Answer:
Please check the explanation.
Step-by-step explanation:
Given the inequality
-2x < 10
-6 < -2x
Part a) Is x = 0 a solution to both inequalities
FOR -2x < 10
substituting x = 0 in -2x < 10
-2x < 10
-3(0) < 10
0 < 10
TRUE!
Thus, x = 0 satisfies the inequality -2x < 10.
∴ x = 0 is the solution to the inequality -2x < 10.
FOR -6 < -2x
substituting x = 0 in -6 < -2x
-6 < -2x
-6 < -2(0)
-6 < 0
TRUE!
Thus, x = 0 satisfies the inequality -6 < -2x
∴ x = 0 is the solution to the inequality -6 < -2x
Conclusion:
x = 0 is a solution to both inequalites.
Part b) Is x = 4 a solution to both inequalities
FOR -2x < 10
substituting x = 4 in -2x < 10
-2x < 10
-3(4) < 10
-12 < 10
TRUE!
Thus, x = 4 satisfies the inequality -2x < 10.
∴ x = 4 is the solution to the inequality -2x < 10.
FOR -6 < -2x
substituting x = 4 in -6 < -2x
-6 < -2x
-6 < -2(4)
-6 < -8
FALSE!
Thus, x = 4 does not satisfiy the inequality -6 < -2x
∴ x = 4 is the NOT a solution to the inequality -6 < -2x.
Conclusion:
x = 4 is NOT a solution to both inequalites.
Part c) Find another value of x that is a solution to both inequalities.
solving -2x < 10
[tex]-2x\:<\:10[/tex]
Multiply both sides by -1 (reverses the inequality)
[tex]\left(-2x\right)\left(-1\right)>10\left(-1\right)[/tex]
Simplify
[tex]2x>-10[/tex]
Divide both sides by 2
[tex]\frac{2x}{2}>\frac{-10}{2}[/tex]
[tex]x>-5[/tex]
[tex]-2x<10\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x>-5\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-5,\:\infty \:\right)\end{bmatrix}[/tex]
solving -6 < -2x
-6 < -2x
switch sides
[tex]-2x>-6[/tex]
Multiply both sides by -1 (reverses the inequality)
[tex]\left(-2x\right)\left(-1\right)<\left(-6\right)\left(-1\right)[/tex]
Simplify
[tex]2x<6[/tex]
Divide both sides by 2
[tex]\frac{2x}{2}<\frac{6}{2}[/tex]
[tex]x<3[/tex]
[tex]-6<-2x\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x<3\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:3\right)\end{bmatrix}[/tex]
Thus, the two intervals:
[tex]\left(-\infty \:,\:3\right)[/tex]
[tex]\left(-5,\:\infty \:\right)[/tex]
The intersection of these two intervals would be the solution to both inequalities.
[tex]\left(-\infty \:,\:3\right)[/tex] and [tex]\left(-5,\:\infty \:\right)[/tex]
As x = 1 is included in both intervals.
so x = 1 would be another solution common to both inequalities.
SUBSTITUTING x = 1
FOR -2x < 10
substituting x = 1 in -2x < 10
-2x < 10
-3(1) < 10
-3 < 10
TRUE!
Thus, x = 1 satisfies the inequality -2x < 10.
∴ x = 1 is the solution to the inequality -2x < 10.
FOR -6 < -2x
substituting x = 1 in -6 < -2x
-6 < -2x
-6 < -2(1)
-6 < -2
TRUE!
Thus, x = 1 satisfies the inequality -6 < -2x
∴ x = 1 is the solution to the inequality -6 < -2x.
Conclusion:
x = 1 is a solution common to both inequalites.
