In Exercises 25-38, solve each equation.

x/4 =2 +(x-3)/3

Question

In Exercises 25-38, solve each equation.

x/4 =2 +(x-3)/3

Accepted Answer

Hello, today we're gonna be solving the following equation. So what we are given is X over five, equal to four plus X minus 2/6. Now, for simplicity, we're gonna represent four as a rational number and we can do that by putting it over one. So the easiest way to approach this problem is to go ahead and eliminate the denominators. In order to do that, we're going to look for a lowest common multiple and the lowest common multiple is going to be a lowest common factor between the three denominators, which in this case is 51 and six. So what is going to be the lowest common multiple between 51 and six? Well, in this case here, that lowest common multiple is going to be 30. So what we're going to do is we're going to multiply each term by 30 in order to eliminate the denominators. So if we write that out, that is going to give us X over five, multiplied by 30/1. And that is going to equal to 4/1 multiplied by 30/1 plus X minus 2/6, multiplied by 30/1 Now, what we're going to do is we're going to cross divide the numerators and denominators. So, in the first term, we have the denominator of five and we have 30 in the numerator. Both of those are multiples of five. And if we divide each one by five, we can reduce the denominator to one and reduce 30 to 6 for the second term. There's nothing further we can do to simplify this. So we're just gonna multiply four and 30 together. And in the last term, we have six in the denominator and 30 in the numerator, both are divisible by six. So if we divide each term by six, that is going to reduce the denominator to one and give us five in the numerator. So that means that this is going to leave us with six times X which is six X equal to four times 30 which is 100 and plus five times the quantity X minus two. Now, what we're going to do is we're going to just open up the parentheses. And in order to do that, we're going to distribute five into X minus two, that is going to give us six X is equal to 100 and 20 plus five X minus 10. And we can further combine the light terms on the right hand side by combining 100 and 20 and negative 10 together. That is going to give us six X is equal to 100 and plus five X. Now what we're going to do in order to solve for X is move all the X terms to the left hand side and keep all the constants on the right. And there's only one thing we need to do. We're going to distract five X to both sides of the equation. And that is going to leave us with X is equal to 100 and 10, which is going to be the solution to this given equation. And with that being said, the answer to this problem is going to be B so I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

In Exercises 25-38, solve each equation. x/4 =2 +(x-3)/3

Key Concepts

Solving Linear Equations

Common Denominators

Distributive Property

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