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Let’s check out your problem:
Solve for
x
x
x
:
\newline
−
1
+
10
(
x
−
1
)
=
−
(
−
3
x
−
7
)
−
x
-1+10(x-1)=-(-3 x-7)-x
−
1
+
10
(
x
−
1
)
=
−
(
−
3
x
−
7
)
−
x
\newline
Answer:
x
=
x=
x
=
View step-by-step help
Home
Math Problems
Algebra 1
Solve linear equations: mixed review
Full solution
Q.
Solve for
x
x
x
:
\newline
−
1
+
10
(
x
−
1
)
=
−
(
−
3
x
−
7
)
−
x
-1+10(x-1)=-(-3 x-7)-x
−
1
+
10
(
x
−
1
)
=
−
(
−
3
x
−
7
)
−
x
\newline
Answer:
x
=
x=
x
=
Distribute and Simplify:
First, distribute the
10
10
10
into the parentheses on the left side of the equation.
\newline
−
1
+
10
(
x
−
1
)
=
−
(
−
3
x
−
7
)
−
x
-1 + 10(x - 1) = -(-3x - 7) - x
−
1
+
10
(
x
−
1
)
=
−
(
−
3
x
−
7
)
−
x
\newline
−
1
+
10
x
−
10
=
3
x
+
7
−
x
-1 + 10x - 10 = 3x + 7 - x
−
1
+
10
x
−
10
=
3
x
+
7
−
x
Combine Like Terms:
Now, combine like terms on both sides of the equation.
\newline
(
10
x
−
10
+
1
)
=
(
3
x
−
x
+
7
)
(10x - 10 + 1) = (3x - x + 7)
(
10
x
−
10
+
1
)
=
(
3
x
−
x
+
7
)
\newline
10
x
−
9
=
2
x
+
7
10x - 9 = 2x + 7
10
x
−
9
=
2
x
+
7
Move Terms:
Next, move the variable terms to one side and the constant terms to the other side.
10
x
−
2
x
=
7
+
9
10x - 2x = 7 + 9
10
x
−
2
x
=
7
+
9
8
x
=
16
8x = 16
8
x
=
16
Divide and Solve:
Finally, divide both sides by
8
8
8
to solve for
x
x
x
.
\newline
8
x
8
=
16
8
\frac{8x}{8} = \frac{16}{8}
8
8
x
=
8
16
\newline
x
=
2
x = 2
x
=
2
More problems from Solve linear equations: mixed review
Question
Solve for x.
\newline
(
3
4
)
x
=
12
(\frac{3}{4})x= 12
(
4
3
)
x
=
12
\newline
x
=
x =
x
=
______
Get tutor help
Posted 1 year ago
Question
Solve for x.
\newline
−
5
9
x
=
15
-\frac{5}{9}x= 15
−
9
5
x
=
15
\newline
x
=
x =
x
=
______
Get tutor help
Posted 1 year ago
Question
How many solutions does the following equation have?
\newline
5
x
+
8
−
7
x
=
−
4
x
+
1
5x+8-7x=-4x+1
5
x
+
8
−
7
x
=
−
4
x
+
1
\newline
Choose
1
1
1
answer:
\newline
(A) No solutions
\newline
(B) Exactly one solution
\newline
(C) Infinitely many solutions
Get tutor help
Posted 1 year ago
Question
How many solutions does the following equation have?
\newline
−
2
z
+
10
+
7
z
=
16
z
+
7
-2z+10+7z=16z+7
−
2
z
+
10
+
7
z
=
16
z
+
7
\newline
Choose
1
1
1
answer:
\newline
(A) No solutions
\newline
(B) Exactly one solution
\newline
(C) Infinitely many solutions
Get tutor help
Posted 10 months ago
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
Choose
1
1
1
answer:
\newline
(A) No solutions
\newline
(B) Exactly one solution
\newline
(C) Infinitely many solutions
Get tutor help
Posted 1 year ago
Question
How many solutions does the following equation have?
\newline
−
9
(
x
+
6
)
=
−
9
x
+
108
-9(x+6)=-9x+108
−
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
Get tutor help
Posted 1 year ago
Question
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
Get tutor help
Posted 1 year ago
Question
How many solutions does the following equation have?
\newline
−
4
x
−
7
+
10
x
=
−
7
+
6
x
-4x-7+10x=-7+6x
−
4
x
−
7
+
10
x
=
−
7
+
6
x
\newline
Choose
1
1
1
answer:
\newline
(A) No solutions
\newline
(B) Exactly one solution
\newline
(C) Infinitely many solutions
Get tutor help
Posted 1 year ago
Question
How many solutions does the following equation have?
\newline
−
17
(
y
−
2
)
=
−
17
y
+
64
-17(y-2)=-17y+64
−
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
Get tutor help
Posted 1 year ago
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
Get tutor help
Posted 1 year ago
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