Find ‘y’ and check. Twenty minus seven ‘y’ equals six ‘y’ minus six. I want to let all “y\’s” be on one side of the equation only, and if they are more than 1, then I will divide the equation by their number – the coefficient before \’y\’. I can choose to leave ‘y’ on the right or left. In fact, I can do it this way. First … maybe it will be easier if I do it on the right and I’ll show you in a second. Now I put all the ‘y’ of the right side, but how will you get rid of all the ‘y’ that are on the left? I have minus seven ‘y’s, or let’s say -7y. Then I can add 7y to each side of the equation, and then adding 7y the left side will be twenty minus seven ‘y’ plus seven ‘y’ which we cross out because their sum is zero, and we are left with only twenty. On the right side we have six ‘y’ plus seven ‘y’, which is thirteen ‘y’. And then we’re left with minus six. And I only want ‘y’ or multiples of ‘y’ on the right, so you have to get rid of those minus six here. Let’s then add six on both sides of the equation. Let me add six to the right too. And then I have thirteen ‘y’ minus six plus six, that’s exactly thirteen ‘y’. And we removed the minus. Then I have twenty plus six, that’s twenty-six. So I have twenty-six is \u200b\u200bequal to thirteen ‘y’. And ‘y’ is only on one side of the equation. On the right, although it will be multiplied by thirteen. So if i want ‘y’ on the right, one ‘y’ on the right, I can divide the two sides into thirteen. In other words, we can multiply both sides of the equation by one thirteenth (one divided by thirteen), the same thing. So let’s divide the two sides of this equation into thirteen. On the right, thirteen times ‘y’ divided by thirteen, that’s just ‘y’. So here is ‘y’. And then twenty-six divided by thirteen right two. Thus the solution is obtained – ‘y’ is equal to two. Now I want to go back to say why I made it easier by putting ” y ” on the right side of the equation. Thus I have a plus value of the coefficient before ‘y’. If I had taken out ‘y’ on both sides of the equation, I would have minus thirteen on both sides, which is also the right way However, let me show you what I say. I could solve the equation like this: Twenty

Algebra