Solve the following equations.(C) 0.4∣3y+2∣=3.60.4\left|3y+2\right|=3.6

y = { 7 3 , − 11 3 } y=\left\lbrace\frac73,-\frac{11}{3}\right\rbrace y = { 3 7 , − 3 11 }

Solve the following equations. ( C ) 0.4 ∣ 3 y + 2 ∣ = 3.6 0.4\left|3y+2\right|=3.6

Welcome back. For this problem, we'd like to solve the equation 0.4 times the absolute value of three y plus two equals 3.6. We know that the first step of solving an absolute value equation is to isolate the absolute value, so we have to get rid of this 0.4 that's being multiplied by that absolute value. But it's difficult to get rid of because it's a decimal, so we can borrow a trick we saw with linear equations and clear that decimal first. We can see that on both sides of the equation, we have a decimal with exactly one decimal place, zero point four and three point six. So we can clear both of those decimals by multiplying both sides of the equation by 10. On the left side, we get 10 times 0.4 times the absolute value of three y plus two, and on the right side, we get 10 times 3.6. When we multiply by 10, we move the decimal point one place to the right. So when we do 10 times 0.4 times the absolute value of three y plus two, ten times 0.4 moves the decimal point one place to the right and gets us four times the absolute value of three y plus two. On the right side of the equation, multiplying by 10 again moves the decimal point one place to the right, and we get 36. Now that we no longer have any decimals, it's much easier to clear this four so that we can isolate our absolute value on the left side of the equation. We're going to divide both sides of the equation by four, again canceling this four on the left and leaving us with absolute value of three y plus two. And on the right, we get 36 divided by four, which is nine. Now we have our equation in the form absolute value equals positive number, so we can rewrite this one equation as two equations without an absolute value. The first one takes the thing in the absolute value, three y plus two, and sets it equal to a, which is nine. And the second takes that three y plus two and sets it equal to negative a, which is negative nine. Now, we just solved both equations separately, starting with three y plus two equals nine. We can subtract two from both sides of the equation to get rid of the plus two on the left, leaving us with three y equals seven, and then we can divide both sides of the equation by three to cancel the coefficient of three on the left, leaving us with y equals seven thirds. That's our first solution. Now, let's solve our second equation, where we have three y plus two is equal to negative nine. Again, we can subtract two from both sides of the equation to cancel our plus two on the left, leaving us with three y is equal to negative 11. And then we can divide both sides of the equation by three to cancel our coefficient of three on the left, leaving us with y equals negative 11 thirds. That's our second solution. You can plug in either of these values for y to the original equation, and you'll see that you get a true statement, meaning that our solution set includes both solutions, seven thirds, and negative 11 thirds. Great work.

Solve the following equations. ( C ) 0.4 ∣ 3 y + 2 ∣ = 3.6 0.4\left|3y+2\right|=3.6

y = { 7 3 , − 11 3 } y=\left\lbrace\frac73,-\frac{11}{3}\right\rbrace y = { 3 7 , − 3 11 }

y = { 14 , − 38 3 } y=\left\lbrace14,-\frac{38}{3}\right\rbrace y = { 14 , − 3 38 }

y = { − 7 3 , 11 3 } y=\left\lbrace-\frac73,\frac{11}{3}\right\rbrace y = { − 3 7 , 3 11 }

y = { − 14 , 38 3 } y=\left\lbrace-14,\frac{38}{3}\right\rbrace y = { − 14 , 3 38 }

2. Linear Equations and Inequalities

Absolute Value Equations

Intermediate Algebra

2. Linear Equations and Inequalities

Absolute Value Equations

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Multiple Choice

Solve the following equations.
(C) $0.4 ∣ 3 y + 2 ∣ = 3.6$

$y=\left\lbrace14,-\frac{38}{3}\right\rbrace$

$y=\left\lbrace\frac73,-\frac{11}{3}\right\rbrace$

$y=\left\lbrace-\frac73,\frac{11}{3}\right\rbrace$

$y=\left\lbrace-14,\frac{38}{3}\right\rbrace$

Verified step by step guidance

Start with the given equation: \$0.4\left|3y + 2\right| = 3.6$.

Isolate the absolute value expression by dividing both sides of the equation by 0.4: $\left|3y + 2\right| = \frac{3.6}{0.4}$.

Simplify the right side to find the value that the absolute value equals: $\left|3y + 2\right| = \text{some positive number}$.

Recall that if $\left|A\right| = B$, where $B > 0$, then $A = B$ or $A = -B$. Apply this to get two separate equations: \$3y + 2 = B$ and $3y + 2 = -B$.

Solve each linear equation for $y$ by isolating $y$: subtract 2 from both sides, then divide by 3.

4:48m

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