Solve the inequality. Express the solution set in interval notation and graph. 13(x+1)≥15(3+2x)\frac13\left(x+1\right)\ge\frac15\left(3+2x\right)31(x+1)≥51(3+2x)

( − ∞ , − 4 ] \left(-\infty,-4]\right. ( − ∞ , − 4 ]

Solve the inequality. Express the solution set in interval notation and graph. 1 3 ( x + 1 ) ≥ 1 5 ( 3 + 2 x ) \frac13\left(x+1\right)\ge\frac15\left(3+2x\right) 3 1 ( x + 1 ) ≥ 5 1 ( 3 + 2 x )

Hey everyone, we want to solve this linear inequality and then express our answer in interval notation and graph it. And here I have 1 third times x plus 1 is greater than or equal to 1 5th times 3 plus 2x. So since I have fractions here the first thing I need to do is to clear those by multiplying by the greatest common denominator. And here, since my denominators are 35, my greatest common denominator is going to be 15. So I'm going to multiply my entire inequality by 15. Remember, we need to distribute to every single term in our inequality. So I have 15 over 3 times x plus 1 is greater than or equal to 15 over 5 having that greatest common denominator times 3 plus 2 x. Okay. So let's simplify this a little bit. Well, 15 divided by 3 is just 5, so this is 5 x plus 1 is greater than or equal to 15 over 5 gives me 3 times 3+2x. Okay, now we need to go ahead and distribute these numbers into our parenthesis here on both sides of my inequality. So starting on that left side, 5 times x gives me 5 x and then 5 times 1 is 5. So this is 5x+5 is greater than or equal to 3 times 3 which is 9 and then 3 times 2x which is 6x. Okay, now we can go ahead and get all of my, variables on one side, all of my constants on the other. So I'm gonna go ahead and move my 6x to this side, move my 5 over here. So subtracting 6x from both sides cancels on that right side and then subtracting 5 from both sides, canceling on the left. So 5x minus 6x is going to give me negative one x, and that's greater than or equal to 9 minus 5, which is 4. Okay. Now, I want to go ahead and isolate x, which I can do by dividing by negative one. But remember, whenever I divide by a negative, that means I need to go ahead and flip my inequality symbol. So this is going to become x is less than or equal to 4 divided by negative 1 which is negative 4 and this is my solution. Don't forget to flip that inequality symbol anytime you divide or multiply it by a negative. Okay. So let's go ahead and graph this. Well, if x is less than or equal to negative 4, that means that negative 4 is going to get a closed circle because it is included in that solution set and then it's anything less than negative 4 all the way to infinity, so I will indicate that with an arrow. Now if I want to write this in interval notation, I'm going to have a negative infinity because it's anything going in that direction that's less than negative 4, and then a negative 4 enclosed in a square bracket. And I'm all done. I'll see you in the next video.

Solve the inequality. Express the solution set in interval notation and graph. 1 3 ( x + 1 ) ≥ 1 5 ( 3 + 2 x ) \frac13\left(x+1\right)\ge\frac15\left(3+2x\right) 3 1 ( x + 1 ) ≥ 5 1 ( 3 + 2 x )

( − ∞ , − 4 ] \left(-\infty,-4]\right. ( − ∞ , − 4 ]

( − 4 , ∞ ) \left(-4,\infty\right) ( − 4 , ∞ )

[ 4 , ∞ ) \left[4,\infty\right) [ 4 , ∞ )

[ − 4 , ∞ ) \left[-4,\infty\right) [ − 4 , ∞ )

1. Equations & Inequalities

Linear Inequalities

College Algebra

1. Equations & Inequalities

Linear Inequalities

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Multiple Choice

Solve the inequality. Express the solution set in interval notation and graph.
$\frac13\left(x+1\right)\ge\frac15\left(3+2x\right)$

$\left(-4,\infty\right)$

A number line with a graph

$\left[4,\infty\right)$

A number line with a graph

$\left[-4,\infty\right)$

A number line with a graph

$\left(-\infty,-4]\right.$

Verified step by step guidance

Start by distributing the constants in the inequality: $13(x+1) \geq 15(3+2x)$. This becomes $13x + 13 \geq 45 + 30x$.

Next, move all terms involving $x$ to one side of the inequality. Subtract $13x$ from both sides to get $13 \geq 45 + 17x$.

Subtract 45 from both sides to isolate the term with $x$: $13 - 45 \geq 17x$, which simplifies to $-32 \geq 17x$.

Divide both sides by 17 to solve for $x$: $-\frac{32}{17} \geq x$. This can be rewritten as $x \leq -\frac{32}{17}$.

Express the solution in interval notation. Since $x$ is less than or equal to $-\frac{32}{17}$, the interval is $(-\infty, -\frac{32}{17}]$. The graph of this solution is a line extending to the left from $-\frac{32}{17}$ with a closed circle at $-\frac{32}{17}$, indicating that $-\frac{32}{17}$ is included in the solution set.

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