Absolute Value Equations - College Algebra

Card 0 of 9

Question

Solve the following

$\left | 3x-5\right |=20$

Answer

We have to set up two equations, which are.

Now lets solve for x in each equation.

$3x=25\rightarrow x=\frac{25}{3}$

$3x=-15\rightarrow x=-5$

So the solutions are and $x=\frac{25}{3}$

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Question

Solve:

Answer

We need to set up two equations since we are dealing with absolute value

$\x-10=20\x-10=-20$

We solve for , in each equation to get the solutions.

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Question

Solve:

Answer

We need to set up two equations since we are dealing with absolute value

$\x-45=2\x-45=-2$

We solve for , in each equation to get the solutions.

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Question

Solve the following:

$\left | x-2 \right |>1$

Answer

To solve, you must split the absolute value into the two following equations.

and

Now, solve for x and right it in interval form.

$x-2>1\Rightarrow x>3$

$x-2<-1\Rightarrow x<1$

$(-\infty,1)\cup (3,\infty)$

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Question

Solve the following equation for .

$\left |x^2-8 \right |=1$

Answer

We first need to get rid of the absolute value symbol to solve the equation. TO break this absolute value, we assign two values to the right hand side, as shown below.

We now proceed to solve each equation independently.

Starting with the first equation, we get

$x=\pm3$

Now for the second equation,

$x=\pm\sqrt{7}$

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Question

Simplify the radical:

$\frac{2\sqrt{20}}{\sqrt{4}}$

Answer

To simplify the radical, break the radical in the numerator down into its factors. When doing so, the radical in the bottom will call with one from the numerator.

$\frac{2\sqrt{20}}{\sqrt{4}}=\frac{2\sqrt{4}\sqrt{5}}{\sqrt{4}}=2{\sqrt{5}}$

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Question

Solve the following equation:

$-3-\left | 8x-6\right |=3$

Answer

To solve for first isolate the absolute value portion on one side of the equation and all other constants on the other side.

$-3-\left | 8x-6\right |=3$

$-\left | 8x-6\right |=6$

$\left | 8x-6\right |=-6$

Recall that the absolute value can come from either a negative or positive value therefore two possible equations are set up.

$x=\frac{12}{8}=\frac{3}{2}: :or: :x=0$

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Question

Solve the equation:

$-5\left | 4+2x\right |=-100$

Answer

To solve for first isolate the absolute value portion on one side of the equation and all other constants on the other side.

$-5\left | 4+2x\right |=-100$

$\left | 4+2x\right |=20$

Since absolute values can come from either negative or positive values, two equations need to be set up and solved for.

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Question

Solve for x:

$\left | x-7 \right | = 3$

Answer

To solve $\left | x-7\right | = 3$ we need to find a way to reverse the operation of taking the absolute value. What we need to do is think about what the absolute value operation does to an expression. Since it makes everything positive, $\left | x-7\right | = \left | (-1) (x-7)\right |$ . So actually solving the original equation comes down to solving the following two equations:

So we get the two solutions as:

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