Isolating Variables Using Fractions It may be necessary to use factoring to isolate a variable in an equation.. Quadratic Equations A quadratic equation is an equation in which the great
Trang 1Removing a Common Factor
If a polynomial contains terms that have common factors, you can factor the polynomial by using the reverse
of the distributive law
Example:
In the binomial 49x3+ 21x, 7x is the greatest common factor of both terms Therefore, you can divide 49x3+ 21x by 7x to get the other factor.
49x3
7
+
x
21x
= 49
7
x x
3
+ 2 7
1
x x
= 7x2+ 3
Thus, factoring 49x3+ 21x results in 7x(7x2+ 3)
Isolating Variables Using Fractions
It may be necessary to use factoring to isolate a variable in an equation
Example:
If ax – c = bx + d, what is x in terms of a, b, c, and d?
■ The first step is to get the “x” terms on the same side of the equation:
ax – bx = c + d
■ Now you can factor out the common “x” term on the left side:
x(a – b) = c + d
■ To finish, divide both sides by a – b to isolate x:
x(
a
a
–
–
b
b)
= c
a
+ – b d
■ The a – b binomial cancels out on the left, resulting in the answer:
x = c a + – bd
Quadratic Trinomials
A quadratic trinomial contains an x2term as well as an x term; x2– 5x + 6 is an example of a quadratic
trinomial Reverse the FOIL method to factor
■ Start by looking at the last term in the trinomial, the number 6 Ask yourself, “What two integers, when multiplied together, have a product of positive 6?”
■ Make a mental list of these integers:
■ Next, look at the middle term of the trinomial, in this case, the negative 5x Choose the two factors
from the above list that also add up to negative 5 Those two factors are: –2 and –3
■ Thus, the trinomial x2– 5x + 6 can be factored as (x – 3)(x – 2).
■ Be sure to use FOIL to double check your answer The correct answer is:
(x – 3)(x – 2) = x2– 2x – 3x + 6 = x2– 5x + 6
Trang 2Algebraic Fractions
Algebraic fractions are very similar to fractions in arithmetic.
Example:
Write 5x–
1
x
0
as a single fraction.
Solution:
Just like arithmetic, you need to find the LCD of 5 and 10, which is 10 Then change each fraction into an equivalent fraction that has 10 as a denominator:
5
x–
1
x
0
5
x(
(
2 2
) )
– 1
x
0
1
2 0
x– 1
x
0
1
x
0
Reciprocal Rules
There are special rules for the sum and difference of reciprocals Memorizing this formula might help you
be more efficient when taking the GRE test:
■ If x and y are not 0, then 1x+ 1yx
x
+
y y
.
■ If x and y are not 0, then 1x– 1yy
x
–
y x
.
Quadratic Equations
A quadratic equation is an equation in which the greatest exponent of the variable is 2, as in x2+ 2x – 15 =
0 A quadratic equation had two roots, which can be found by breaking down the quadratic equation into two simple equations You can do this by factoring or by using the quadratic formula to find the roots
Zero-Product Rule
The zero-product rule states that if the product of two or more numbers is 0, then at least one of the numbers is 0.
Example:
Solve for x.
(x + 5)(x – 3) = 0
Using the zero-product rule, it can be determined that either x + 5 = 0 or that x – 3 = 0.
x + 5 – 5 0 – 5 or x – 3 + 3 0 + 3
Trang 3Solving Quadratic Equations by Factoring
Example:
x2+ 4x = 0 must be factored before it can be solved: x(x + 4) = 0, and
the equation x(x + 4) = 0 becomes x = 0 and x + 4 = 0.
–4 = –4
x = 0 and x = –4
■ If a quadratic equation is not equal to zero, you need to rewrite it
Example:
Given x2– 5x = 14, you will need to subtract 14 from both sides to form
x2– 5x – 14 = 0 This quadratic equation can now be factored by using the zero-product rule.
Therefore, x2– 5x – 14 = 0 becomes (x – 7)(x + 2) = 0 and using the zero-product rule,
you can set the two equations equal to zero
Solving Quadratic Equations Using the Quadratic Formula
The standard form of a quadratic equation is ax2+ bx + c = 0, where a, b, and c are real numbers (a 0) To use the quadratic formula to solve a quadratic equation, first put the equation into standard form and
iden-tify a, b, and c Then substitute those values into the formula:
x =
For example, in the quadratic equation 2x2– x – 6 = 0, a = 2, b = –1, and c = –6 When these values are
substituted into the formula, two answers will result:
x =
x =
x =1 47
x = 1 +47or 1 –47
x = 2 or x = –46or –23
1 49
4
–(–1) (–1) 4(2)(– 2 – 6)
2(2)
–b b c2– 4a
2a
Trang 4Quadratic equations can have two real solutions, as in the previous example Therefore, it is important
to check each solution to see if it satisfies the equation Keep in mind that some quadratic equations may have only one or no solution at all
Check:
2x2– x – 6 = 0
2(2)2– (2) – 6 = 0 or 2(–23)2– (–23) – 6 = 0
2(4) – 8 = 0 2(94) – 412= 0
Therefore, both solutions are correct
Systems of Equations
A system of equations is a set of two or more equations with the same solution Two methods for solving a
system of equations are substitution and elimination.
Substitution
Substitution involves solving for one variable in terms of another and then substituting that expression into the second equation
Example:
2p + q = 11 and p + 2q = 13
■ First, choose an equation and rewrite it, isolating one variable in terms of the other It does not matter which variable you choose:
2p + q = 11 becomes q = 11 – 2p
■ Second, substitute 11 – 2p for q in the other equation and solve:
p + 2(11 – 2p) = 13
p + 22 – 4p = 13
22 – 3p = 13
22 = 13 + 3p
9 = 3p
p = 3
Trang 5■ Now substitute this answer into either original equation for p to find q:
2p + q = 11
2(3) + q = 11
6 + q = 11
q = 5
Thus, p = 3 and q = 5.
Elimination
Elimination involves writing one equation over another and then adding or subtracting the like terms so that
one letter is eliminated
Example:
x – 9 = 2y and x – 3 = 5y
■ Rewrite each equation in the formula ax + by = c.
x – 9 = 2y becomes x – 2y = 9 and x – 3 = 5y becomes x – 5y = 3.
■ If you subtract the two equations, the “x” terms will be eliminated, leaving only one variable:
Subtract:
3
3y= 6
3
y = 2
■ Substitute 2 for y in one of the original equations and solve for x.
x – 9 = 2y
x – 9 = 2(2)
x – 9 = 4
x – 9 + 9 = 4 + 9
x = 13
■ The answer to the system of equations is y = 2 and x = 13.
Inequalities
Linear inequalities are solved in much the same way as simple equations The most important difference is that
when an inequality is multiplied or divided by a negative number, the inequality symbol changes direction
x – 2y = 9
–(x – 5y = 3)
Trang 6–30 –15
Solving Linear Inequalities
To solve a linear inequality, isolate the letter and solve the same as you would in a first-degree equation Remember to reverse the direction of the inequality sign if you divide or multiply both sides of the equation
by a negative number
Example:
If 7 – 2x 21, find x.
■ Isolate the variable:
7 – 2x 21
–7 –7
–2x 14
■ Because you are dividing by a negative number, the inequality symbol changes direction:
–
–
2
2
x
–
14
2
x –7
■ The answer consists of all real numbers less than –7
Solving Combined (or Compound) Inequalities
To solve an inequality that has the form c ax + b d, isolate the letter by performing the same operation
on each member of the equation
Example:
If –10 –5y – 5 15, find y.
■ Add five to each member of the inequality:
–10 + 5 –5y – 5 + 5 15 + 5
–5 –5y 20
■ Divide each term by –5, changing the direction of both inequality symbols:
–
–
5
5
–
–
5
5
y
–
20 5
= 1 y –4
■ The solution consists of all real numbers less than 1 and greater than – 4
... inequality, isolate the letter and solve the same as you would in a first-degree equation Remember to reverse the direction of the inequality sign if you divide or multiply both sides of the equation...■ Isolate the variable:
7 – 2x 21
? ?7 ? ?7
–2x 14
■ Because you are dividing by a negative number, the inequality symbol changes... in terms of the other It does not matter which variable you choose:
2p + q = 11 becomes q = 11 – 2p
■ Second, substitute 11 – 2p for q in the other equation