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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]
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