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orcompound inequality because the absolute value is greater than the given value.
.LHS /(- 1).=.RHS /(- 1).
Distribute (x-3)
Distribute x
Identity Property of Multiplication
Add terms
LHS+4=RHS+4
Is the first term a perfect square? | x^2= x^2 âś“ |
Is the last term a perfect square? | 1= 1^2 âś“ |
Is the middle term twice the product of 1 and x? | 2x=2* 1* x âś“ |
x-3 > 4 or x-3< - 4 Let's isolate x in both of these cases before writing the solution set.
The solution to this type of compound inequality is the combination of the solution sets. First Solution Set:& x>7 Second Solution Set:& x<- 1 Combined Solution Set:& x<- 1 or x>7 This inequality tells us that all values greater than 7 or less than - 1 will satisfy the inequality. Now, we can check our solution by substituting four values of x into the original inequality.
Value of x | Substitute | Simplify |
---|---|---|
- 2 | 10| - 2-3|? > 40 | 50 > 40 |
0 | 10| 0-3|? > 40 | 30 ≯ 40 |
6 | 10| 6-3| ? > 40 | 30 ≯ 40 |
8 | 10| 8-3| ? > 40 | 50 > 40 |
As we can see, values of x greater than 7 or less than - 1 are indeed the solution of the given inequality.
Write as a power
sqrt(LHS)=sqrt(RHS)
Calculate root
LHS+5=RHS+5