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Let’s check out your problem:
There are
3
3
3
consecutive integers with a sum of
42
42
42
. What are the integers?
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Math Problems
Algebra 1
Consecutive integer problems
Full solution
Q.
There are
3
3
3
consecutive integers with a sum of
42
42
42
. What are the integers?
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\_\_\_\_\_
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,
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Define First Integer:
Let
x
x
x
be the first integer.
\newline
Consecutive integers increase by
1
1
1
each time.
\newline
Consecutive integers:
x
x
x
,
x
+
1
x+1
x
+
1
,
x
+
2
x+2
x
+
2
Set Up Equation:
Set up the equation that represents the sum of
3
3
3
consecutive integers equal to
42
42
42
.
\newline
Equation:
x
+
(
x
+
1
)
+
(
x
+
2
)
=
42
x + (x+1) + (x+2) = 42
x
+
(
x
+
1
)
+
(
x
+
2
)
=
42
Simplify Equation:
Simplify the left side of the equation
x
+
(
x
+
1
)
+
(
x
+
2
)
=
42
x + (x+1) + (x+2) = 42
x
+
(
x
+
1
)
+
(
x
+
2
)
=
42
.
x
+
(
x
+
1
)
+
(
x
+
2
)
=
42
x + (x+1) + (x+2) = 42
x
+
(
x
+
1
)
+
(
x
+
2
)
=
42
(
x
+
x
+
x
)
+
(
1
+
2
)
=
42
(x + x + x) + (1+2) = 42
(
x
+
x
+
x
)
+
(
1
+
2
)
=
42
3
x
+
3
=
42
3x + 3 = 42
3
x
+
3
=
42
Isolate Variable Term:
Isolate the variable term in the equation
3
x
+
3
=
42
3x + 3 = 42
3
x
+
3
=
42
.
\newline
3
x
+
3
=
42
3x + 3 = 42
3
x
+
3
=
42
\newline
3
x
+
3
−
3
=
42
−
3
3x + 3 - 3 = 42 - 3
3
x
+
3
−
3
=
42
−
3
\newline
3
x
=
39
3x = 39
3
x
=
39
Solve for x:
Solve for x.
\newline
3
x
=
39
3x = 39
3
x
=
39
\newline
3
x
3
=
39
3
\frac{3x}{3} = \frac{39}{3}
3
3
x
=
3
39
\newline
x
=
13
x = 13
x
=
13
Determine Consecutive Integers:
Determine the
3
3
3
consecutive integers that add up to
42
42
42
.
\newline
Three consecutive integers:
\newline
x
x
x
,
x
+
1
x+1
x
+
1
,
x
+
2
x+2
x
+
2
\newline
13
13
13
,
13
+
1
13+1
13
+
1
,
13
+
2
13+2
13
+
2
\newline
13
13
13
,
14
14
14
,
42
42
42
0
0
0
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4
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x
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\newline
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x
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\newline
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z
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7
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−
2
z
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10
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7
z
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16
z
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7
\newline
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1
1
1
answer:
\newline
(A) No solutions
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(B) Exactly one solution
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Question
How many solutions does the following equation have?
\newline
7
(
y
−
8
)
=
7
y
+
42
7(y-8)=7y+42
7
(
y
−
8
)
=
7
y
+
42
\newline
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1
1
1
answer:
\newline
(A) No solutions
\newline
(B) Exactly one solution
\newline
(C) Infinitely many solutions
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How many solutions does the following equation have?
\newline
−
9
(
x
+
6
)
=
−
9
x
+
108
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−
9
(
x
+
6
)
=
−
9
x
+
108
\newline
Choose
1
1
1
answer:
\newline
(A) No solutions
\newline
(B) Exactly one solution
\newline
(C) Infinitely many solutions
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How many solutions does the following equation have?
\newline
−
6
(
x
+
7
)
=
−
4
x
−
2
-6(x+7)=-4x-2
−
6
(
x
+
7
)
=
−
4
x
−
2
\newline
Choose
1
1
1
answer:
\newline
(A) No solutions
\newline
(B) Exactly one solution
\newline
(C) Infinitely many solutions
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How many solutions does the following equation have?
\newline
−
4
x
−
7
+
10
x
=
−
7
+
6
x
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−
4
x
−
7
+
10
x
=
−
7
+
6
x
\newline
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1
1
1
answer:
\newline
(A) No solutions
\newline
(B) Exactly one solution
\newline
(C) Infinitely many solutions
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\newline
−
17
(
y
−
2
)
=
−
17
y
+
64
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−
17
(
y
−
2
)
=
−
17
y
+
64
\newline
Choose
1
1
1
answer:
\newline
(A) No solutions
\newline
(B) Exactly one solution
\newline
(C) Infinitely many solutions
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Question
How many solutions does the following equation have?
\newline
9
z
−
6
+
7
z
=
16
z
−
6
9z-6+7z=16z-6
9
z
−
6
+
7
z
=
16
z
−
6
\newline
Choose
1
1
1
answer:
\newline
(A) No solutions
\newline
(B) Exactly one solution
\newline
(C) Infinitely many solutions
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