Exercise 22 Page 227

How many cases do you have after you remove the absolute value?

t=0 or t=10

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Before we can solve this equation, we need to isolate the absolute value expression using the Properties of Equality.

4|5-t|=20

DivEqn

.LHS /4.=.RHS /4.

|5-t|=5

An absolute value measures an expression's distance from a midpoint on a number line. |5-t|= 5

This equation means that the distance is 5, either in the positive direction or the negative direction. |5-t|= 5 ⇒ l5-t= 5 5-t= - 5 To find the solutions to the absolute value equation, we need to solve both of these cases for t.

| 5-t|=5

lc 5-t ≥ 0:5-t = 5 & (I) 5-t < 0:5-t = - 5 & (II)

lc5-t=5 & (I) 5-t=- 5 & (II)

(I), (II): LHS-5=RHS-5

lc- t=0 & (I) - t=- 10 & (II)

(I), (II):LHS * (- 1)=RHS* (- 1)

lt_1=0 t_2=10

Both t_1=0 and t_2=10 are solutions to the equation.

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Let's check the solutions by substituting them into the original equation. We will start with t_1=0.

4|5-t|=20

Substitute

t= 0

4|5- 0|? =20

▼

Evaluate left-hand side

SubTerm

Subtract term

4|5|? =20

AbsPos

|5|=5

4(5)? =20

Multiply

20=20

Since we got an identity the solution is correct. Now, let's substitute t_2=10.

4|5-t|=5

Substitute

t= 10

4|5- 10|? =20

▼

Evaluate left-hand side

SubTerm

Subtract term

4|- 5|? =20

AbsNeg

|-5|=5

4(5)? =20

Multiply

20=20

Again, we got an identity. The solution t=10 is also correct.

Exercise 22 Page 227

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