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Penyelesaian - Absolute value equations

Hasil Akhir Langkah-langkah Terma dan topik

Exact form:

x = - \frac{13}{5}, - \frac{11}{3}

x=-\frac{13}{5} , -\frac{11}{3}

Mixed number form:

x = - 2 \frac{3}{5}, - 3 \frac{2}{3}

x=-2\frac{3}{5} , -3\frac{2}{3}

Decimal form:

x = - 2.6, - 3.667

x=-2.6 , -3.667

Lihat langkah Learn more with Tiger Try this:

Penjelasan langkah demi langkah

1. Rewrite the equation with one absolute value terms on each side

$| x + 1 | + 4 | x + 3 | = 0$

Add $- 4 | x + 3 |$ to both sides of the equation:

$| x + 1 | + 4 | x + 3 | - 4 | x + 3 | = - 4 | x + 3 |$

Permudahkan aritmetik

$| x + 1 | = - 4 | x + 3 |$

2. Rewrite the equation without absolute value bars

Use the rules:
$| x | = | y |$ → $x = \pm y$ and $| x | = | y |$ → $\pm x = y$
to write all four options of the equation
$| x + 1 | = - 4 | x + 3 |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| x + 1 \| = - 4 \| x + 3 \|$
$x = + y$	$(x + 1) = - 4 (x + 3)$
$x = - y$	$(x + 1) = - 4 (- (x + 3))$
$+ x = y$	$(x + 1) = - 4 (x + 3)$
$- x = y$	$- (x + 1) = - 4 (x + 3)$

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:

$\| x \| = \| y \|$	$\| x + 1 \| = - 4 \| x + 3 \|$
$x = + y$ , $+ x = y$	$(x + 1) = - 4 (x + 3)$
$x = - y$ , $- x = y$	$(x + 1) = - 4 (- (x + 3))$

3. Solve the two equations for x

11 additional steps

$(x + 1) = - 4 \cdot (x + 3)$

Expand the parentheses:

$(x + 1) = - 4 x - 4 \cdot 3$

Permudahkan aritmetik:

$(x + 1) = - 4 x - 12$

Add to both sides:

$(x + 1) + 4 x = (- 4 x - 12) + 4 x$

Kumpulkan sebutan sejenis:

$(x + 4 x) + 1 = (- 4 x - 12) + 4 x$

Permudahkan aritmetik:

$5 x + 1 = (- 4 x - 12) + 4 x$

Kumpulkan sebutan sejenis:

$5 x + 1 = (- 4 x + 4 x) - 12$

Permudahkan aritmetik:

$5 x + 1 = - 12$

Subtract from both sides:

$(5 x + 1) - 1 = - 12 - 1$

Permudahkan aritmetik:

$5 x = - 12 - 1$

Permudahkan aritmetik:

$5 x = - 13$

Divide both sides by :

$\frac{(5 x)}{5} = \frac{- 13}{5}$

Permudahkan pecahan:

$x = \frac{- 13}{5}$

16 additional steps

$(x + 1) = - 4 \cdot (- (x + 3))$

Expand the parentheses:

$(x + 1) = - 4 \cdot (- x - 3)$

$(x + 1) = - 4 \cdot - x - 4 \cdot - 3$

Kumpulkan sebutan sejenis:

$(x + 1) = (- 4 \cdot - 1) x - 4 \cdot - 3$

Darabkan pekali:

$(x + 1) = 4 x - 4 \cdot - 3$

Permudahkan aritmetik:

$(x + 1) = 4 x + 12$

Subtract from both sides:

$(x + 1) - 4 x = (4 x + 12) - 4 x$

Kumpulkan sebutan sejenis:

$(x - 4 x) + 1 = (4 x + 12) - 4 x$

Permudahkan aritmetik:

$- 3 x + 1 = (4 x + 12) - 4 x$

Kumpulkan sebutan sejenis:

$- 3 x + 1 = (4 x - 4 x) + 12$

Permudahkan aritmetik:

$- 3 x + 1 = 12$

Subtract from both sides:

$(- 3 x + 1) - 1 = 12 - 1$

Permudahkan aritmetik:

$- 3 x = 12 - 1$

Permudahkan aritmetik:

$- 3 x = 11$

Divide both sides by :

$\frac{(- 3 x)}{- 3} = \frac{11}{- 3}$

Hapuskan tanda negatif:

$\frac{3 x}{3} = \frac{11}{- 3}$

Permudahkan pecahan:

$x = \frac{11}{- 3}$

Pindahkan tanda negatif dari penyebut ke pembilang:

$x = \frac{- 11}{3}$

4. List the solutions

$x = - \frac{13}{5}, - \frac{11}{3}$
(2 solution(s))

5. Graph

Each line represents the function of one side of the equation:
$y = | x + 1 |$
$y = - 4 | x + 3 |$
The equation is true where the two lines cross.

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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.

Penyelesaian - Absolute value equations

Penjelasan langkah demi langkah

1. Rewrite the equation with one absolute value terms on each side

2. Rewrite the equation without absolute value bars

3. Solve the two equations for x

4. List the solutions

5. Graph

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Terma dan topik

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