Solve a linear equations with coefficients represented by letters for a specific variable

In today's lesson, we will be solving linear equations with coefficients, represented by letters for a specific variable. First, let's review what coefficients are. Coefficients are the numbers that come before the variable. So in this first example, three is our coefficient. In our second example, 5 is our coefficient. In our third example, negative 14 is our coefficient. But what if we're asking to solve for variables with coefficients represented by letters? That means that coefficients are letters. So in this first example, a is the coefficient. And the second example, Y is a coefficient. And in the third example, negative R is the coefficient. So let's work out some examples. This first example is asking us to solve for X, meaning isolate X so first look locate where X is in this equation. Here it is. So now we need to undo the operations that are happening to acts. So I'm going to add 5 to both sides. Notice I write the plus 5 underneath the 7 because it is a like term. So negative 5 plus 5 cancels out because it becomes zero. And let's bring down what we have left. A, X equals B plus 12. Now we've got to figure out what our next step is. So let's kind of think about this example. We have 7 X equals B plus 12, we would divide both sides by 7 because that is the coefficient. But now we have a coefficient that is a letter. So we need to divide by that letter, which in this case is a so we're going to divide both sides by a over a becomes one, so we're left with one X equals B plus 12 divided by a and we're done. We isolated X let's try another one. Again, we're asked to solve for X so let's locate X in the equation. There it is again. And we're going to undo the operations. Here we're going to minus three B from both sides. Minus three B notice I haven't written the minus three B underneath the 5 because they are not like terms. So plus three B minus three being canceled out becoming zero. Minus AX equals 5 minus three B again, we need to divide by the coefficient. And in this case, it is negative a we're going to divide by negative a over here, negative a over a, cancels out, becomes one, and we're left with X equals 5 minus three B divided by negative a now sometimes we need to rewrite the fraction. We can always rearrange the numerator of the fraction, so it can become X equals negative three B plus 5, divided by negative a and maybe we don't want the negative in the denominator. So in that case, we need to multiply both the numerator and the denominator by negative one. And that would give us three B minus 5 divided by A.