Okay, here's a video that's going to review all main concepts you learned in grade 9 math in under 60 minutes. You're going to learn all of grade 9 in under an hour. So this is a great video if you're doing exam review or if you're about to take the course and you just want to know what you're going to learn. Now in this video, I'm not going to go in-depth into any of the topics.
I'm going to go through a couple examples of each topic, but if you want in-depth explanations and proofs of each topic, go to jensenmath.ca and there's video lessons for each topic there. So I'm going to divide this video into three parts based on the three main units you do in grade 9. So the first part is going to be on algebra. And in algebra you learn about polynomials, you learn how to collect like terms, you learn distributive property, you learn exponent laws, and the next part is solving equations. So I'm going to go through that part first, and watch the second video if you want to learn about linear relations, and the third video if you want to review the geometry section. So watch all three videos and you'll have reviewed all of the main concepts of grade 9 math.
in under an hour. So let's get started with the first topic right away and I'm gonna go quickly just so I can get this done in under an hour. So the first main topic you would have learned in grade 9 math is exponent laws. So to do exponent laws, first main thing you have to remember, you have to know what a power is. So let's say we have 5 squared.
We have to know that that means 5 times 5, right? This is the base, 5 is the base, 2 is the exponent, and this is a power. So the exponent tells you how many of the base, so how many factors of the base are being multiplied together.
So there are some exponent laws. You would have learned the product of powers rule. So if you have two powers being multiplied together, the dot means multiply, so two powers being multiplied together have the same base, that's important the base has to be the same, there's a rule, you keep the base the same and you add the exponents.
That equals x to the a plus b. So you can rewrite it as a single power by keeping the base the same, adding the exponents. What if you have two powers for the same base being divided by each other? There's a similar rule for that one. Oops, similar rule.
You keep the base the same, and this time you subtract the exponents. That's x to the a minus b. You would have learned the power of a power rule, so if you have an exponent on top of an exponent, you keep the base and you multiply the exponents, x to the a times b. What if you have a power where the base is a quotient, so a power of a quotient? You have to remember that this exponent gets put onto the numerator and the denominator, so this is equal to a to the x over b to the power of x.
What if we have a power where the base is a product of numbers or variables? So this exponent outside of the brackets gets put onto each of the factors inside the brackets. So this would be equal to a to the x times b to the x. And A couple more rules you would have learned that anything to the power of 0 is equal to 1. And the negative exponent rule, if you have a power with a negative exponent, you can't leave a final answer with a negative exponent in this section, so what you have to do is rewrite this for the positive exponent by writing the reciprocal of it. So it would be 1 over x to the power of positive a.
So I'm going to go through a couple examples that review all of these rules. These are the main exponent rules you would have learned in grade 9. So let's practice a few of them without variables first, then I'll throw some variables into the mix. So here we have powers that all have the same base, right?
They all have a base of 4. The first two are being multiplied together, so what I can do is rewrite it as a single power by adding the exponents. 3 plus 5 is 8. And then I have to divide that by 4 squared, so I can rewrite that. Since they have the same base and they're being divided, I keep the base the same, and I subtract the exponents. So I end up with 4 to the power of 6. And then after I've written it as a single power, I can evaluate it.
4 to the power of 6 is 4096. Number 4 to the power of 6 means 4 times 4 times 4 times 4 times 4 times 4. It doesn't mean 4 times 6. Okay, power of a quotient. And we have power of a power here as well. So let's start off with the power of a power rule.
So I have an exponent on top of an exponent. I can rewrite this by multiplying these powers together. 2 times 3 is 6. And notice I put the negative sign into the numerator. I could have put it into the denominator if I wanted to, but don't put it into both. I decided to put it into the numerator to make it easier for me here.
So here I have negative 2 over 3 to the power of 6. So what I have now is power of a quotient. So remember for power of a quotient, the exponent, if the base is a quotient, the exponent of the power goes on to the numerator and the denominator. So this equals negative 2 to the 6 over 3 to the 6. And then I have to, let me move this down.
And then I have to evaluate each of those powers. So negative 2 to the power of 6 is 64, and 3 to the power of 6 is 729. So that's fully simplified, I can't reduce that fraction any further. Let's look at these two here.
So I have productive powers here, and the powers have the same base, so I can use my exponent law to simplify by keeping the same base and adding the exponents. 4 plus negative 7, be careful with your integers, 4 plus negative 7 is negative 3. but I can't leave that as my final answer with the negative exponent. What I have to do is rewrite that with a positive exponent by writing the reciprocal of it. So 1 over 5 to the power of positive 3. And now I can evaluate 5 to the power of 3. That gives me 125. So 1 over 125 is my final answer there. How about here?
I have quotient of powers. So what I have to do, I can rewrite it as a single power by keeping the base and subtracting the exponents. 7 minus 7 is 0. And remember, anything to the exponent of 0 is equal to 1. And you should notice up here, when you have something divided by itself, that answer is always going to be 1 as well.
So think of it either way. Let's throw some variables into the mix. So I have...
a product of powers here. I have these three powers being multiplied together. They're all the same base, so I can rewrite as a single power. By keeping the same base and adding the exponents, 3 plus 4 plus 5, that's 12. a to the 12, I can't evaluate that.
Since a is a variable, so we leave it like that. There's our final answer. What we have here is power over product.
So the base of the power is a product of 4x squared and y to the 5. So what I have to do is put this outside exponent onto... each of the factors in the product onto the 4, onto the x squared, and on y to the 5. So I have to do 4 cubed. I'll put in brackets just so it looks nicer here.
I have to do x squared cubed, and I have to do y to the 5 cubed. Right? This outside exponent goes onto each of the factors of the product of the base.
And now I can evaluate. 4 cubed is 64. x to the 2 to the 3, well, that's power by power. I know I have to multiply the exponents there, so that's x to the 6. and y to the 5 to the 3 power of a power rule again y to the 15 and that's done.
Here I have product of some powers and some coefficients here so let's start off with the 5 times 4. Always start with your coefficients we can multiply 5 times 4 just as you always would it's 20. Now you look for powers that have the same base I have an m to the 5 and an m squared those have the same base so I know when I multiply powers that have the same base I keep the base the same and I add the exponents. 5 plus 2 is 7, so I have m to the 7. Also, I have two powers of n that are being multiplied. So I can simplify that by writing it as a single power by adding the exponents.
And keep in mind this n has an exponent of 1, even though you don't see an exponent there. So 1 plus 4 is 5. And that is fully simplified there. That question is done. Let's look at a quotient of powers here.
Same thing, start with the coefficients. Start with your 36 divided by 27 and divide 36 by 27 exactly as you always would. But we don't want a decimal answer, we just want to reduce it, right? 27 doesn't go into 36 evenly, but I can reduce that fraction by finding a number that goes into both of them evenly.
And in fact, 9 goes into both 36 and 27. 9 goes into 36 four times, and it goes into 27 three times. So I can reduce 36 over 27 to 4 over 3. Now I look at my variables. So I have some powers with the same basis.
I have an x cubed and an x to the 6th that are being divided by each other. So I know I can simplify that, writing it as a single power, if I subtract the exponents. 3 minus 6 is negative 3. And make sure you put your quotients into the numerator here.
We'll take care of that negative in a second. Next, we also have two powers that have a base of y. y to the 9 divided by y to the 4. Rewrite it as a single power by keeping the base. and subtracting the exponents, 9 minus 4 is 5. And remember, always put your quotients into the numerator.
Now we have to do something with this answer. We can't leave our answer with a power that has a negative exponent. This x to the negative 3 is a power with a negative exponent. We can take care of that negative by writing the reciprocal of it.
And what's going to happen is, this x to the negative 3, not the 4, just the x to the negative 3, just that power, if we bring it into the denominator, it'll make the exponent positive. So what we have for our final answer, the 4 and the y to the 5 stay in the numerator, the 3 stays in the denominator, and we bring the power of x to the negative 3 to the denominator, and it makes it a power of positive 3. So there's my final answer there. Let's do one more. If you can do this example, you can be confident that you understand all of the exponent laws. So for this example, start with just the numerator.
Forget about the denominator for now. Let's look at just the numerator and simplify that. I have an 8 times a 4, that's 32. I have, now look for powers of the same base, I have a b cubed times a b to the 1. Keep the base, add the exponents.
Remember, product rule, when multiplying powers of the same base, you add the exponents. And now I have a d to the 1 times a d to the 2, that's d to the 3. If we look in the denominator, let's just leave this 2 out front for now. And let's just do this power of a product rule here, where this exponent of 2 has to go on each factor.
of the base. So that's the one, the 2, the b, and the d. So we have to square the 2, which makes it a 4. We have to square the b, which makes it a b squared, and square the d, which makes it a d squared. So now I have 2 times 4 b squared d squared.
So I can simplify that. 2 times 4 and make it an 8. So b to the 4, leave the numerator for now. 2 times 4 is 8. And then I have b squared, d squared. And last step, let's do our dividing.
32 divided by 8. Divide those just like you would divide any old numbers. 32 divided by 8, 8 goes into 32 four times. So my answer, 4. Now do b to the 4 divided by b to the 2. When dividing powers the same base, keep the base.
Subtract the exponents. d to the 3 divided by d to the 2. Once again, keep the base, subtract the exponents, it's a d to the 1. If it's a 1, you don't have to write the 1. Alright, that's it for exponent law review. Let's move on to polynomials quickly.
So basically you have to remember what a term is. A term is an expression formed with a product of numbers and variables. So for example, 2x, that's a term. It's a product of a number, 2, and a variable, x. And what a polynomial is, it's an expression...
consisting of one or more of these terms that are connected by addition or subtraction operators. So for example, I could have the polynomial 4x squared plus 3x plus 1. That's a polynomial where we have three terms, term 1, 2, 3. They're separated by addition or subtraction signs. Now we can classify a polynomial based on how many terms it has by name. If a polynomial only has one term, like this 2x up here, we call that a monomial. If a polynomial has two terms, we call it a binomial.
If it has three terms, we call it a trinomial. And if it has anything more than that, there's no special name for it. Like if it has four terms, we just simply call it a four-term polynomial.
If it had five, we would call it a five-term polynomial, and so on. So one more thing you'll have to be able to do is state the degree of a term and the degree of a polynomial. Now to state the degree of a term by looking at one individual term, you can state the degree of it just by adding up, so finding the sum of, the exponents on all of the variables in that term. So if we look at this first example here, 3x squared y, this is one term.
So this is a monomial. There's nothing else added or subtracted from it. It's one term, has two variables, so to find the degree of this term, we add the exponents on the variables. Keep in mind this y has a 1. So on the x and on the y we add the exponents 2 plus 1 is 3, so this, this right here, this term right here is degree 3. If we look at this next example here, what we have here is three different terms being added together, so we call this a trinomial. We could find the degree of each term, like this first term here is degree 5, this first term here is degree 4, because that has an exponent of 1. and this third term is degree 6. So if we want to figure out the degree of the entire polynomial, all we have to do is figure out the degree of the highest degree term in this polynomial.
We don't add all these degrees together, we just pick the term that has the highest degree. So in this case, it would be the third term, it's degree 6. So what we say is the entire polynomial is degree 6. The degree of the polynomial is equal to the degree of the highest degree term in the polynomial. Here we have a binomial, right, two terms. 1, 2, separated by a subtraction sign.
The first term is degree 7, right? 1 plus 5 plus 1 is 7. This term is degree 6. So to find the degree of the entire polynomial, it's equal to the degree of the highest degree term. This term is the higher degree, so the degree of the entire polynomial is 7. Let's look at collecting like terms now.
So first of all, what are like terms? Like terms... Our terms have the exact same variables with the exact same exponents. For example, these two terms They have the exact same variables They both have an X and a Y and on those variables are the exact same Exponents on the X is a 2 and on the Y is a 1 on both of them So those are like terms the fact that the coefficients are different do not make them not like terms The coefficients don't matter when deciding if they're like terms or not. You just look at the variables and the exponents So if we look at these two here, these are not like terms.
Why? because this one has a y to the power of 1, this one has a y to the power of 2, so they don't have the same variables for the same exponents, so they are not like terms. Why do we have to be able to know what like terms are?
Because we can collect like terms together. For example, here, I have four terms that are being added or subtracted from each other. When adding or subtracting terms, we can collect them together by adding or subtracting the coefficients only. and keeping the variable the same.
This is different than exponent laws. With exponent laws, we were using them when we were multiplying powers or dividing powers. But now we're going to be adding or subtracting terms, and we can collect like terms together.
So what we want to first do is group the terms that are like terms together and make sure the sign that is to the left of the term stays with it. So for example, I have a negative 2x and a negative 5x. Those are like terms, so let's write those beside each other, keeping the signs that are to the left of them. And I also have...
a positive 7y and a negative 9y. Let's write those beside each other because those are like terms. Now I can collect them together.
I can collect the negative 2x minus 5x together because they're like terms by just adding or subtracting the coefficients only. So negative 2 minus 5, that's negative 7, and then you keep the variable exactly the same. Don't change the exponent on the variable at all, it stays exactly the same.
Don't get this confused with exponent laws. Now, let's look at this group of like terms, the 7y minus 9y. 7 minus 9 is negative 2, so I write negative 2y.
So that is fully complete. I can't collect these two terms together because they are not like terms, so that expression is fully simplified. Let's look at the next expression here. It's a little bit longer.
Find the groups of like terms. I have a 3x squared y and an 8x squared y. Those both have the exact same variables for the same exponent, so I'm going to write those beside each other.
3x squared y. 8x squared y. I have a 4y and a negative 1y.
So I'll write my positive 4y and my negative 1y. Remember the coefficient is 1 if you don't see it. And I also have two constant terms.
Constant term means the term without a variable. Those are like terms. I have a positive 7 and a positive 80. Write those beside each other. Now collect your like terms. So 3x squared y plus 8x squared y.
Just add the coefficients only. 11. x squared y. Don't change the exponents on the variables at all when you're adding or subtracting like terms together. I have a positive 4y minus 1y, that's positive 3y, and I have positive 7 plus 80, that's positive 87. That expression is fully simplified. None of these three things are like terms with each other, so I can't collect them together.
Don't try and go any further than you can. And notice the order I wrote these in. You should do it in this order. The highest degree term goes first, and then it goes in descending order.
This term is degree 3, right? 2 plus 1 is 3. This term is degree 1, and a constant term is degree 0. So the degree should go in descending order. Okay, let's look at distributive property.
So basically for multiplying a monomial by a polynomial, in this case a binomial, this is how you do it. Everything inside the brackets gets multiplied by the term out front of the brackets, and that gets rid of the brackets. So if I want to do a times x plus y, I have to do a times x right here, and I have to do a times y. here. So for example if I have 5 times 4x plus 2, I need to multiply the 4x and the 2, both of them, by the 5 that's out front.
So 5 times 4x is 20x, 5 times 2 is positive 10. There's my solution, those can't be collected together because they're not like terms. So let's practice distributed properties. So here I'm doing a monomial multiplied by a trinomial. So everything in the brackets needs to be multiplied by the term out front.
Be careful with your signs. and then that will get rid of the bracket. So I have to do negative 3 times 2x squared.
Well, negative 3 times 2 is negative 6, so I have negative 6x squared. I have to do negative 3 times negative 5x. Don't forget this sign belongs to this 5x, so negative times negative is positive 15x.
And I have to do negative 3 times positive 4, that's negative 12. And I can't collect any of these three terms together because they are not like terms. How about here? I'm going to have to multiply the x plus 3. by the 3 out front. I'm going to have to multiply this x and this 1 by the 2 out front, and that will get rid of the brackets. So 3 times x is 3x, 3 times 3 is positive 9, 2 times x, positive 2x, 2 times 1, positive 2. Now I can collect like terms here because I have a 3x and a 2x, put those together, that's 5x, and I have a 9 and a 2, put those together, that's positive 11. Let's do one more example here for collecting like terms.
I have 4k times k and times negative 3. Now keep in mind this k inside the brackets, think of that as a 1k if you want. So when multiplying 4k by 1k, you multiply the 4 and the 1 together, that's 4. Multiply the k and the k together. When multiplying powers to the same base, you keep the base, add the exponents. There's a 1 on both of them, 1 plus 1 is 2. 4k times negative 3 is negative 12k, right? Positive times negative is negative.
Over here, I have to do negative 2 times k squared. that's negative 2k squared. I have to do negative 2 times negative 3k, negative times negative is positive 6k, and I also have to do negative 2 times positive 4, that gives me negative 8. Here, if you don't see a number in front of the brackets, there's an invisible 1 there, so I have to do negative 1 times k squared, negative 1 times negative 5, so that gives me negative 1k squared, and negative 1 times negative 5 is positive 5. Collect your like terms.
Do the highest degree terms first. So I have a 4k squared, a negative 2k squared, a negative 1k squared. Let's collect those together. 4 minus 2 is 2, minus 1 is 1. So I have 1k squared. I'll just write that as k squared.
Next, I have a negative 12k, a positive 6k. Negative 12 plus 6 is negative 6. So I'll write negative 6k. And lastly, my constant terms. I've got a negative 8. plus 5. Negative 8 plus 5 is negative 3, so I'll write minus 3, and that's done. I can't collect any of those three terms together because they're not like terms.
You probably would have learned this section before distributive property. We learned adding and subtracting polynomials. I think it's easier to do this section after you know distributive property.
So basically, if you have a set of brackets and you don't see a number out front, there's a 1 there. Here, there's a 1 there. There's a 1 there.
Now, to get rid of the brackets, you multiply everything in the brackets by the number out front. So 1 times x and 1 times negative 6, that's not going to change anything. That's just going to give us x minus 6. But here, if there's a negative 1 out front, multiplying both by negative 1, it's just going to change the sign of both terms in the brackets.
Negative 1 times 2 is negative 2. Negative 1 times negative 5x is positive 5x. So what happens is both of these terms change signs. The signs change.
Here I have positive 1 times x, positive 1 times 4. That's not going to change anything. Multiplying things by 1 doesn't change anything. Collect your like terms.
I have a 1x plus 5x plus 1x. That is 7x. And now my constant terms, I have negative 6, take away 2, plus 4. Negative 6, take away 2 is negative 8. Plus 4 is negative 4. So I have minus 4. And that expression is fully simplified. Can't collect those together because they're not like terms. Okay, let's move on to the last thing you would have learned in the first unit in grade 9. You would have learned how to solve equations.
And basically, solving an equation means to figure out what value of the variable makes the equation true. So here we have x plus 4 equals 7. That means something plus 4 is equal to 7. Now, you can probably guess the answer is 3 because 3 plus 4 is 7, and that's the right answer. But for more complicated questions, like when we get to questions like here, here, you're going to want to be able to solve these algebraically using...
probably the balanced method is the method your teacher would have taught you first. So basically you're allowed to solve this equation, figure out the value of x that makes it true, by isolating the variable by moving all of the other numbers away from the x onto the other side of the equation until you have x by itself, and then you'll have your answer. So to isolate the x, we want to move everything away from it.
So right now there's a plus 4 on the same side of the equation with it. We can remove that plus 4 by subtracting 4. As long as we do the same thing to both sides of the equation, that keeps the equation balanced. And we're allowed to do anything we want to the equation as long as we do the same thing to both sides. I've subtracted 4 from both sides, so the equation is still equal.
Both sides are still equal to each other because I've done the same thing to both sides. And look what happens if we subtract 4 from both sides. On the left, I have 4 minus 4, that's 0. So all that's left on the left side of the equation is now the x.
On the left side of the equation, I have 7 minus 4. And 7 minus 4. is 3. So 3 is the correct answer. And don't forget you can plug this answer back into your equation to check if it worked. Is 3 plus 4 equal to 7?
Yes, you have the right answer. Now you should notice that a trick to figure out how to isolate the variable, right now this g5 is being subtracted from it. You just do the opposite of subtracting 5, which is adding 5. So if we add 5, don't forget to do it to both sides of the equation. Whenever you do it to one side, you have to do it to the other.
If I add 5 to both sides of the equation, on the left I have negative 5 plus 5, that's 0, so all that's left on the left side is g. And on the right I have negative 3 plus 5, negative 3 plus 5 is 2. So that's my final answer, g is equal to 2. And you can check your answer by plugging it back into your original equation. Is 2 minus 5 equal to negative 3?
Yes, so we have the right answer. Here, this is different, because it's not 5 plus u, it's 5 times u. We have 5u, so to isolate the u, that is currently being multiplied by 5, we do the opposite of multiplying by 5, which is dividing by 5. And remember, you have to balance the equation.
Whatever you do to one side, you have to do to the other. And then on the left side of the equation, we have 5 divided by 5, which is 1, so those cancel out. So what you have left, you have just a u on the left side, and on the right, we have negative 20 over 5. And what is negative 20 divided by 5? It is negative 4. Don't forget, you can check your answer.
5 times negative 4 is negative 20, so it's the right answer. Okay, let's move on to two-step equations. First thing you want to do is you want to isolate the term that has the variable. So we want to isolate the 7y by moving this positive 8 to the other side by subtracting 8. So what we're going to do, subtract 8 from both sides of the equation, and on the left we have 8 minus 8, that's 0, it's gone.
So all we have left on the left side of the equation is 7y. On the right we have 15 minus 8. and 15 minus 8 is 7. So now we have 7y equals 7. Right now the y is being multiplied by 7. To separate a coefficient from a variable like this, we have to divide both sides by the 7, and these 7s on the left cancel out because 7 divided by 7 is 1. So all we have left is y equals 7 over 7, and 7 over 7 is 1. So there's our final answer. You can double check. If we plug 1 into this equation, 8 plus 7 times 1 is 15. We have the right answer.
Okay, what if we have an equation where we have more than one term that has a variable? Well, you want to start by getting both of those terms to the same side of the equation. Now, I want to point something out here. For these previous questions, what you might have noticed you could have done, instead of using balanced method, you could have thought of just moving things to the other side of the equation by doing the opposite operation, like we can move this plus 4 over by making it a minus 4. Right, you can move something to the other side of the equation as long as you apply the opposite operation. Right, and that would give us what we had, 7 minus 4, x is 3. So let's look here.
We want to get all the terms with a variable on the same side of the equation. So I want to move, actually I'm going to move the other way. I'm going to bring the negative 8x to the right side of the equation, and get all the terms that don't have a variable to the other side. So I'm going to bring the plus 15 to this side by applying the opposite operation. So that just means we're going to change the sign of the term.
So... On the left side of the equation, I'm going to leave the negative 5, and I'm going to bring this plus 15 over, and it becomes a minus 15. On the right side of the equation, I'm going to leave the 2x, and I'm going to bring this negative 8x over, which is going to change the sign of the term, and becomes a plus 8x. Now collect my like terms, and then isolate the variable.
Right now the x is being multiplied by 10, the opposite of multiplying by 10. Divide by 10. So divide both sides by 10 to keep it balanced. The 10s cancel. And what I have is negative 20 over 10 equals x. Negative 20 divided by 10. I know that's negative 2. So negative 2 is my answer.
And you could plug that back in and check your answer. Here, if you have brackets, let's start off by getting rid of the brackets by using your distributive property that you would have learned previously. So start by getting rid of the brackets by multiplying the term out front by all the terms inside the brackets.
Notice I didn't distribute to this negative 8 because it's not in the brackets. So I have 4x plus 12 equals 2x plus 12 minus 8. Now we want to, I'm going to simplify this 12 minus 8 if you don't mind. Positive 12, take away 8, that's positive 4. So I'm just going to simplify that quickly. I'm going to get all the terms of the variable on the same side. So I'm going to bring this.
positive 2x to the left becomes a minus 2x. I'm going to bring the constant terms to the right, so the positive 4 stays on the right. Bring the plus 12 over becomes a minus 12. Collect, and then isolate the variable.
The x is being multiplied by 2, so divide both sides by 2. These 2s cancel because 2 divided by 2 is 1. And what I have, I have just an x on the left. On the right, I have negative 8 divided by 2, which is negative 4. And you could, don't forget, you could double chance. check your answer by plugging it into the equation here to make sure you have the right answer.
Okay, here let's look at fractions. So I have c divided by 3. Well, don't forget, if we want to move that divided by 3 to the other side, what's the opposite of dividing by 3? Well, the opposite of dividing by 3 is multiplying by 3. So I'm going to multiply both sides of the equation, because we have to keep it balanced, by 3. So I rewrote the equation, but multiplied both sides by 3. Why did I do that? Because here I have a 3 divided by a 3. that equals 1, so they cancel out, so all I have left is c equals 3 times 2, which is 6. And we can check, is 6 divided by 3 equal to 2? Yes, we have the right answer.
What if it looks like this? Whenever you see a fraction in an equation, you can get rid of the fraction by multiplying both sides of the equation by whatever the denominator is. So I'm going to multiply the left side by 4, and I'm going to multiply the right side by 4. Whatever you do to one side, you have to do to the other.
And why did I choose 4? Because 4 divided by 4 is 1, so they cancel out. So what I have is 1 times x minus 3, which is just x minus 3. And on the other side, 4 times negative 2, which is negative 8. Move this minus 3 to the other side, or think of balance method. Opposite of subtracting 3 is adding 3, so add 3 to both sides.
On the left, negative 3 plus 3 is 0, so I just have x equals negative 8 plus 3. Negative 8 plus 3 is negative. 5. Let's do another question with multiple fractions. So we have multiple fractions here. If you have more than one fraction, you can get rid of both of them at the same time if we multiply both equations by a common denominator. So what's a common multiple between 3 and 5?
So if we were to count by 3s and count by 5s, what's the first number they would have in common? It would be 15. So what we're going to do is multiply both sides by 15. Remember, whatever you do to one side, you have to do to the other side to keep it balanced. So I multiplied both sides by 15 and you'll see how that works right here. On the left I have 15 divided by 3, 15 divided by 3 is 5. On the right I have 15 divided by 5, 15 divided by 5 is 3. So now my fractions are gone. So I have 5 times 1 times x plus 4, so I'll just write 5 times x plus 4. And here I have 3 times 1 times x plus 2, so 3 times 1 is 3, so 3 times x plus 2. And now you know how to solve it from here.
I'll just do it very quickly. Get rid of your brackets using distributive property. 5x plus 20 equals 3x plus 6. Get all the terms of the variable on the same side.
I'll bring them to the left. So I'll bring the 3x over becomes a minus 3x. Leave the constant of 6. Bring the plus 20 over becomes a minus 20. I have 2x equals negative 14. whoops, minus 20, 2x equals negative 14, and then isolate the x by dividing both sides by 2, because that x is being multiplied by 2 right now, make sure it stays balanced, those twos cancel, and I have x equals negative 14 over 2, which is equal to negative 7. So there's my final answer, x equals negative 7. What I have here for this next example, I have a special case that we can use here.
Now, we could multiply both sides by 30 to get rid of the brackets. That works fine, but I have a shorter way. If we have fraction equals fraction, and that's it.
Nothing else added off at the end here or anywhere else. If you have fraction equals fraction, you can use a shortcut. You can use what's called cross multiplication. You can multiply the denominator of one side by the numerator of the other.
So 6 times x minus 4 is equal to, on the other side of the equation, write the product of the denominator of the other fraction by the numerator of the other one. So 5 times x minus 3. And then solve from here. Distributive property 6x minus 24 equals 5x minus 15. Get all the terms of the variable on one side. 6x minus 5x equals negative 15 plus 24. And we get x equals 9. There's our final answer. You can plug back in and check.
Now cross multiplication is nice, it's a nice shortcut, but make sure you only use it when you have fraction equals fraction. That's the only time it works. This, if you can do this question, you're good with solving equations. So with this question here, what we need to do is get rid of all three fractions, and on one side of the equation we have multiple terms.
So we're gonna have to find a common denominator between 2, 3, and 4. and that would be 12. So we're going to have to multiply both sides of the equation by 12 if we want to get rid of the fractions. So I'm going to multiply both sides by 12, and what happens over here, since I have more than one term on this side, this 12 is going to get multiplied by both terms. So what it's going to look like, I'm going to have to do 12 times x plus 1 over 2 plus 12 times 2x plus 3 over 3 equals 12 times x over 4. And then from there we can simplify 12 divided by 2 is 6, 12 divided by 3 is 4, and 12 divided by 4 is 3. So I have no more fractions. I have 6 times x plus 1 plus 4 times 2x plus 3 equals 3 times x, which I'll just write as 3x. And then you're good solving from here.
Distributive property, 8x plus 12 equals 3x. I'm going to collect some like terms on the left here for 6x plus 8x is 14x, 6 plus 12 is 18. I'm going to get all the terms of the x on the same side, so I'm going to bring the positive 3x over, it becomes a negative 3x, bring the plus 18 over, it becomes a negative 18, so I have 11x equals negative 18, and what I have here, divide both sides. by 11 to get the x by itself and what I have is x equals negative 18 over 11 and that's a fine answer a fraction is a perfectly fine answer just make sure it can't be reduced any further x is negative 18 over 11. Okay last thing probably you would have done in the solving equations is look at some word problems I'm going to do a very quick one just to remind you how to do it So two friends are collecting pop can tabs. Natasha has 250 more than Kristen. Together they have 880 pop can tabs.
How many pop can tabs has each friend collected? So the question wants to know how many pop can tabs does Natasha have? How many pop can tabs does Kristen have? So that's what the question is asking us. So we want to start by making a polynomial expression for each of those people.
Natasha has 250 more than Kristen. We don't know how many Kristen has. So we use a variable for Kristen, and we know Natasha has 250 more than x, so x plus 250 represents that scenario. And then we write an equation using these two expressions. Together, they have 880. So I know if I add those two expressions together, so if I do x plus x plus 250, I know that that should equal 880. Now I can solve this equation, 1x plus 1x is 2x.
Bring that 250 to the other side, becomes a minus 250. So what I have is 2x equals 630, and then divide both sides by 2 to get rid of that coefficient of 2. I have x equals 315. So what does that tell us? It tells us Kristen has 315, and Natasha, remember, is 250 more than x, and we know x is 315. So what I have is 5. 65 and I know together those two numbers add to 880 so I have the correct answer here. Okay, that's it for algebra Watch the next two parts watch part two linear relations in part three geometry geometry will be the shortest section This is the longest section of grade nine So make sure you watch the next two sections and you'll have learned all of grade nine in under one hour