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Big Ideas Math Algebra 1, 2015

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Big Ideas Math Algebra 1, 2015 View details

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Exercise 15 Page 97

Solve for x and consider all possible values of a, b, c, and d, including 0, positive values, and negative values.

When a=c and d>b: All real numbers.

When a=c and d No solution.

When a>c: x

When a x>d-b/a-c.

Practice makes perfect

Before we can think about the possible solution sets for x, let's simplify our inequality a little bit so that x is as isolated as possible.

ax+b

LHS-b

ax

LHS-cx

ax-cx

Factor out x

x(a-c)

We cannot go any further just yet. What if a-c=0? What if a-c is a negative number? There are actually four cases we need to consider for this inequality. First case: &a = c and d > b Second case: &a = c and d < b Third case: &a > c Fourth case: &a < c Let's look at these cases one at a time.

First case

In the first case, we should consider when a=c and d>b. When a=c we have a-c=0.

x(a-c)

(a-c)= 0

x* 0

Zero Property of Multiplication

0

With the assumption that d>b we must have that d-b is always a positive number. Therefore, we have 0<"a positive number" which is always true. The inequality would then hold true for all real numbers.

Second case

In the second case, we should consider when a=c and d

x(a-c)

(a-c)= 0

x* 0

Zero Property of Multiplication

0

With the assumption that dThird case In the third case, we should consider when a>c. When a>c, we must have that a-c will always be a positive number in which case we can simply divide and keep the inequality symbol as it is.
x(a-c)
DivIneq
.LHS /(a-c).<.RHS /(a-c).
x
Our solution set is then all values such that x< d-ba-c.
Fourth case
In the fourth case, we should consider when a
x(a-c)
DivNegIneq
Divide by (a-c) and flip inequality sign
x>d-b/a-c
Our solution set is then all values such that x> d-ba-c.

Subchapter links

Exercises p.97