How many solutions does each of the following equations have? 1 2 ( x + 4 ) − 1 3 = 1 6 x − 2 \frac12\left(x+4\right)-\frac13=\frac16x-2

Welcome back. In this problem, we've been given the equation one half times x plus four minus one third equals one sixth x minus two. And we want to see if this equation has one solution, infinite solutions, or no solution. We're going to have to solve the equation to answer this question, so let's get into it. We can see that the equation has fractions. We have one half, minus one third, and one sixth. So we need to find the LCD of those fractions so we can multiply by that LCD on both sides of the equation. Our denominators are two, three, and six, so the LCD is going to be six. So when we multiply by six on both sides of the equation, we get six times one half times x plus four minus one third, on the left, equals six times one sixth x minus two, on the right. Let's distribute that six on both sides of the equation. Starting with on the left, six times one half times x plus four. Six times one half is three, so we get three times x plus four. And then we have six times negative one third, and six times one third is two, so we get minus two. On the right, we have six times one sixth x, and six times one sixth is one, so we get one x, which we can also just write as x. And then we have six times minus two, which is minus 12. Next, we have to simplify our equation. So let's distribute that three on the left side. We have three times x, which is three x, and then three times four, which is 12, and we still have that minus two on the left. And then on the right, we still have x minus 12. Next, we have to combine our like terms on the left side of the equation. We have plus 12 and minus two, which combine to 10, so we get three x plus 10 on the left side of the equation equals x minus 12 on the right. Now we have to get variable terms on one side of the equation and constants on the other, and we currently have both variable terms and constant terms on both sides of the equation. So let's try to get the variables to the left and the constants to the right, starting with the constants. We have to get rid of the plus 10 on the left side of the equation, which we can do by subtracting 10 on both sides. This is gonna cancel the plus 10 on the left, leaving us with three x equals x minus 12 minus 10, which is minus 22. Next, we want to get rid of the x on the right, which we can do by subtracting x from both sides of the equation. It gets rid of the x on the right, so we get minus 22 on the right, and we have three x minus x, which is two x on the left. Next, we wanna isolate our x, which we can do by dividing by two on both sides of the equation. This is gonna cancel our coefficient of two on the left, leaving us with just x. And then we have negative 22 divided by two on the right, which is negative 11. So we end up getting the solution x equals negative 11, which means if we plug in that one solution, negative 11, to our equation, we get a true statement. But if we tried to plug in anything else, we wouldn't get a true statement. That means that we only have the one solution, x equals negative 11. Great work.

2. Linear Equations and Inequalities

Solving Linear Equations

Beginning Algebra

2. Linear Equations and Inequalities

Solving Linear Equations

Struggling with Beginning Algebra?

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Multiple Choice

How many solutions does each of the following equations have?
$\frac{1}{2} (x + 4) - \frac{1}{3} = \frac{1}{6} x - 2$

One solution

Infinite solutions

No solution

Verified step by step guidance

Start by writing down the given equation: $\frac{1}{2}(x + 4) - \frac{1}{3} = \frac{1}{6}x - 2$.

Distribute $\frac{1}{2}$ to both terms inside the parentheses: $\frac{1}{2} \times x$ and $\frac{1}{2} \times 4$, which gives $\frac{1}{2}x + 2$.

Rewrite the equation with the distributed terms: $\frac{1}{2}x + 2 - \frac{1}{3} = \frac{1}{6}x - 2$.

Combine the constants on the left side: $2 - \frac{1}{3}$, by finding a common denominator and subtracting.

Next, get all the $x$ terms on one side and constants on the other by subtracting $\frac{1}{6}x$ from both sides and subtracting the combined constant from both sides. Then simplify to see if you get a true statement, a contradiction, or an identity, which will tell you if there is one solution, no solution, or infinite solutions.

6:21m

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How many solutions does each of the following equations have?12(x+4)−13=16x−2\(\frac\)12\(\left\)(x+4\(\right\))-\(\frac\)13=\(\frac\)16x-2

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How many solutions does each of the following equations have?
$\frac{1}{2} (x + 4) - \frac{1}{3} = \frac{1}{6} x - 2$