Exercise 22 Page 227

Before we can solve this equation, we need to isolate the absolute value expression using the Properties of Equality. An absolute value measures an expression's distance from a midpoint on a number line. |5-t|= 5This 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.

Checking Our Answers

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

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

Multiply

20=20

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