Chapter Four — Linear Equations in One Variable
The Humongous Book of Algebra Problems
56
Adding and Subtracting to Solve an Equation
Add to/subtract from both sides
Note: Problems 4.1–4.3 refer to this statement: “Five more than a number is equal to thirteen.”
4.1 Express the statement as an algebraic equation in terms of x.
According to Problem 3.2, the statement “more than” in this context indicates
a sum, so “five more than a number” is expressed as x + 5. The phrase “is equal
to” indicates the presence of an equal sign, so the algebraic equivalent of “five
more than a number is equal to thirteen” is x + 5 = 13.
Note: Problems 4.1–4.3 refer to this statement: “Five more than a number is equal to thirteen”
4.2 Solve the equation generated in Problem 4.1 for x.
To solve an equation for a variable, you must isolate that variable on one side of
the equal sign, usually the left side. The equation x + 5 = 13 has two things left
of the equal sign, x and the number 5. If you subtract 5 from the left side of the
equation to eliminate it, you must subtract 5 from the right side of the equation
as well to maintain the equality of the statement.
When five is subtracted from both sides of the equation, isolating x left of the
equal sign, you are left with x = 8, the solution to the equation.
Note: Problems 4.1–4.3 refer to this statement: “Five more than a number is equal to thirteen.”
4.3 Verify that the solution to Problem 4.2 is correct.
To verify that x = 8 is the solution to the equation x + 5 = 13, substitute 8 into the
equation for x.
Because substituting x = 8 into the equation produces a true statement (13 = 13),
x = 8 is the correct solution.
Or 5 + x.
That works,
too.
“Isolating” a
variable means
that it, alone,
is on one side of
the equal sign and
everything else is on
the other side. After
you isolate x, you end
up with “x =” or “= x,”
an equation with
only x on either
the left or
right side.
If you
add some-thing
to (or subtract
something from) one
side of an equation,
do the same thing to
the other side. The two
sides only stay equal
if you treat them
exactly the same
way.
There’s only one correct solution. When
equations have one unique variable and the
highest power of that variable is 1, you only get
one solution.
Chapter Four — Linear Equations in One Variable
The Humongous Book of Algebra Problems
57
4.4 Express the following statement as an equation and solve it: –6 equals the
difference of a number and 19.
The word “difference” indicates subtraction, so the difference of a number
and 19 is expressed as x – 19. According to the statement, –6 is equal to that
difference, so the corresponding equation is –6 = x – 19.
To solve the equation, you must isolate x. In this case, it is simpler to isolate x
right of the equal sign by adding 19 to both sides of the equation.
The solution to the equation is 13 = x, or x = 13. According to the symmetric
property described in Problem 1.37, those solutions are equivalent.
Note: Problems 4.5–4.6 refer to the equation 2x + 1 = x – 3.
4.5 Solve the equation for x.
To isolate x, all the variable terms (the terms containing x) should be moved
to the left side of the equation and all the constant terms (the terms with no
variables) should be moved to the right side of the equation. In other words,
move x on the right side of the equation by subtracting x from both sides, and
move 1 on the left side of the equation by subtracting 1 from both sides.
The solution to the equation is x = –4.
Note: Problems 4.5–4.6 refer to the equation 2x + 1 = x – 3.
4.6 Verify the solution generated in Problem 4.5.
Substitute x = –4 into the equation 2x + 1 = x – 3. Ensure that you replace the x’s
on both sides of the equation with –4.
Because substituting x = –4 into the equation produces a true statement
(–7 = –7), x = –4 is the correct solution.
You want to
get x by itself,
but x has a –19
next to it. Get rid of
that negative 19 with
a positive 19. Add 19
to both sides.
When you
subtract things,
write them under
their like terms. In
other words, when you
subtract x from the
right side, write –x
underneath 2x, and
when you subtract
1 from the left
side, write –1
underneath –3.
2x and –x are
like terms, but 2x – x
does NOT equal 2! You
don’t cancel out the
variables. Remember,
–x is the same as –1x,
so 2x – 1x = 1x. Two of
something (2x) minus
one of those things (1x
or x) leaves behind one
thing (1x or x).
Chapter Four — Linear Equations in One Variable
The Humongous Book of Algebra Problems
58
4.7 Solve the equation 5 + 6(x – 1) = 11 + 5x for x and verify the solution.
Before you attempt to isolate x, simplify the left side of the equation using the
distributive property.
Further simplify the left side of the equation by combining like terms 5 and –6:
6x + 5 – 6 = 11 + 5x
6x – 1 = 11 + 5x
Isolate x on the left side of the equation by moving the variable terms left of the
equal sign and moving the constant terms right of the equal sign.
The solution to the equation is x = 12. Verify the solution by substituting it into
the original equation.
Because substituting x = 12 into the equation produces a true statement
(71 = 71), x = 12 is the correct solution.
4.8 Solve the equation and verify the solution: 4(x + 7) – 18 = 3(x + 10) + 6.
Apply the distributive property to simplify both sides of the equation.
To remove 3x from the right side of the equation, subtract it from both sides.
Check out
Problem 3.23
if you need
to review the
distributive
property.
“Moving”
terms is a by–
product of
getting rid of
terms you don’t
want. You don’t want
–1 on the left side, so
you zap it by adding 1.
However, you have to do
the same thing to both
sides of the equation,
so that banished 1
moves over to the right
side of the equation,
where it’s combined
with the constant
term 11 that’s
already
there.
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