I think of a number. I divide it by seven. My answer is 2.5. What number did I start with?

To solve this equation first multiply both sides of the equation by 7

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Solving algebraic fractions (equal to x + a)

﻿Hi everyone. I'm Mr. Lund, and in this lesson, we're going to be solving equations with algebraic fractions that are equal to an unknown and a number we do know. Hi everyone. I am going to solve 3b over five equals 12. How do I solve? Well, the first thing I need to do is multiply both sides of the equation by five. That finds me that the numerator of 3b is equal to 60. By dividing both sides by three, I can find b is equal to 20. I have solved my equation. Look at this example, how is it different? Do you see that there are variables on both sides of this equation? To solve an equation like this, multiply both sides by the denominator. Again, the denominator is five. So I'm going to multiply both sides of the equation by five. That finds me that the numerator of 3b is equal to 60 minus 5b. Remember to multiply both terms on the right side of the equation. By adding 5b to both sides of the equation, I discover that 6b is equal to 60. And dividing both sides by eight helps me solve the equation and say that b is equal to 7.5. You can check your answer by substituting it back into the original equation. Let's have one more example for good look. Let's solve 4b over seven is equal to 2b plus nine. What do you think my first step would be? The first step I'm going to take is to multiply both sides of the equation by seven, that finds that the numerator of 4b is equal to 14b plus 63. Remember to multiply both terms on the right side of the equation. By subtracting 14b from both sides of the equation, I could find that negative 10b is equal to 63, Dividing both sides by negative 10, finds me that b is equal to negative 6.3. Check your solution by substituting the value back into the original equation. Here are some questions for you to try. Pause the video and return to check your answers. Here's the solutions to questions one and two. Slow and steady wins the race with these more complex examples. Remember, if you are multiplying through by a denominator of a fraction on one side, and you have two terms on the other side, then you have to multiply both of those two terms. That is usually where pupils make a mistake. Oh, not only pupils, teachers as well. Let's try question three now. Pause the video and return to check your answers. Here is a solution to question number three. Now, I don't think that Mo would ever say to Dora, "I've got 4m divided by three sweets." But mathematically, We can use this to create a question that lets us view solving equations with fractions in a different way. So here is some more examples for you to try. Pause the video and return to check your answers. Here is the solutions for questions number four and five. If you want to check whether your solutions are correct, substitute your values back into the original equation and that should tell you whether they're right or not. Well done for getting this far. Pause the video and return to check your answers. Here is the solutions for question number six. Really well done for getting this far, lot of work involved. Hopefully as you're going through the worksheets, you're building up your fluency and your work is getting faster.

What would the output be?

In this lesson, we will solve equations with an algebraic fraction equal to a number and an unknown value.

Lesson overview:Solving algebraic fractions (equal to x + a)

Core Content

Solving equations 2 (Simple algebraic fractions):