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Penyelesaian - Absolute value equations

Exact form: a=2,-13
a=2 , -\frac{1}{3}
Decimal form: a=2,0.333
a=2 , -0.333

Other Ways to Solve

Absolute value equations

Penjelasan langkah demi langkah

1. Rewrite the equation without absolute value bars

Use the rules:
|x|=|y|x=±y and |x|=|y|±x=y
to write all four options of the equation
|5a3|=|a+5|
without the absolute value bars:

|x|=|y||5a3|=|a+5|
x=+y(5a3)=(a+5)
x=y(5a3)=(a+5)
+x=y(5a3)=(a+5)
x=y(5a3)=(a+5)

When simplified, equations x=+y and +x=y are the same and equations x=y and x=y are the same, so we end up with only 2 equations:

|x|=|y||5a3|=|a+5|
x=+y , +x=y(5a3)=(a+5)
x=y , x=y(5a3)=(a+5)

2. Solve the two equations for a

11 additional steps

(5a-3)=(a+5)

Subtract from both sides:

(5a-3)-a=(a+5)-a

Kumpulkan sebutan sejenis:

(5a-a)-3=(a+5)-a

Permudahkan aritmetik:

4a-3=(a+5)-a

Kumpulkan sebutan sejenis:

4a-3=(a-a)+5

Permudahkan aritmetik:

4a3=5

Add to both sides:

(4a-3)+3=5+3

Permudahkan aritmetik:

4a=5+3

Permudahkan aritmetik:

4a=8

Divide both sides by :

(4a)4=84

Permudahkan pecahan:

a=84

Cari faktor sepunya terbesar bagi pembilang dan penyebut:

a=(2·4)(1·4)

Faktorkan keluar dan hapuskan faktor sepunya terbesar:

a=2

12 additional steps

(5a-3)=-(a+5)

Expand the parentheses:

(5a-3)=-a-5

Add to both sides:

(5a-3)+a=(-a-5)+a

Kumpulkan sebutan sejenis:

(5a+a)-3=(-a-5)+a

Permudahkan aritmetik:

6a-3=(-a-5)+a

Kumpulkan sebutan sejenis:

6a-3=(-a+a)-5

Permudahkan aritmetik:

6a3=5

Add to both sides:

(6a-3)+3=-5+3

Permudahkan aritmetik:

6a=5+3

Permudahkan aritmetik:

6a=2

Divide both sides by :

(6a)6=-26

Permudahkan pecahan:

a=-26

Cari faktor sepunya terbesar bagi pembilang dan penyebut:

a=(-1·2)(3·2)

Faktorkan keluar dan hapuskan faktor sepunya terbesar:

a=-13

3. List the solutions

a=2,-13
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|5a3|
y=|a+5|
The equation is true where the two lines cross.

Mengapa belajar ini

Learn more with Tiger

We encounter absolute values almost daily. For example: If you walk 3 miles to school, do you also walk minus 3 miles when you go back home? The answer is no because distances use absolute value. The absolute value of the distance between home and school is 3 miles, there or back.
In short, absolute values help us deal with concepts like distance, ranges of possible values, and deviation from a set value.