In other words, the equation we just solved should really look like this on your scratch paper: Check by Plugging Back In Always check your answer by plugging it back into the original e
Trang 1Equations as Balanced Scales
Algebra is really common sense once you start
think-ing of every equation as a balanced scale The terms
on either side of the equals sign must be “balanced,”
just like the weights on a scale
The laws of equations come from the
common-sense rules for keeping a scale balanced
Imag-ine that you are the keeper of a scale, and you
must keep it balanced What would you do if
someone took weights from one side of the
scale? You’d remove an equal weight from the
other side, of course This is a law of equality:
anything you do to one side of the equation, you
must do to the other to maintain the balance
Example:
If 12x − 8 = 28, then what is the value of 3x − 2?
Don’t worry about solving for x, because that’s not what
the question is asking for Notice that the expression
you are given, 12x− 8, is 4 times the expression you are
looking for, 3x− 2 So to turn the given expression into
the one you want, divide by 4 Of course, you must do
the same to the other side to keep the balance:
Solving as Unwrapping
Solving simple algebraic equations is basically the
same thing as unwrapping a present (And it’s just as
fun, too, right? Okay, maybe not.) Wrapping a present
involves a sequence of steps: 1 Put the gift in the box 2
Close the box 3 Wrap the paper around the box 4 Put
the bow on Here’s the important part: unwrapping the
present just means inverting those steps and
revers-ing their order: 1 Take off the bow 2 Unwrap the
paper 3 Open the box 4 Take out the gift.
Example:
Solve for x: 5x2− 9 = 31
The problem is that x is not alone on the left side; it is
“wrapped up” like a gift How is it wrapped? Think of
the order of operations for turning x into 5x2− 9:
1 Square it: x2
2 Multiply by 5: 5x2
3 Subtract 9: 5x2− 9
12 8
4
28
x
x
So to “unwrap” it, you reverse and invert the steps:
1 Add 9: (5x2− 9) + 9 = 5x2
2 Divide by 5: 5x2/5 = x2
3 Find the square roots (both of them!): = ±x
If you perform these steps to both sides, 5x2− 9 = 31 transforms into
Watch Your Steps
To solve that last equation, we had to perform three operations Many equations, as you probably know, require more steps to solve It is very important that you keep track of your steps so that you can check your work if you need to In other words, the equation
we just solved should really look like this on your scratch paper:
Check by Plugging Back In
Always check your answer by plugging it back into the original equation to see if it works Remember that solving an equation means simply finding the value of each unknown that makes the equation true
Example:
Are solutions to 5x2− 9 = 31? Plug them in:
− 9 = 5(8) − 9 = 40 − 9 = 31 (Yes!)
5 8
2
±
( )± 8
( ) :
5 9 31 1
2
Step Add 9 Step 2 (Divide by 5):
+ +
=
9 9
5x2 40 5 Step 3 (Simplify):
x2
5
40 5
=
x2 =8 Step 4 (Square rooot): x= ± 8
x= ± 8
± x2
Lesson 1: Solving Equations
There’s a lot of detail to learn and understand to do well on the SAT For more tools and resources that will help, visit our Online Practice Plus at www.MHPracticePlus.com/SATmath.
Trang 2Concept Review 1: Solving Equations
1 Explain the laws of equality
2 Are there any operations you can perform to both sides of an equation that will not yield a true equation?
Explain
Show your steps and check your work in solving the following equations
Solve the following equations for the given expression by performing one operation on both sides.
7 If , then 10x+ 12 = Operation: Solution:
8 If 18y + 12 = 7, then 6y + 4 = Operation: Solution:
5
2x + =3 7
x+ = −2 x
3
5 2
2
3
2
2
x
x
=
Trang 3SAT Practice 1: Solving Equations
1. If 5d + 12 = 24, then 5d − 12 =
(A) −24 (B) −12 (C) 0
(D) 12 (E) 24
2. What number decreased by 7 equals 5 times the
number?
3. If , then y+ 5 =
4. If 2x2− 5x = 9, then 12x2− 30x =
(A) −54 (B) −6 (C) 18
(D) 36 (E) 54
( p+ 2)2= ( p − 5)2
5. The equation above is true for which of the
fol-lowing values of p?
(A) −2 and 5
(B) 2 and −5
(C) 0 and 1
(D) 1.5 only
(E) 3.5 only
2
5
2
2
y
y
=
4 7
−4
7
−5 7
−7 5
−7
4
6. If what is the value of x?
7. The product of x and y is 36 If both x and y are
integers, then what is the least possible value of
x − y?
(A) −37 (B) −36 (C) −35 (D) −9 (E) −6
8. The graph of y = f(x) contains the points
(−1, 7) and (1, 3) Which of the following could
be f(x)?
I f(x) = 5x − 2
II f(x) = x2− 2x + 4
III f(x) = −2x + 5
(A) I only (B) I and II only (C) I and III only (D) II and III only (E) I, II, and III
9. If , which of the following gives all
possible values of x?
(A) 9 only (B) −9 and 9 (C) 81 only (D) 81 and −81 (E) 961
10. For all positive values of m and n,
2m−2n
2
3 2
m n
+
3 2 2
+ n
m
2
m n
−
3
3 2
x
m nx− = ,
20− x=11
−7 5
−24 7
−25 2
5 7
5 1
x+ = ,
1
2
3
4
5
7
8
9
6
1
0
2
3
4
5
7
8
9
6
1 0 2 3 4 5 7 8 9 6
1 0 2 3 4 5 7 8 9 6
Trang 4Multiply by 3: 2x2= 3x2
Subtract 2x2: 0 = x2
Take square root: 0 = x
6
Cross-multiply: 2x + 4 = 3x − 15
7 Operation: Multiply both sides by 4
Solution: 10x+ 12 = 28
8 Operation: Divide both sides by 3
Solution: 6y+ 4 = 7/3
x+ = −2 x
3
5 2
2 3
2 2
x x
=
Concept Review 1
1 The laws of equality say that whatever you do to
one side of an equation you must do to the other
side, to maintain the balance
2 Yes Dividing by 0 and taking the square root of a
negative number are “undefined” operations in the
real numbers Be careful, then, when dividing both
sides of an equation by an unknown, to check that
the unknown could not possibly be 0
Commutative law: 14x − 12 = 3x
Add 12, subtract 3x: 11x= 12
Take square root:
x− = ±4 5
Answer Key 1: Solving Equations
SAT Practice 1
2 A Translate into an equation: x − 7 = 5x
You can also “test” the choices and see which one
works, but that’s probably more time-consuming
3 5 It’s easiest to solve the equation for y, then
add 5
Multiply by 5: 2y2= 5y2
Subtract 2y2: 0 = 3y2
Take the square root: 0 = y
Multiply by 6: 12x2− 30x = 54
5 D Plugging in and checking is perhaps easiest
here, but you could do the algebra too:
(p+ 2)2= (p − 5)2
FOIL: p2+ 4p + 4 = p2− 10p + 25
Subtract p2: 4p + 4 = −10p + 25
Subtract 4: 4p = −10p + 21
Divide by 14: p= 1.5
2 5
2 2
y y
=
6 A
Multiply by 5x: 25 + 7x = 5x
Divide by −2: −25/2 = x
7 C Guess and check here If x = −36 and y = −1,
or x = 1 and y = 36, then x − y = −35.
8 E Just plug in the points (−1, 7) and (1, 3) to the equations, and confirm that the points “satisfy” all three equations
9 C
Subtract 20:
Multiply by −1:
Square both sides: x= 81
Multiply by m − nx: 3x = 2(m − nx)
Distribute on right: 3x = 2m − 2nx
Factor left side: x(3 + 2n) = 2m
Divide by (3 + 2n): x m
n
= 2+
3 2
3x
m nx−
x= 9
− x= −9
20− x=11
5 7
5 1
Trang 5A system is simply a set of equations that are true at
the same time, such as these:
Although many values for x and y “satisfy” the first
equation, like (4, 0) and (2, −3) (plug them in and
check!), there is only one solution that works in both
equations: (6, 3) (Plug this into both equations and
check!)
The Law of Substitution
The law of substitution simply says that if
two things are equal, you can always substitute
one for the other
Example:
The easiest way to solve this is to substitute the
sec-ond equation (which is already “solved” for y) into the
first, so that you eliminate one of the unknowns Since
y = x + 1, you can replace the y in the first equation
with x+ 1 and get
3x + (x + 1)2= 7 FOIL the squared binomial: 3x + x2+ 2x + 1 = 7
Combine like terms: x2+ 5x + 1 = 7
Subtract 7: x2+ 5x − 6 = 0
Factor the left side: (x + 6)(x − 1) = 0
Apply the Zero Product Property: x = −6 or x = 1
Plug values back into 2nd equation:
y= (−6) + 1 = −5
or y= (1) + 1 = 2 Solutions: (−6, −5) and (1, 2) (Check!)
Combining Equations
If the two equations in the system are alike
enough, you can sometimes solve them more
easily by combining equations The idea is
sim-ple: if you add or subtract the corresponding
sides of two true equations together, the result
should also be a true equation, because you are
adding equal things to both sides This strategy
can be simpler than substitution
1
2
+ =
= +
⎧
⎨
⎩
3 2 12
3 15
− = + =
Example:
Add equations: 5x = 30 Divide by 5: x= 6
Plug this back in and solve for y:
2(6) − 5y = 7
Special Kinds of “Solving”
Sometimes a question gives you a system, but rather than asking you to solve for each unknown, it simply asks you to evaluate another expression Look carefully at what the question asks you to evaluate, and see whether there is a simple way of combining the equations (adding, subtracting, multiplying, dividing) to find the expression
Example:
If 3x − 6y = 10 and 4x + 2y = 2, what is the value of 7x − 4y?
Don’t solve for x and y! Just notice that 7x − 4y equals (3x − 6y) + (4x + 2y) = 10 + 2 = 12.
“Letter-Heavy” Systems
An equation with more than one unknown, or a system with more unknowns than equations, is
“letter-heavy.” Simple equations and systems usually have just one solution, but these “letter-heavy” equations and systems usually have more than one solution, and you can often easily find solutions simply by “plugging in” values
Example:
If 2m + 5n = 10 and m ≠ 0, then what is the value
of You can “guess and check” a solution to the equation
pretty easily Notice that m = −5, n = 4 works If you
plug these values into the expression you’re evaluating, you’ll see it simplifies to 2
4
10 5
m n
− ?
3 5 23
− = + =
⎧
⎨
⎩
Lesson 2: Systems
Adding the correspond-ing sides will eliminate
the y’s from the system
Trang 6Concept Review 2: Systems
1 What is a system?
2 What are two algebraic methods for solving systems?
3 How do you check the solution of a system?
Solve the following systems by substitution, and check your solution
Solve the following systems by combination, and check your solution
Give three different solutions to each of the following “letter-heavy” systems
a + b = 7
x= _ y= _ a= _ b= _ c= _
x= _ y= _ a= _ b= _ c= _
x= _ y= _ a= _ b= _ c= _
3 3 9
4 3 2
− = − + =
5 5 3
2 5
+ =
=
Trang 7SAT Practice 2: Systems
6. If and , then which of the
following expresses m in terms of y?
(A)
(B)
(C)
(D)
(E)
7. The sum of two numbers is 5 and their difference
is 2 Which of the following could be the differ-ence of their squares?
(A) −17 (B) −3 (C) 3 (D) 10 (E) 21
8. If 7x + 2y − 6z = 12, and if x, y, and z are
pos-itive, then what is the value of (A) 1/12
(B) 1/6 (C) 1/4 (D) 5/12 (E) 7/12
35
17 70
8 35
9 70
2 35
a
c=
4 3
1 7
b
c=
a b
3 5
=
2
7 2
+ +
z
18 6
3
− y
y
2
y
y
2
y3
18
18
3
y
m5 y2
6
=
m y
=
1. If 3x + 2y = 72 and y = 3x, then x =
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10
2. The difference of two numbers is 4 and their sum
is −7 What is their product?
(A) −33.0
(B) −28.0
(C) −10.25
(D) 8.25
(E) 10.5
3. If 4m − 7n = 10 and 2m + 2n = 4, what is the value
of 2m − 9n?
4. If 9p = 3a + 1 and 7p = 2a − 3, then which of the
following expresses p in terms of a?
5. The cost of one hamburger and two large sodas is
$5.40 The cost of three hamburgers and one large
soda is $8.70 What is the cost of one hamburger?
(A) $1.50 (B) $1.95 (C) $2.40
(D) $2.50 (E) $2.75
a+ 4 2
7
9
a
2 63
a
2 3 9
a−
3 1
7
a+
1
2
3
4
5
7
8
9
6
1
0
2
3
4
5
7
8
9
6
1
0
2
3
4
5
7
8
9
6
1 0 2 3 4 5 7 8 9 6
Trang 8Answer Key 2: Systems
+ a + b = 12
Add the equations: 2a= 17
Plug in to find b: (8.5) − b = 5
Subtract 8.5: −b = −3.5
Multiply by −1: b= 3.5
8 (10/3, −6) −3x − 5y = 20
−(−3x − 4y = 14)
Subtract the equations: −y = 6
Multiply by −1: y= −6
Plug in to find x: −3x − 5(−6) = 20
Simplify: −3x + 30 = 20
Divide by −3: x= 10/3
9 (−12, 15) Add the equations to get
Combine fractions:
Multiply by 12: 7x= −84 Divide by 7: x= −12
Plug in and solve for y: y= 15
10 There are many solutions Here are a few: (0, 8); (20, 0); (10, 4); (5, 6)
11 There are many solutions Here are a few: (1, 6, 0); (3, 4, −4); (2, 5, 31); (7, 0, 24)
7
12x = −7
3+ = −4 7
Concept Review 2
1 Any set of equations that are true at the same time
2 Substitution and combination
3 Plug the solutions back into the equations and
check that both equations are true
4 (−10, −8) Substitute: 3(y − 2) − 4y = 2
Distribute: 3y − 6 − 4y = 2
Combine: −y − 6 = 2
Multiply by −1: y= −8
Plug in and solve for x: x = y − 2 = (−8) −2 = −10
5 (4, 3) and (2, −3) Substitute: x2− 2(3x − 9) = 10
Distribute: x2− 6x + 18 = 10
Subtract 10: x2− 6x + 8 = 0
Factor: (x − 4)(x − 2) = 0
(Look over Lesson 5 if that step was tough!)
Zero Product Property: x = 4 or x = 2
Plug in and solve for y: y = 3x − 9 = 3(4) − 9 = 3
or = 3(2) − 9 = −3
So the solutions are x = 4 and y = 3 or
x = 2 and y = −3.
6 (2, 5) Substitute:
Simplify: 2n + 5 = 3n Subtract 2n: 5 = n Plug in to find m: m=2( )=
5 5 2
5 2
5n 5 3n
⎛
⎝⎜ ⎞⎠⎟ + =
SAT Practice 2
1 C Substitute: 3x + 2(3x) = 72
Divide by 9: x= 8
2 D Translate into equations: x − y = 4
x + y = −7
Add the equations: 2x = −3
Substitute: −1.5 + y = −7
(−1.5)(−5.5) = 8.25
3 6 Subtract them:
2m − 9n = (4m − 7n) − (2m + 2n)
= 10 − 4 = 6
4 E Subtracting gives 2p = a + 4
Divide by 2: p = (a + 4)/2
5 C Translate: h + 2s = 5.40
3h + s = 8.70
Multiply 2nd eq by 2: 6h + 2s = 17.40
−(h + 2s = 5.40)
Subtract 1st equation: 5h= 12.00
6 A Divide the first equation by the second:
Simplify:
7 D Translate: x + y = 5 and x − y = 2
Although you could solve this system by combin-ing, it’s easier to remember the “difference of squares” factoring formula:
x2− y2= (x + y)(x − y) = (5)(2) = 10
m
= ×3 62 =183
m
y
6 5
2
3 6
= ÷
Trang 98 B This is “letter-heavy,” so you can guess and
check a solution, like (2, 11, 4), and evaluate
can also just add 6z to both sides of the equation
to get 7x + 2y = 12 + 6z, then substitute:
2
7 2
2
12 6
2
6 2
1 6
+
+z =( ++ )= ( )++ =
z z
z z
2 4
7 2 2 11
6 36
1 6
+( ) ( )+ ( )= =
9 B Multiply the equations:
Multiply by 3
2
3 35
3 2
9 70 : a
⎝⎜
⎞
⎠⎟=
a b
b c
a c
2
4 3
2 3
3 5
1 7 3
⎛
⎝⎜
⎞
⎠⎟
⎛
⎝⎜
⎞
⎠⎟ = =
⎛
⎝⎜
⎞
⎠⎟
⎛
⎝⎜
⎞
⎠⎟=335
Trang 10Lesson 3: Working with Exponentials
What Are Exponentials?
An exponential is simply any term with these three
parts:
If a term seems not to have a coefficient or
ex-ponent, the coefficient or exponent is always
assumed to be 1!
Examples:
2x means 2x1 y3means 1y3
Expand to Understand
Good students memorize the rules for working
with exponentials, but great students understand
where the rules come from They come from
simply expanding the exponentials and then
collecting or cancelling the factors
Example:
What is (x5)2in simplest terms? Do you add
ex-ponents to get x7? Multiply to get x10? Power to
get x25?
The answer is clear when you expand the exponential
Just remember that raising to the nth power simply
means multiplying by itself n times So (x5)2= (x5)(x5)
= (x䡠x䡠x䡠x䡠x)(x䡠x䡠x䡠x䡠x) = x10 Doing this helps you to see
and understand the rule of “multiplying the powers.”
Adding and Subtracting Exponentials
When adding or subtracting exponentials, you
can combine only like terms, that is, terms with
the same base and the same exponent When
adding or subtracting like exponentials,
re-member to leave the bases and exponents alone
Example:
5x3+ 6x3+ 4x2= (5x3+ 6x3) + 4x2= x3(5 + 6) + 4x2=
11x3+ 4x2
Notice that combining like terms always involves the
Law of Distribution (Chapter 7, Lesson 2)
Multiplying and Dividing Exponentials
You can simplify a product or quotient of
exponen-tials when the bases are the same or the exponents are the same
If the bases are the same, add the exponents (when multiplying) or subtract the exponents (when dividing) and leave the bases alone
(5m5)(12m2) = (5)(m)(m)(m)(m)(m)(12) (m)(m)
= (5)(12)(m)(m)(m)(m)(m)(m)(m)
= 60m7
If the exponents are the same, multiply (or divide) the bases and leave the exponents alone
Example:
(3m4)(7n4) = (3)(m)(m)(m)(m)(7)(n)(n)(n)(n)
= (3)(7)(mn)(mn)(mn)(mn)
= 21(mn)4
Raising Exponentials to Powers
When raising an exponential to a power, multi-ply the exponents, but don’t forget to raise the coefficient to the power and leave the base alone
Example:
(3y4)3= (3y4)(3y4)(3y4)
= (3y䡠y䡠y䡠y)(3y䡠y䡠y䡠y)(3y䡠y䡠y䡠y)
= (3)(3)(3)(y䡠y䡠y䡠y)(y䡠y䡠y䡠y)(y䡠y䡠y䡠y)
= 27y12
5 12 3
5 12 12 12 12 12
3 3 3 3
5 5
( ) ( ) = ( )( )( )( )( ) ( )( )( )( ))( )
= ⎛
⎝⎜
⎞
⎠⎟
⎛
⎝⎜
⎞
⎠⎟
⎛
⎝⎜
⎞
⎠⎟
⎛
⎝⎜
⎞
⎠⎟
3
5 12 3
12 3
12 3
12 3 1 12 3
5 45
⎛
⎝⎜
⎞
⎠⎟
= ( )
6 3
6 3
7 4
p p
p p p p p p p
p p p p
= ( )( )( )( )( )( )( ) ( )( )( )( ) = 22p3
Coefficient
Base
Exponent
4 x3