Answer:
Please check the explanation.
Step-by-step explanation:
Given the inequality
-2x < 10
-6 < -2x
<u>Part a) Is x = 0 a solution to both inequalities</u>
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.
<u>Part b) Is x = 4 a solution to both inequalities</u>
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.
<u>solving -2x < 10</u>
Multiply both sides by -1 (reverses the inequality)
Simplify
Divide both sides by 2
<u>solving -6 < -2x</u>
-6 < -2x
switch sides
Multiply both sides by -1 (reverses the inequality)
Simplify
Divide both sides by 2
Thus, the two intervals:
The intersection of these two intervals would be the solution to both inequalities.
and
As x = 1 is included in both intervals.
so x = 1 would be another solution common to both inequalities.
<h3>SUBSTITUTING x = 1</h3>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.
