + 6 = 9
3 (1) + 6 = 9
9 = 9
These examples show that you can tamper with the equation in any way you want,
as long as you commit the same tampering on both sides. If you follow this rule,
you can manipulate the question how you want without affecting the value of its
variables.
Solving an Equation with One Variable
To solve an equation with one variable, you must isolate that variable.
Isolating a variable means manipulating the equation until the variable is the
only thing remaining on one side of the equation. Then, by definition, that
variable is equal to everything on the other side, and you have successfully
“solved for the variable.”
For the quickest results, take the equation apart in the reverse order of
operations. That is, first add and subtract any extra terms on the same side as
the variable. Then, multiply and divide anything on the same side of the
variable. Next, raise both sides of the equation to a power or take their roots
according to any exponent attached to the variable. And finally, do anything
inside parentheses. This process is PEMDAS in reverse (SADMEP!). The idea is to
“undo” everything that is being done to the variable so that it will be isolated
in the end. Let’s look at an example:
In this equation, the variable x is being squared, multiplied by 3, added
to 5, etc. We need to do the opposite of all these operations in order to
isolate x and thus solve the equation.
First, subtract 1 from both sides of the equation:
Then, multiply both sides of the equation by 4:
Next, divide both sides of the equation by 3:
(3x2 + 5 ) × 3 ÷ 3 = 240 ÷ 3
(3x2 + 5 ) = 80
Now, subtract 5 from both sides of the equation:
(3x2 + 5 -5 ) = 80 - 5
3x2 = 75
Again, divide both sides of the equation by 3:
3x2 ÷ 3 = 75 ÷ 3
3x2 = 25
Finally, take the square root of each side of the equation:
We have isolated x to show that x = ±5.
Sometimes the variable that needs to be isolated is not conveniently located.
For example, it might be in a denominator or an exponent. Equations like these
are solved the same way as any other equation, except that you may need
different techniques to isolate the variable. Let’s look at a couple of
examples:
Solve for x in the equation 1 / x + 2 = 4.
The key step is to multiply both sides by x to extract the variable from
the denominator. It is not at all uncommon to have to move the variable from
side to side in order to isolate it.
Remember, performing an operation on a variable is mathematically no different
than performing that operation on a constant or any other quantity.
Here’s another, slightly more complicated example:
This question is a good example of how it’s not always simple to isolate a
variable. (Don’t worry about the logarithm in this problem—we’ll review these
later on in the chapter.) However, as you can see, even the thorniest problems
can be solved systematically—as long as you have the right tools. In the next
section, we’ll discuss factoring and distributing, two techniques that were used
in this example.
So, having just given you a very basic introduction to solving equations, we’ll
reemphasize two things:
- Do the same thing to both sides.
- Work backward (with respect to the order of operations).
Now we get into some more interesting tools you will need to solve certain
equations.
Distributing and Factoring
Distributing and factoring are two of the most important techniques in algebra.
They give you ways of manipulating expressions without changing the expression’s
value. So it follows that you can factor or distribute one side of the equation
without doing the same for the other side of the equation.
The basis for both techniques is the following property, called the distributive
property:
a × ( b + c + ......) = a × b + a × c + ....
Similarly:
a × ( - b - c - ......) = - a × b - a × c - ....
a can be any kind of term, from a variable to a constant to a combination
of the two.
Distributing
tor into an expression within parentheses, you simply
multiply each term inside the parentheses by the factor outside the parentheses.
For example, consider the expression 3y(y2 – 6):
3
y(
y2 – 6) = 3
y3 - 18
y
If we set the original, undistributed expression equal to another expression,
you can see why distributing facilitates the solving of some equations. Solving
3y (y2 – 6) = 3y3 + 36 looks quite
difficult. But if you distribute the 3y, you get:
3y3 - 18 y = 3y3 + 36
Subtracting 3y3 from both sides gives us:
- 18 3y= 36 y
- y = 2
Factoring
Factoring an expression is essentially the opposite of distributing. Consider
the expression 4x3 – 8x2 + 4x, for
example. You can factor out the GCF of the terms, which is 4x:
4x3 – 8x + 4x = 4x(x2 - 2 + 1)
The expression simplifies further:
4x(x2 - 2 + 1) = 4x(x - 1) 2
See how useful these techniques are? You can group or ungroup quantities in an
equation to make your calculations easier. In the last example from the previous
section on manipulating equations, we distributed and factored to solve
an equation. First, we distributed the quantity log 3 into the sum of x
and 2 (on the right side of the equation). We later factored the term x
out of the expression x log 2 – x log 3 (on the left side of the
equation).
Distributing eliminates parentheses, and factoring creates them. It’s your job
as a Math IC mathematician to decide which technique will best help you solve a
problem.
Let’s see a few examples:
Combining Like Terms
After factoring and distributing, there are additional steps you can take to
simplify expressions or equations. Combining like terms is one of the simpler
techniques you can use, and involves adding or subtracting the coefficients of
variables that are raised to the same power. For example, by combining like
terms, the expression:
x2 - x3 + 4 x2 + 3x3
can be simplified to:
x3 (- 1 + 3) + x2 ( 1 + 4)= 2x3 + 5x2
by adding the coefficients of the variable x3 together and the
coefficients of x2 together.
Generally speaking, when you have an expression in which one variable is raised
to the same power in different terms, you can factor out the variable and add or
subtract the coefficients, combining them into one coefficient and therefore
combining the “like” terms into one term. A general formula for combining like
pairs looks something like this:
axk + bxk + cxk + =xk( a + b + c)
Next to display next topic in the chapter.
Mathematics Practice Questions
Video Lessons and 10 Fully Explained Grand Tests
Large number of solved practice MCQ with explanations. Video Lessons and 10 Fully explained Grand/Full Tests.