Let's review from our last lesson. I want you to try summing this equation for y.
Negative 10 3rds is correct.We start by multiplying each set of our equation by multiplying 1 times 5 is 5 and at 3 times 5 plus y is 15 plus 3y. We subtract 15 from both sides and then divide by 3 to get y is equal to negative 10 thirds.
So you know how to solve for a variable using numbers. First we multiplied those sides by the quantity 5 plus y, we distributed the positive 3 here, and then we subtracted 15 and divided by 3. But what if we had an equation like this one and we wanted to solve for y? Notice the setup of these two equations are nearly identical. It turns out that we can solve for y by following nearly the same exact steps here. Instead of using numbers, though, we'll have variables. First we'll multiply each side of our equations by x plus y. Then we'll distribute z to both of these terms, to get zx plus zy equals one x. When we're at this step, we want to isolate y. Remember we wanted to solve for y or this variable and get it by itself. So we should isolate this on one side of the equation by subtracting zx from both sides. We'll get zy is equal to x minus z times x. These are not like terms so we can't combine them. The variables are different. And finally to isolate y we divide both sides by z. This gives us with one y equal to x minus z times x all divided by z. This is pretty amazing, we actually solved an equation for a variable in terms of other variables. So, in this work we really multiplied by x plus y first Then we distributed the positive z to each of these terms. Next we subtracted zx from both sides of our equation to get the y alone on one side. And to solve for y, or to get one y, we divided by z, which gave us our final answer. See if you can look back at this work and this work, to draw comparisons between the two. They're not that much different. We're really just trying to get 1 variable solved in terms of others. When you're doing the comparison, think about z as being 3. And think about x as being 5. This is a skill that might come in handy with other courses. Especially when you're trying to solve for certain variables. Maybe you don't want to find this variable, z. Maybe you want to find x or y. These 2 equations are actually the same. They just look different. This equation and this equation are actually equivalent. They just look different in from. This one is solve for z, and this one is solve for y. If I knew the values of x and y, I could quickly compute the value of z. But, maybe, instead, I really want to find the value of y. If I know the values of x and z, I can quickly find the value of y by plugging into this equation.
This could come in handy with something like a temperature conversion. This equation relates degree Celsius to Fahrenheit. So if I told you that the temperature was 32 degrees in Fahrenheit, then you could quickly plug that in to tell me that the Celsius would be, well, 0 since 32 minus 32 would equal 0. But what if I told you the temperature was 32 degrees Celsius. Would you be able to tell me the degrees Fahrenheit? You could plug in 32 into C and then use your equation solving skills to figure this out but I wouldn't want to do this work every single time for a new degree Celsius. I want an equation that just has F solve for. So let's figure that out together. I want you to use your knowledge of solving equations to solve for F here. Type your final answer in this box. Remember, the equation here should contain the variables C and F. And since you're solving for F, your equation should start with F equals. Good luck on this one.
F would equal 9C divided by 5 plus 32. Nice algebra skills for figuring that one out. We start by multiplying both sides of our equation by 9. This will create a fraction. And keep in mind we don't multiply these 2 terms by this 9. This whole term is distributed to both of them. So we simplify or cancel the 9s here first. Next we distribute 5 to F and to negative 32. We add a 160 to both sides, to isolate 5 F. And finally we divide by the coefficient of F, which is 5 on both sides of our equation. We divide 5 into 9 C to get 9 C 5ths. And we divide 5 into a 160 to get positive 32. Here's our equation. Now there was a different way you could've gotten to the same result by dividing both sides here by 5, first. Then you would have just add the 32. I think that would've been a little bit quicker.
Let's consider our first equation again with x, y and z. This time let's solve our equation for x. We can think about cross multiplying or by multiplying both sides by x plus y. We distribute the z to get zx plus zy equals 1x. And now here's where we might get stuck trying to solve for x. Notice that x is on different sides of the equation. We need to move all the x terms to one side of the equation, so that way we can isolate it on one side. Any term that doesn't contain x should be on the other side of the equation. So let's do some rearranging. What would be an equivalent equation for this one, out of these choices?
Well, we want to get x on one side of the equation. So let's subtract x from both sides. I'll get negative x plus zx plus zy equals 0. Remember, I can't combine this negative x with either this term or this term since the variables are completely different. To isolate these two terms that have x, I subtract zy from both sides. This gives me negative x plus zx equals negative zy. Now, this equation doesn't quite match any of these. So there must be something else I can do. Well, let's multiply this equation through by a negative 1. If we multiply every term by a negative 1, we really change all of these signs. This will become positive, this will become negative and this will become positive. So x minus zx is equal to zy. Well, that would be this one here, x minus zx equal to zy. This equation has just been flipped over. This was the tough way to do it. Let me show you a quicker way. We know we want to get these x terms alone on one side of the equation. So if we just subtract zx from both sides, we get there a lot quicker. Zy is equal to x minus zx. This is where your algebra skills can come into play. If you think a little bit forward you can predict what you're going to get. Try and choose the simpler method. Having two x terms isolated on the right side of the equation is the same as having x terms isolated on the left side of the equation. This one just happens to be much easier.
Now that we have this equation, let's think about what to do next. Remember, our goal here was to solve for x. So we have x on one side of the equation. In fact it's in two terms. So how can we isolate x from these two terms? If our variable appears in two terms, we need to factor in order to isolate it. Both of these terms share a factor of x, which means I'll be left with x times the quantity 1 minus z. A quick distribution let's me know that this step is correct. Finally, we divide both sides by the number 1 minus z to get our final answer. 1x, or just x, is equal to zy divided by 1 minus z. Here, we actually solve for x. So solving for a variable is very similar to solving for numbers. We want to isolate the variable on one side of the equation, and sometimes that might involve factoring. Then we just divide by the coefficient, or everything that's not x, to get x alone.
I want you to try a problem very similar to the one we just did. Try solving this equation for t. Keep in mind that you may have to factor. First, get the t's isolated on 1 side of the equation, and then see if you need a factor, or just divide by a coefficient. Good luck.
First we multiply by the LCD, rs on both sides of our equation. This gives us t on the left side and, and the quantity t plus r times r on the right side. Next we distribute the rt to these terms to get this equations. We subtract rt from both sides to get t minus rt equals r squared. Remember, we want to solve for t so let's get it on one side. I'm going to rewrite our equation up here and now we have to factor. We want to get one t out of these two terms. If we factor our t, we're left with t times 1 minus r equals r squared. Finally we divide by 1 minus r to get t is equal to r squared divided by 1 minus r. Fantastic work if you got this. Now I know this is pretty challenging so don't worry if you don't have the hang of it yet. Let's try working through a couple more and see if you pick it up.
Try manipulating this equation to solve for r.
R would equal a plus s divided by s minus a fantastic algebra skills for giving that one correct. If you stumble a bit then lets see how we do this, there was a factoring part to this so that might have been tricky first we multiply both sides by the LCD r, r minus 1 we distribute the s to get sr minus s is equal to ar plus a we also have to distribute the a here Now, I'm at this point. And this is where I need to think back for what I'm solving for. I'm solving for r. So let's get these r terms on 1 side of our equation. And these other variable terms here. So we'll have sr minus ar minus s equals a. We get this from subtracting ar from both sides of our equation. If we add s to both sides. We'll have the r terms on one side of the equation. And now we want to factor. We take r out from each of these terms, and then divide each side of this equation by s minus a. This gives us our final answer. And remember that these two equations are actually the same. A plus s in the numerator is the same as s plus a.
Try solving this equation for t. Remember what you're doing about halfway through. Think about if you need to factor or rearrange terms to get t's on one side of the equation. Good luck.
This would be the answer for t. Fantastic work if you found it. We start by multiplying both sides by the denominator, s plus t. This would give us 3s squared plus 3st on the left and 4rt on the right. We just had to distribute this 3s to each of these terms. Next we subtract 3st from both sides of the equation. This allows us to get nothing but t terms on 1 side. Now that we have terms of t on 1 side, and everything else on the other. We can factor a t out from these terms. We'll have t times the quantity, 4r minus 3s. We divide both sides of our equation by this quantity to get t alone. These reduce to 1, leaving us with t equals this expression. I hope you're starting to see some patterns when solving these equations.
Here's our fifth problem. Try solving it for v.
v Would equal w divided by at minus 1. Great effort for getting that one correct. When we multiplied both sides by the lowest common denominator, or vt, we can see that these two factors will cancel. We're left with avt equals v plus w. We subtract v from both sides to solve for v, or really to isolate it on one side of the equation. Now that we're here, we have to factor this V out. When we factor the V out, we'll have V times AT minus one, and then we can divide by this quantity to get one V alone. V would equal this expression. Keep in mind, this is simplifying the one. I hope you're becoming a pro at these.
Try solving this expression for a. This will be our last practice problem for this section. Good luck and try your best.
For a, we'll get this expression. Great work if you got that one correct. We can do this problem by cross multiplying. So, we'll have negative 7 times 2a minus we're multiplying both sides by the least common denominator, negative 42. It would be the product of these two numbers. We know negative 42 divided by 6 equals negative 7. So we're really left with a negative 7 in our numerator, and we multiply it by 2a minus 3. Something similar happens on the right, negative distribute this negative 7 to each of these terms. So we'll have negative 14a, plus 21 equals 12ab. Remember what we're doing here, we're solving for a, so let's get this a term over to the right side, so we have two a terms together. We add 14a to both sides to get this equation. These terms both share an a, which is what we are solving for. So we factor the a out, leaving us with 12b times 14. Finally, we divide both sides by this expression to solve for 1a. These two factors reduce to 1, leaving us with a equals 21 divided by 12b plus