Sat - MC Grawhill part 32 pot

300 MCGRAW-HILL’S SAT
Equations 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:5x
2
− 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 5x
2
− 9:
1. Square it: x
2
2. Multiply by 5: 5x
2
3. Subtract 9: 5x
2
− 9
12 8
4
28
4
327
x
x
−
=−=
So to “unwrap” it, you reverse and invert the steps:
1. Add 9: (5x
2

− 9) + 9 = 5x
2
2. Divide by 5: 5x
2
/5 = x
2
3. Find the square roots (both of them!): =±͉x͉
If you perform these steps to both sides, 5x
2
− 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 5x
2
− 9 = 31? Plug them in:
− 9 = 5(8) − 9 = 40 − 9 = 31 (Yes!)

58
2
±
()
± 8
():
5931
1
2
x −=
Step Add 9
Step 2
(Divide by 5)
:
++
=
99
540
2
x
5
Step 3 (Simplify):
x
2
5
40
5
=
x
2

8=
Step 4 (Square r
ooot): 8x =±
x =± 8.
± x
2
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.
CHAPTER 8 / ESSENTIAL ALGEBRA I SKILLS 301
Concept 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.
3. 9x − 12 + 5x = 3x 4. (x − 4)
2
= 5
5. 6.
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
2
37
x
+=
xx+
=
−2

3
5
2
2
3
2
2
x
x=
302 MCGRAW-HILL’S SAT
SAT 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?
(A) (B) (C)
(D) (E)
3. If , then y + 5 =
4. If 2x
2
− 5x = 9, then 12x
2
− 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?
(A) (B) −7 (C)
(D) (E) 7

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 − 2͉
II. f(x) = x
2
− 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,
if then x =
(A) (B)
(C) (D)
(E)
3
22mn−
2
32

m
n+
32
2
+ n
m
23
2
m
n
−
22
3
mn−
3
2
x
mnx−
= ,
20 11−=x
−
7
5
−
24
7
−
25
2
57

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
CHAPTER 8 / ESSENTIAL ALGEBRA I SKILLS 303
5.
Multiply by 3: 2x
2
= 3x
2
Subtract 2x
2
:0 = x
2
Take square root: 0 = x
6.
Cross-multiply: 2x + 4 = 3x − 15
Subtract 2x: 4 = x − 15
Add 15: 19 = x
7. Operation: Multiply both sides by 4

Solution: 10x + 12 = 28
8. Operation: Divide both sides by 3
Solution: 6y + 4 = 7/3
xx+
=
−2
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.
3. 9x − 12 + 5x = 3x
Commutative law: 14x − 12 = 3x
Add 12, subtract 3x: 11x = 12
Divide by 11: x = 12/11
4. (x − 4)
2

= 5
Take square root:
Add 4:
x =±45
x −=±45
Answer Key 1: Solving Equations
SAT Practice 1
1. C 5d + 12 = 24
Subtract 24: 5d − 12 = 0
2. A Translate into an equation: x − 7 = 5x
Subtract x: −7 = 4x
Divide by 4: −7/4 = x
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: 2y
2
= 5y
2
Subtract 2y
2
:0 = 3y
2
Divide by 3: 0 = y
2
Take the square root: 0 = y
Add 5: 5 = y + 5
4. E 2x
2

− 5x = 9
Multiply by 6: 12x
2
− 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: p
2
+ 4p + 4 = p
2
− 10p + 25
Subtract p
2
:4p + 4 =−10p + 25
Subtract 4: 4p =−10p + 21
Add 10p: 14p = 21
Divide by 14: p = 1.5
2
5
2
2
y
y=
6. A
Multiply by 5x: 25 + 7x = 5x
Subtract 7x: 25 =−2x

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
10. D = 2
Multiply by m − nx: 3x = 2(m − nx)
Distribute on right: 3x = 2m − 2nx
Add 2nx: 3x + 2nx = 2m
Factor left side: x(3 + 2n) = 2m
Divide by (3 + 2n):
x
m
n
=
+
2
32
3x
mnx−
x = 9
−=−x 9
20 11−=x
57
5

1
x
+=
304 MCGRAW-HILL’S SAT
Systems
A 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 + x
2
+ 2x + 1 = 7
Combine like terms: x
2

+ 5x + 1 = 7
Subtract 7: x
2
+ 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.
37
1
2
xy
yx
+=
=+
⎧
⎨
⎩
3212

315
xy
xy
−=
+=
Example:
Add equations: 5x = 30
Divide by 5: x = 6
Plug this back in and solve for y:
2(6) − 5y = 7
Simplify: 12 − 5y = 7
Subtract 12: − 5y =−5
Divide by 5: y = 1
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−
?
25 7
3523
xy
xy
−=
+=
⎧
⎨
⎩
Lesson 2: Systems
Adding the correspond-
ing sides will eliminate
the y’s from the system

CHAPTER 8 / ESSENTIAL ALGEBRA I SKILLS 305
Concept 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.
4. 3x − 4y = 25.x
2
− 2y = 10 6.
x = y − 2 y = 3x − 9
Solve the following systems by combination, and check your solution.
7. a − b = 58.−3x − 5y = 20 9.
a + b = 12 −3x − 4y = 14
Give three different solutions to each of the following “letter-heavy” systems.
10. 2x + 5y = 40 11. −2a + 5b + c = 10
a + b = 7
x = _______ y = _______ a = _______ b = _______ c = _______
x = _______ y = _______ a = _______ b = _______ c = _______
x = _______ y = _______ a = _______ b = _______ c = _______
xy
xy
33
9
43
2
−=−
+=
553
2
5

mn
mn
+=
=
306 MCGRAW-HILL’S SAT
SAT 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
9. If and , then
(A) (B) (C)

(D) (E)
9
35
17
70
8
35
9
70
2
35
a
c
=
4
3
1
7
b
c
=
a
b2
3
5
=
2
72
+
+

z
xy
?
18
6
3
− y
y
2
y
y
2
y
3
18
18
3
y
m
y
5
2
6
=
m
y
6
3
=
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?
(A) (B) (C)
(D) (E)
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
23
9
a−
31
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
CHAPTER 8 / ESSENTIAL ALGEBRA I SKILLS 307
Answer Key 2: Systems
7. (8.5, 3.5) a − b = 5
+ a + b = 12
Add the equations: 2a = 17
Divide by 2: a = 8.5
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
Subtract 30: −3x =−10
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
12
7
x
=−
xx
34
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

Add 6: −y = 8
Multiply by −1: y =−8
Plug in and solve for x: x = y − 2 = (−8) −2 =−10
5. (4, 3) and (2, −3) Substitute: x
2
− 2(3x − 9) = 10
Distribute: x
2
− 6x + 18 = 10
Subtract 10: x
2
− 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
52
5

2
5
53nn
⎛
⎝
⎜
⎞
⎠
⎟
+=
SAT Practice 2
1. C Substitute: 3x + 2(3x) = 72
Simplify: 9x = 72
Divide by 9: x = 8
2. D Translate into equations: x − y = 4
x + y =−7
Add the equations: 2x =−3
Divide by 2: x =−1.5
Substitute: −1.5 + y =−7
Add 1.5: y =−5.5
(−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
Divide by 5: h = 2.40
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:
x
2
− y
2
= (x + y)(x − y) = (5)(2) = 10
m
y
yy
=× =
3618
23
m
m
y
y
6
5
2
3
6

=÷
308 MCGRAW-HILL’S SAT
8. B This is “letter-heavy,” so you can guess and
check a solution, like (2, 11, 4), and evaluate
the expression: . Or you
can also just add 6z to both sides of the equation
to get 7x + 2y = 12 + 6z, then substitute:
2
72
2
12 6
2
62
1
6
+
+
=
+
+
()
=
+
+
()
=
z
xy
z
z

z
z
24
72 211
6
36
1
6
+
()
()
+
()
==
9. B Multiply the equations:
Multiply by
3
2
3
35
3
2
9
70
:
a
c
=
⎛
⎝

⎜
⎞
⎠
⎟
=
a
b
b
c
a
c2
4
3
2
3
3
5
1
7
3
⎛
⎝
⎜
⎞
⎠
⎟
⎛
⎝
⎜
⎞

⎠
⎟
==
⎛
⎝
⎜
⎞
⎠
⎟
⎛
⎝
⎜
⎞
⎠
⎟
=
335
CHAPTER 8 / ESSENTIAL ALGEBRA I SKILLS 309
Lesson 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 2x
1
y
3

means 1y
3
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 (x
5
)
2
in simplest terms? Do you add ex-
ponents to get x
7
? Multiply to get x
10
? Power to
get x
25
?
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 (x
5
)
2
= (x
5

)(x
5
)
= (x⅐x⅐x⅐x⅐x)(x⅐x⅐x⅐x⅐x) = x
10
. 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:
5x
3
+ 6x
3
+ 4x
2
= (5x
3
+ 6x
3
) + 4x
2
= x
3
(5 + 6) + 4x
2

=
11x
3
+ 4x
2
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.
(5m
5
)(12m
2
) =(5)(m)(m)(m)(m)(m)(12) (m)(m)
= (5)(12)(m)(m)(m)(m)(m)(m)(m)
= 60m
7
If the exponents are the same, multiply (or divide)
the bases and leave the exponents alone.
Example:
(3m
4
)(7n
4
) = (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:
(3y
4
)
3
= (3y
4
)(3y
4
)(3y
4
)
= (3y⅐y⅐y⅐y)(3y⅐y⅐y⅐y)(3y⅐y⅐y⅐y)
= (3)(3)(3)(y⅐y⅐y⅐y)(y⅐y⅐y⅐y)(y⅐y⅐y⅐y)
= 27y
12
512
3
51212121212
3333
5
5
()

()
=
()()()()()
()()()(
))( )
=
⎛
⎝
⎜
⎞
⎠
⎟
⎛
⎝
⎜
⎞
⎠
⎟
⎛
⎝
⎜
⎞
⎠
⎟
⎛
⎝
⎜
⎞
⎠
⎟

3
5
12
3
12
3
12
3
12
3
112
3
54
5
⎛
⎝
⎜
⎞
⎠
⎟
= ()
6
3
6
3
7
4
p
p
ppppppp

pppp
=
()()()()()()()
()()()()
= 22
3
p
Coefficient
Base
Exponent
4
x
3