Solve the following equations with 2 absolute values.(A) ∣3x+4∣=∣−2x+7∣\left|3x+4\right|=\left|-2x+7\right|

x={35,−11}x=\left\lbrace\frac35,-11\right\rbrace

Solve the following equations with 2 absolute values. ( A ) ∣ 3 x + 4 ∣ = ∣ − 2 x + 7 ∣ \left|3x+4\right|=\left|-2x+7\right|

Welcome to the practice. In this problem, we'd like to solve the equation absolute value of three x plus four equals the absolute value of negative two x plus seven. Because we have that the absolute value of one expression is equal to the absolute value of another expression, we can use the rule that we saw in the concept video to take this one equation with two absolute values and split it into two equations without any absolute value. Our first equation will just be the two expressions from the absolute values equal to each other. So that first equation is going to be three x plus four from the absolute value on the left is equal to negative two x plus seven from the absolute value on the right. For our second equation, we're going to take the expression from the absolute value on the left, which is three x plus four, and set it equal to the negative version of the expression from the absolute value on the right, which is negative two x plus seven. Now that we have our two equations, we just solve both of them separately to get our solutions. Let's start with that first equation, three x plus four equals negative two x plus seven. We want constants on one side and variables on the other, so let's start by trying to get rid of this plus four on the left, which we can do by subtracting four from both sides of the equation. That will cancel the plus four on the left leaving us with three x. And on the right, we have negative two x plus seven minus four, which is plus three. Next, we wanna get rid of this minus two x on the right, which we can do by adding two x to both sides of the equation. On the left, we get three x plus two x, which is five x. And on the right, minus two x and plus two x cancel, leaving us with three. And finally, we can divide both sides of the equation by five to get rid of our coefficient of five on the left, leaving us with x equals three fifths. So that's our first solution. Let's solve our second equation to see if we have any more solutions. Now our left side is all set, so let's copy it over three x plus four. But on the right side, we do need to distribute that negative sign, which again is just like multiplying everything in the parenthesis by negative one. So first, we have negative one times negative two x, which is positive two x. Then we have negative one times seven, which is negative seven. Now that all of our distributing is all set, it's all about getting variables on one side and constants on the other. Let's get rid of this plus four on the left in the same way we did with our first equation by just subtracting four from both sides of the equation. It will cancel our plus four on the left leaving us with three x, and on the right, we'll have two x minus seven minus four, which is minus 11. Next, we wanna get rid of this two x on the right, which we can do by subtracting two x from both sides of the equation. On the left, we get three x minus two x, which is just one x or x. And then two x and minus two x cancel on the right, leaving us with minus 11. So our other solution is x equals negative 11, and we can put the two of them together in a solution set, including that first solution of three fifths and our second solution of negative 11. So now we have the solutions to our equation. Great work.

Solve the following equations with 2 absolute values.(A) ∣3x+4∣=∣−2x+7∣\left|3x+4\right|=\left|-2x+7\right|

x={35,−11}x=\left\lbrace\frac35,-11\right\rbrace

x={3,−11}x=\left\lbrace3,-11\right\rbrace

x={35,3}x=\left\lbrace\frac35,3\right\rbrace

x={−35,3}x=\left\lbrace-\frac35,3\right\rbrace

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Multiple Choice

Solve the following equations with 2 absolute values.
(A) $∣ 3 x + 4 ∣ = ∣ - 2 x + 7 ∣$

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

$x = {3, - 11}$

$x = {\frac{3}{5}, 3}$

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

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Verified step by step guidance

Set up two separate equations based on the property of absolute values: 1) $3x + 4 = -2x + 7$ 2) $3x + 4 = -(-2x + 7)$, which simplifies to $3x + 4 = 2x - 7$.

Solve the first equation $3x + 4 = -2x + 7$ by isolating $x$: Add \$2x$ to both sides: $3x + 2x + 4 = 7$ Simplify: $5x + 4 = 7$ Subtract 4 from both sides: $5x = 3$ Divide both sides by 5: $x = \frac{3}{5}$.

Solve the second equation $3x + 4 = 2x - 7$ by isolating $x$: Subtract \$2x$ from both sides: \(3x - 2x + 4 = -7$ Simplify: \)x + 4 = -7$ Subtract 4 from both sides: $x = -11$.

Check both solutions in the original equation to ensure they satisfy the absolute value equality. Both $x = \frac{3}{5}$ and $x = -11$ should work, so the solution set is $\left\{ \frac{3}{5}, -11 \right\}$.

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Struggling with Beginning & Intermediate Algebra?

Solve the following equations with 2 absolute values.(A) ∣3x+4∣=∣−2x+7∣\(\left\)|3x+4\(\right\)|=\(\left\)|-2x+7\(\right\)|

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Solve the following equations with 2 absolute values.
(A) $∣ 3 x + 4 ∣ = ∣ - 2 x + 7 ∣$