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 10 or 20
2 Grade
Adding and Subtracting up to 100
Addition and Subtraction within 10 or 20
3 Grade
Addition and Subtraction up to 1000
Multiplication up to 5
Multiplication Table
Rounding
Dividing
Perimeter
4 Grade
Adding and Subtracting
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
6 Grade
Percents
Signed Numbers
The Coordinate Plane
Equations
Polynomials
Symplifying Expressions
Polynomial Expressions
Factoring
7 Grade
Angles
Parametric Linear Equations
Word Problems
Exponentiation
Roots
Quadratic Equations
Vieta's Formulas
Progressions
Arithmetic Progressions
Geometric Progression
Progressions
Number Sequences
Reciprocal Equations
Logarithms
Logarithmic Expressions
Logarithmic Equations
Extremal value problems
Trigonometry
Geometry
Intercept Theorem
Law of Sines
Law of Cosines
Probability
Limits of Functions
Properties of Triangles
Pythagorean Theorem
Inverse Trigonometric Functions
Analytic Geometry
Easy
Normal
Difficult
Parametric Linear Equations - Problems with Solutions
Problem 1
Find the value of
a
, for which the equation [tex]ax=1[/tex] has no solutions.
Solution:
If we can divide by
a
, there is always the solution [tex]x=\frac{1}{a}[/tex]. Let's check the case where we cannot divide by
a
, in other words
a=0
. We get the equation [tex]0x=1[/tex], which has no solutions. Therefore the answer to the problem is
a=0
.
Problem 2
Find the value of the real parameter
a
, for which the equation [tex](a-2)x=(a-2)^2[/tex] has any
x
for solution.
Solution:
For an equation to have any
x
as solution, it must be of the form
0x=0
. Which means that [tex](a-2)=(a-2)^2=0[/tex], or
a=2
.
Problem 3
Find the value of the parameter
b
, for which the equation
0x=b-7
has at least one solution
x
.
Solution:
For any
x
, the value of the left side is zero. We get
b-7=0
, or
b=7
. By substituting, we get
0x=0
, which has infinitely many solutions.
Problem 4
Solve the equation for
a=3
:
[tex]x+a=2a+1[/tex]
Solution:
We subtract
a
from both sides of the equation to get
[tex]x=a+1[/tex]. We substitute
a=3
:
[tex]x=3+1=4[/tex]
Problem 5
Find the value of
a
, for which the equation
[tex]\frac{1}{a+5}x=a+7[/tex] is not defined.
Solution:
If the equation is to be not defined, then a denominator must be zero. The only denominator is [tex]a+5[/tex], so we get [tex]a=-5[/tex]
Problem 6
Solve the equation [tex](a^3+1)x=109+a[/tex] for [tex]a=3[/tex]
Solution:
We directly substitute:
[tex](3^3+1)x=109+3[/tex]
[tex]28x=112[/tex]
[tex]x=4[/tex]
Problem 7
Solve the equation [tex](a^4-250)x=a^3+2[/tex] for
a=4
.
Solution:
We substitute:
[tex](4^4-250)x=4^3+2[/tex]
[tex](256-250)x=64+2[/tex]
[tex]6x=66[/tex]
[tex]x=11[/tex]
Problem 8
Solve the parametric linear equation [tex](a^3+9)x=237+10a[/tex] for
a=-2
.
Solution:
We substitute:
[tex]((-2)^3+9)x=237+10\cdot(-2)[/tex]
[tex](-8+9)x=237-20[/tex]
[tex]x=217[/tex]
Problem 9
Solve the equation [tex]5ax=85[/tex] for
a=17
.
Solution:
We directly substitute:
[tex]5\cdot17x=85[/tex]
[tex]85x=85[/tex]
[tex]x=1[/tex]
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
Google+
Facebook
Author
Copyright © 2005 - 2018.