
Jordan K. answered 08/28/15
Tutor
4.9
(79)
Nationally Certified Math Teacher (grades 6 through 12)
Hi Erica,
We need to understand at the outset that absolute value equations will always yield two answers, because the quantity within the absolute value bars could actually be either positive or negative.
Let's begin by assuming the quantity between the absolute value bars to be positive. In this case, we just drop the absolute value bars and solve normally:
a + 3 = 2/5
(5)(a) + 5(3) = (5)(2/5)
5a + 15 = 2
5a = 2 - 15
5a = -13
a = -13/5 (1st answer)
Now let's assume the quantity between the absolute value bars to be negative. In this case, we first need to distribute a minus sign over each term within the absolute value bars, i.e. multiply each term within the absolute value bars by negative 1, and then we can solve normally:
(-1)(a) + (-1)(3) = 2/5
-a - 3 = 2/5
(5)(-a) + (5)(-3) = (5)(2/5)
-5a - 15 = 2
-5a = 2 + 15
-5a = 17
a = -17/5 (2nd answer)
We can check both our answers by plugging each one into our absolute value equation and see if both sides match:
¦a + 3¦ = 2/5
¦-13/5 + 3¦ = 2/5
¦-13/5 + (5/5)(3)¦ = 2/5
¦-13/5 + 15/5¦ = 2/5
¦2/5¦ = 2/5
2/5 = 2/5 (1st answer checks)
¦a + 3¦ = 2/5
¦-17/5 + 3¦ = 2/5
¦-17/5 + (5/5)(3)¦ = 2/5
¦-17/5 + 15/5¦ = 2/5
¦-2/5¦ = 2/5
¦-17/5 + 3¦ = 2/5
¦-17/5 + (5/5)(3)¦ = 2/5
¦-17/5 + 15/5¦ = 2/5
¦-2/5¦ = 2/5
2/5 = 2/5 (2nd answer checks)
Thanks for submitting this problem and glad to help.
God bless, Jordan.