Solve the following equations with 2 absolute values.(B) ∣2x−34∣=∣x+12∣\left|\frac{2x-3}{4}\right|=\left|x+\frac12\right|

x = { − 5 2 , 1 6 } x=\left\lbrace-\frac52,\frac16\right\rbrace

Solve the following equations with 2 absolute values. ( B ) ∣ 2 x − 3 4 ∣ = ∣ x + 1 2 ∣ \left|\frac{2x-3}{4}\right|=\left|x+\frac12\right|

Welcome back. For this problem, we'd like to solve the equation absolute value of two x minus three over four equals absolute value of x plus one half. We do have fractions here, but they're inside of our absolute values. So we need to start by dealing with the absolute values, which means breaking up our one equation with two absolute values into two equations with no absolute values. The first one is just gonna set the two expressions from inside the absolute values equal to each other. So we get two x minus three over four is equal to x plus one half. The second equation is going to take the expression from the absolute value on the left, two x minus three over four, and set it equal to the negative version of the expression inside the absolute value on the right, x plus one half. Now we have our two equations, so we should focus on solving both of them separately, starting with our first equation, two x minus three over four equals x plus one half. We can see that we do have fractions here with denominators four and two. So that means that the LCD is going to be four, and we can clear our fractions by multiplying both sides of the equation by four. So we get four times two x minus three over four equals four times x plus one half. Now what's going on on the left is we were doing four times two x minus three divided by four. We're multiplying and dividing by four, so they cancel, and we're just left with two x minus three. On the right side of the equation, we can distribute that four, getting four x plus four times one half, which is two. Now we wanna try to get variables on one side and constants on the other. Let's try to get variables to the right and constants to the left. To get rid of our plus two on the right, we can subtract two from both sides of the equation, leaving us with two x minus three minus two is minus five. And on the right, plus two minus two cancel, leaving us with four x. Now we wanna get rid of the two x on the left, so we can subtract two x from both sides of the equation, which cancels two x on the left leaving us with minus five, and we get four x minus two x, which is two x. Finally, we divide both sides of the equation by two. So on the right, we end up getting just x, and that's equal to negative five halves. So we got one solution from our first equation. Let's solve our other equation to see if we have other solutions. Our other equation is two x minus three over four is equal to negative x plus one half. Before we try to get rid of our fractions, we should simplify the right side of the equation. So on the left, we still have two x minus three over four, but now on the right, we have negative one times x, which is negative x, and negative one times one half, which is minus one half. Now that we've distributed that negative sign, let's focus on getting rid of the fractions. We still have denominators four and two, so our LCD is still going to be four, and we can multiply both sides of the equation by four. So we get four times two x minus three over four is equal to four times negative x minus one half. Now we get the same left side of our equation as we had in our first equation, so we know that this simplifies down to two x minus three. On the right side, we have four times negative x, which is negative four x, and four times negative one half, which is negative two. Now let's focus on trying to get our variables to the left and constants to the right. We can add three to both sides of the equation to cancel our minus three on the left, leaving us with two x equals negative four x minus two plus three, which is plus one on the right. Next, we wanna get rid of our minus four x on the right so we can add four x on both sides of the equation. On the left side of the equation, we get two x plus four x equals six x, and minus four x plus four x cancel, leaving us with one. And finally, when we divide both sides of the equation by six, we're left with just x on the left equals one sixth on the right, so we get our final solution. We can put those two solutions together in a solution set that includes our first solution of negative five halves and our second solution of one sixth. Now that we've solved our equation, we're all set. Great work.

Solve the following equations with 2 absolute values. ( B ) ∣ 2 x − 3 4 ∣ = ∣ x + 1 2 ∣ \left|\frac{2x-3}{4}\right|=\left|x+\frac12\right|

x = { − 5 2 , 1 6 } x=\left\lbrace-\frac52,\frac16\right\rbrace

x = { − 5 2 , 5 6 } x=\left\lbrace-\frac52,\frac56\right\rbrace

x = { 5 2 , − 1 6 } x=\left\lbrace\frac52,-\frac16\right\rbrace

x = { 5 2 , − 5 6 } x=\left\lbrace\frac52,-\frac56\right\rbrace

9. Inequalities and Absolute Value

Absolute Value Equations

Beginning & Intermediate Algebra

9. Inequalities and Absolute Value

Absolute Value Equations

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Multiple Choice

Solve the following equations with 2 absolute values.
(B) $∣ \frac{2 x - 3}{4} ∣ = ∣ x + \frac{1}{2} ∣$

$x = {- \frac{5}{2}, \frac{5}{6}}$

$x = {\frac{5}{2}, - \frac{1}{6}}$

$x = {- \frac{5}{2}, \frac{1}{6}}$

$x = {\frac{5}{2}, - \frac{5}{6}}$

Verified step by step guidance

Start with the given equation involving absolute values: $\left|\frac{2x - 3}{4}\right| = \left|x + \frac{1}{2}\right|$.

Recall that if $|A| = |B|$, then either $A = B$ or $A = -B$. So, set up two separate equations: 1) $\frac{2x - 3}{4} = x + \frac{1}{2}$ and 2) $\frac{2x - 3}{4} = -\left(x + \frac{1}{2}\right)$.

Solve the first equation $\frac{2x - 3}{4} = x + \frac{1}{2}$ by multiplying both sides by 4 to clear the denominator, then isolate $x$ and simplify.

Solve the second equation $\frac{2x - 3}{4} = -\left(x + \frac{1}{2}\right)$ similarly by multiplying both sides by 4, distributing the negative sign on the right, then isolate $x$ and simplify.

Check each solution by substituting back into the original absolute value equation to ensure both sides are equal, confirming valid solutions.

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