How do you solve #abs(a-5)/8=5#?

Question

Hazel Anglin · Accepted Answer

#|a - 5| = 8 xx 5#

We must take into account two distinct circumstances.

One of two possibilities exists: either the absolute value is positive or negative.

Initially, the affirmative case:

#a - 5 = 40#

#a = 45#

Let's move on to the adverse scenario:

#-(a - 5) = 40#

#-a + 5 = 40#

#-a = 35#

#a = -35#

Hence, the solution set is #{-35, 45}#.

I hope this is useful!

Eva Adamson · Answer

#a= 45" "# or #" "a = -35#

Absolute values can have positive or negative values inside of them, but they are always positive when they emerge from the absolute value sign.

First, multiply both sides by #8#, the opposite of dividing by #8#.

#(a- 5)/ 8 xx 8 = 5 xx 8#

Since #8/8= 1#, you have

# a -5 = 40#

#a-5# has two possible values. Solve for both values.

#(a-5) = + 40" "# or #" " a-5 = -40#

#a-5 = + 40 -># add #5# to both sides

#a- cancel(5) + cancel(5) = + 40 + 5#

Since #-5+ 5 = 0# and #40+5 = +45#, you have

#a= + 45#

#a-5 = -40 -># add #5# to both sides

#a - cancel(5) + cancel(5) = -40 +5#

Once again, #-5 + 5 = 0# and #-40 + 5 = -35#

#a = -35#

Isaiah Anthony · Answer

undefined To solve the equation $ \frac{{\text{abs}(a-5)}}{8} = 5 $:

1. Multiply both sides of the equation by 8 to eliminate the fraction.
2. Solve for $ \text{abs}(a-5) $ by isolating it on one side of the equation.
3. Once $ \text{abs}(a-5) $ is isolated, determine the two possible scenarios: $ a-5 = 8 \times 5 $ and $ -(a-5) = 8 \times 5 $.
4. Solve each scenario separately for $ a $.

The equation becomes:

$$ \text{abs}(a-5) = 8 \times 5 $$

$$ \text{abs}(a-5) = 40 $$

Now, consider both scenarios:

1. $ a - 5 = 40 $:
   - Add 5 to both sides: $ a = 40 + 5 $
   - $ a = 45 $

2. $ -(a - 5) = 40 $:
   - Distribute the negative sign: $ -a + 5 = 40 $
   - Subtract 5 from both sides: $ -a = 40 - 5 $
   - $ -a = 35 $
   - Multiply both sides by -1 to solve for $ a $: $ a = -35 $

Therefore, the solutions for $ a $ are $ a = 45 $ and $ a = -35 $.