MENU
❌
Home
Math Forum/Help
Problem Solver
Practice
Algebra
Geometry
Tests
College Math
History
Games
MAIN MENU
1 Grade
Adding and subtracting up to 10
Comparing numbers up to 10
Adding and subtracting up to 20
Addition and Subtraction within 20
2 Grade
Adding and Subtracting up to 100
Addition and Subtraction within 20
3 Grade
Addition and Subtraction within 1000
Multiplication up to 5
Multiplication Table
Rounding
Dividing
Perimeter
Addition, Multiplication, Division
4 Grade
Adding and Subtracting
Addition, Multiplication, Division
Equivalent Fractions
Divisibility by 2, 3, 4, 5, 9
Area of Squares and Rectangles
Fractions
Equivalent Fractions
Least Common Multiple
Adding and Subtracting
Fraction Multiplication and Division
Operations
Mixed Numbers
Decimals
Expressions
6 Grade
Percents
Signed Numbers
The Coordinate Plane
Equations
Expressions
Polynomials
Symplifying Expressions
Polynomial Vocabulary
Polynomial Expressions
Factoring
7 Grade
Angles
Inequalities
Linear Functions
8 Grade
Linear Functions
Systems of equations
Slope
Parametric Linear Equations
Word Problems
Exponents
Roots
Quadratic Equations
Quadratic Inequalities
Vieta's Formulas
Progressions
Arithmetic Progressions
Geometric Progression
Progressions
Number Sequences
Reciprocal Equations
Logarithms
Logarithmic Expressions
Logarithmic Equations
Extremal value problems
Trigonometry
Numbers Classification
Geometry
Intercept Theorem
Slope
Law of Sines
Law of Cosines
Vectors
Probability
Limits of Functions
Properties of Triangles
Pythagorean Theorem
Matrices
Complex Numbers
Inverse Trigonometric Functions
Analytic Geometry
Analytic Geometry
Circle
Parabola
Ellipse
Conic sections
Polar coordinates
Integrals
Integrals
Integration by Parts
Home
Practice
Linear(Simple) Equations
Easy
Normal
Difficult
Linear(Simple) Equations: Problems with Solutions
Problem 1
Find the solution
n
to the equation
n + 2 = 6
Solution:
By substracting 2 from both sides, we get
n + 2 - 2 = 6 - 2
,
or
n = 4
Problem 2
Solve the equation
z - 5 = 6
.
Solution:
By adding
5
to both sides, we get
z=5+6
, or
z=11
.
Problem 3
Solve the equation
5 - t = 0
.
Solution:
We add
t
to both sides of the equation, and we get
5-t+t=0+t
, or
t=5
.
Problem 4
Solve the equation:
n + 7=13
Solution:
By substracting 7 from both sides, we get
n+7-7=13-7
, or
n=6
.
Problem 5
Solve the equation [tex]x+34=82[/tex]
Solution:
[tex]x=82-34=48[/tex]
Problem 6
Solve the linear equation
3-a=2a
.
Solution:
By adding
a
to both sides, we get
3-a+a=2a+a
, or
3a=3
. The we divide both sides by 3 to reach the final answer,
a=1
.
Problem 7
Solve the linear equation [tex]2x-20=10[/tex]
Solution:
[tex]2x=10+20[/tex]
[tex]2x=30[/tex]
[tex]x=15[/tex]
Problem 8
Solve the equation [tex]20 - 2x = 18[/tex]
Solution:
[tex]20 - 2x = 18[/tex]
[tex]20 - 18 = 2x[/tex]
[tex]2=2x[/tex]
[tex]x=1[/tex]
Problem 9
Find
c
, if [tex]5c - 2 = 33[/tex].
Solution:
We add 2 to both sides and get
5c-2+2=33+2
, or
5c=35
. We divide both sides by 5 in order to get
c=7
.
Problem 10
Solve the equation [tex]4x=72[/tex]
Solution:
We divide both sides of the equation with 4:
[tex]\frac{4x}{4}=\frac{72}{4}[/tex]
[tex]x=18[/tex]
Problem 11
Solve the equation [tex]13x-15=24[/tex]
Solution:
[tex]13x=24+15[/tex]
[tex]13x=39[/tex]
[tex]x=3[/tex]
Problem 12
Find the solution b to the equation [tex]\frac{b}{3}=3[/tex].
Solution:
We multiply both sides of the equation by
3
in order to get [tex]\frac{b}{3}\cdot3=3\cdot3[/tex], or
b=9
.
Problem 13
Solve the linear equation [tex]2x+2=40[/tex]
Solution:
[tex]2x=40-2[/tex]
[tex]2x=38[/tex]
[tex]x=19[/tex]
Problem 14
Solve the equation
3x + 1 = 16
.
Solution:
By substracting 1 from both sides, we get
3x+1-1=16-1
, or
3x=15
. We divide both sides by 3 in order to get the value for x:
x=5
.
Problem 15
Solve the equation
2x + 5 = 9
Solution:
By substracting 5 from both sides of the equation, we get
2x+5-5=9-5
, or
2x=4
. Then we divide both sides by 2 to reach the solution, [tex]x=2[/tex].
Problem 16
Solve the equation
m+10=3m
.
Solution:
By substracting
m
from both sides, we get
m+10-m=3m-m
, so
10=2m
or
2m=10
. Dividing both sides of the equation by 2, we find that
m=5
Problem 17
Solve the linear equation [tex]19z=38+6\times 19[/tex]
Solution:
We divide both sides by 19:
[tex]\frac{19z}{19}=\frac{38}{19}+6\cdot\frac{19}{19}[/tex]
[tex]z=2+6[/tex]
[tex]z=8[/tex]
Problem 18
Find the solution
y
to the linear equation
2y+6=y+11
.
Solution:
First, we substract
6
from both sides to get
2y+6-6=y+11-6
, or
2y=y+5
.
Then we substract
y
from both sides to reach the final solution,
y=5
.
Problem 19
Solve the equation [tex]\frac{1}{x}=\frac{1}{5}[/tex]
Solution:
The equation is defined for [tex]x \ne 0[/tex]. We cross multiply and get
[tex]x=5[/tex], which is a valid value for
x
and the solution.
Problem 20
Find the solution
x
to the equation [tex]x-1+x-2+x-3=0[/tex]
Solution:
[tex]x+x+x-1-2-3=0[/tex]
[tex]3x-6=0[/tex]
[tex]3x=6[/tex]
[tex]x=2[/tex]
Problem 21
Find the solution
y
to the equation [tex]y+10=13y-74[/tex]
Solution:
[tex]10+74=13y-y[/tex]
[tex]12y=84[/tex]
[tex]y=7[/tex]
Problem 22
Solve the equation
3c=8c
.
Solution:
We substract
3c
from both sides of the equation to get
3c-3c=8c-3c
, or
5c=0
. The we divide both sides by 5 and the answer is
c=0
.
Problem 23
Solve the linear equation [tex]\frac{3}{8}a=3[/tex]
Solution:
First we multiply both sides by 8 to get free of the denominator. That yields [tex]\frac{3}{8}a\cdot8=3\cdot8[/tex], or
3a=24
. Then we divide both sides by 3 to get
a=8
.
Problem 24
Solve the equation
4x - 9 = 2x + 7
.
Solution:
4x-9=2x+7
2x=16
x=16/2
x=8
Problem 25
If [tex]5x+12=3x-24[/tex], determine the value of
x
.
Solution:
[tex]5x-3x=-24-12[/tex]
[tex]2x=-36[/tex]
[tex]x=-18[/tex]
Problem 26
Find the solution to the equation [tex]\frac{x}{5}=8[/tex].
Solution:
We multiply both sides by 5: [tex]5\cdot\frac{x}{5}=8\cdot5[/tex] in order to get
x=40
.
Problem 27
If
28 = 10y - 3y
, find
y
.
Solution:
We substract on the right side:
28=7y
. Then we divide by 7 in order to get the solution,
y=4
.
Problem 28
-2х - 6 = 2
Solution:
-2х = 2 + 6
-2х = 8
х= -8 : 2
х= -4
Problem 29 sent by Bhoomi Bembde
If 19 + 19x = 3x - 9 find the value of x.
Solution:
19 + 19x = 3x - 9
19x - 3x= -9 - 19
+16x = -28
x= -28 ÷ 16
x= -1.75 or -7/4
Problem 30 sent by Oshin Patel
If x + 7 = 16 then what is the value of 8x - 72 = ?
Solution:
x + 7 = 16
x = 16 - 7
x = 9
8x - 72 = 8 × 9 - 72 = 72 - 72=0
Problem 31
3x + 3 = 7x - 9
Solution:
Subtract 3x from both sides of the equation. 3x + 3 - 3x = 7x - 9 - 3x
3 = 4x - 9
Add 9
3 + 9 = 4x - 9 + 9
12 = 4x
Divide by 4
3 = x
Easy
Normal
Difficult
Submit a problem on this page.
Problem text:
Solution:
Answer:
Your name(if you would like to be published):
E-mail(you will be notified when the problem is published)
Notes
: use [tex][/tex] (as in the forum if you would like to use latex).
Correct:
Wrong:
Unsolved problems:
Contact email:
Follow us on
Twitter
Facebook
Author
Copyright © 2005 - 2020.