The post Solving 3 Variable Systems of Equations by Elimination appeared first on Toronto Tutor Team.

]]>For example, what is the value of x, y and z for 3x + 2y – z = 8, x – 3y + 2z = 7 and x – 2y + z = 4

Let’s number our three equations:

1) 3x + 2y – z = 8

2) x – 3y + 2z = 7

3) x – 2y + z = 4

The first step is to get rid of one of the variables by adding or subtracting two equations.

Since x has the same coefficient in equations 2 and 3, let’s subtract them:

(x – 3y + 2z = 7) – (x – 2y + z = 4)

x – 3y + 2z – x + 2y – z = 7 – 4

-3y + 2y + 2z – z = 3

-y + z = 3

We’ll call this equation 4:

4) -y + z = 3

This equation does not involve variable x, but we still can’t solve it because it has 2 unknowns, so we need 2 unique equations. The way to do this is to find another equation that does not involve variable x through a different combination of equations. We already used equations 2 and 3 to find equation 4, now we’ll use equations 1 and 3.

We can multiply both sides of equation 3 by 3 to get the same coefficients for x, resulting in 3x – 6y + 3z = 12

Now we subtract this from equation 1:

(3x + 2y – z = 8) – (3x – 6y + 3z = 12)

3x + 2y – z – 3x + 6y – 3z = 8 – 12

2y + 6y – z – 3z = -4

8y – 4z = -4

We can actually simplify this by dividing everything by 4 to get

2y – z = -1

We’ll call this equation 5:

5) 2y – z = -1

Now we have 2 equations that only involve y and z, so we can use the elimination method to solve. Since equation 4 has a +z term and equation 5 has a -z term, let’s add them together:

(-y + z = 3) + (2y – z = -1)

-y + z + 2y – z = 3 – 1

-y + 2y = 2

y = 2

Now we can plug the value of y into equation 4 or 5 to solve for z. Let’s plug it into equation 4:

-y + z = 3

-2 + z = 3

z = 5

If you were to plug y into equation 5 you also get z = 5. Now that we have the values of y and z, we can simply plug them into equation 1 or 2 to find x. Let’s use equation 2:

x – 3y + 2z = 7

x – 3(2) + 2(5) = 7

x – 6 + 10 = 7

x + 4 = 7

x = 3

If you want to check that this is correct we can plug our answers into the original equations:

1) 3x + 2y – z = 8

2) x – 3y + 2z = 7

3) x – 2y + z = 4

1) 3(3) + 2(2) – (5) = 8

2) (3) – 3(2) + 2(5) = 7

3) (3) – 2(2) + (5) = 4

Here’s a practice problem to try on your own:

What is the value of x, y and z for 5x – 3y – 2z = -42, 2x + 3y – 7z = -37, and 4x – y + z = -3

The answer is x = -1, y = 7, z = 8

And that’s it for solving 3 variable systems of equations using elimination!

If you prefer to watch a video version of this tutorial:

You can also check out more great math tutorials.

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]]>The post Solving Systems of Equations by Elimination appeared first on Toronto Tutor Team.

]]>The elimination method works by first eliminating one of the variables that exists in the given equations. The way to accomplish this is to multiply or divide the given equations by a constant, and try to one get one of the variables in all equations to have the same coefficient.

For example, if the given equations are 3x + 4y = 5 and x – 2y = 1

Our goal is for one of the variables to have the same coefficient in both equations (doens’t matter which one).

We can multiply both sides of the second equation by 2 to write x – 2y = 1 as 2x – 4y = 2, so the coefficient for variable y will match with the coefficient for y in the first equation.

The next step is to add or subtract the equations to eliminate one of the variables.

We now have the equations:

3x + 4y = 5

2x – 4y = 2

The first equation has a +4y term and the second equation has a -4y term, so if we add the 2 equations together, they will cancel out. The way to add the equations is to simply group the left side terms and right side terms of both equations together.

3x + 4y + 2x – 4y = 5 + 2

3x + 2x = 5 + 2

5x = 7

x = 7/5

Now we can plug this value of x into either of the equations to solve for y!

Let’s look at another example, the same one as the substitution method tutorial:

What is the value of x and y for 3x – 5y = -7, and x + 3y = 13

Let’s number our two equations:

1) 2x – 5y = -7

2) x + 3y = 13

We can get the same coefficient on both sides by multiplying equation 2 by 2, we’ll call this equation 3:

3) 2x + 6y = 26

Now if we subtract equation 1 from equation 3:

(2x – 5y = -7) – (2x + 6y = 26)

2x – 5y – 2x – 6y = -7 – 26

-5y – 6y = -33

-11y = -33

y = 3

Now we can plug this value of y into equation 1:

2x – 5y = -7

2x – 5(3) = -7

2x = 8

x = 4

Just to check, we can also plug the answer for y into equation 2

x + 3y = 13

x + 3(3) = 13

x + 9 = 13

x = 13 – 9

x = 4

It’s great that we get the same value, as expected if we did everything correctly.

We can go through another example, also solved in the substitution tutorial:

What is the value of x and y for 3x + 2y = 11, and 7x – 4y = 10?

Let’s number our equations again:

1) 3x + 2y = 11

2) 7x – 4y = 10

We need one of the variable coefficients to match, and we could do either one, but just looking at the equations, the y coefficients are a lot easier, because 4 is simply 2 doubled (3 and 7 are a lot more annoying to match). We’ll multiply both sides of equation 1 by 2 to get equation 3:

3) 6x + 4y = 22

Now since equation 2 has a -4y term and equation 3 has a +4y term, we simply add them together:

(7x – 4y = 10) + (6x + 4y = 22)

7x – 4y + 6x + 4y = 10 + 22

7x + 6x = 32

13x = 32

x = 32/13

Now we can plug this value of x into equation 1 or 2, let’s do equation 1:

3x + 2y = 11

3(32/13) + 2y = 11

96/13 + 2y = 11

2y = 143/13 – 96/13

2y = 47/13

y = 47/26

We can check if we found the correct values by plugging them into both equations:

1) 3x + 2y = 11

2) 7x – 4y = 10

1) 3(^{33}/_{13}) + 2(^{47}/_{26}) = 11

2) 7(^{33}/_{13}) – 4(^{47}/_{26}) = 10

If you find that this doesn’t hold true, it means we made an error somewhere.

The best way to learn is by practicing, so try out these examples on your own:

What is the value of x and y for 2x + 4y = 22 and 7x – 3y = 26?

Answer: x = 5, y = 3

What is the value of x and y for ^{1}/_{2}x – 6y = -1 and 8x + 3y = 17?

Answer: x = ^{1}/_{2}, y = ^{1}/_{3}

If you can solve these, you’re all set for solving systems of equations by elimination!

If you prefer to watch a video version of this tutorial:

You can also check out more great math tutorials.

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]]>The post Solving System of Equations by Substitution appeared first on Toronto Tutor Team.

]]>For example, consider the system x = 4y + 2, y = 11x + 2

The first equation is solved for x, and the second is solved for y by isolating each of the variables in the equation.

Once we have written our system in this way, we substitute the equation for one variable into the other equation in place of that variable. So we would plug the isolated expression for x into the variable x in the isolated equation for y.

We know x = 4y + 2, so that means y = 11(4y + 2) + 2, and now we can solve this equation for y.

Let’s look at a complete example:

What is the value of x and y for the system 2x – 5y = -7 and x + 3y = 13?

Let’s number our two equations:

1) 2x – 5y = -7

2) x + 3y = 13

We can rearrange equation 2 by moving the 3y to the right side, which gives us equation 3:

3) x = 13 – 3y

Now we can plug equation 3 into equation 1.

2x – 5y = -7

2(13 – 3y) – 5y = -7 This is plugging in 13 – 3y for “x” in equation 1

26 – 6y – 5y = -7

-11y = -33

y = 3

We’ve found y! We’ll call this equation 4:

4) y = 3

Now we can plug equation 4 into equation 1, 2 or 3 to solve for x. Let’s use equation 3:

x = 13 – 3y

x = 13 – 3(3)

x = 13 – 9

x = 4

And that’s it! Just as a sanity check, we can try plugging in equation 4 into equation 1:

2x – 5y = -7

2x – 5(3) = -7

2x – 15 = -7

2x = -7 + 15

2x = 8

x = 4

We get the same answer, which is a great sign that we’ve done things right.

Let’s do one more example:

What is the value of x and y for 3x + 2y = 11, and 7x – 4y = 10?

Let’s number our equations again:

1) 3x + 2y = 11

2) 7x – 4y = 10

We can beging with either equation, let’s rearrange equation 1:

3x + 2y = 11

3x = 11 – 2y

x = ^{11}/_{3} – ^{2}/_{3}y We’ll call this equation 3, x in terms of y

Now let’s plug equation 3 into equation 2:

7x – 4y = 10

7(^{11}/_{3} – ^{2}/_{3}y) – 4y = 10

^{77}/_{3} – ^{14}/_{3}y – 4y = 10

^{77}/_{3} – ^{14}/_{3}y – ^{12}/_{3}y = 10

^{77}/_{3} – ^{26}/_{3}y = 10

^{26}/_{3}y = ^{77}/_{3} – ^{30}/_{3}

^{26}/_{3}y = ^{47}/_{3}

y = ^{47}/_{26} Let’s call this equation 4

Pretty annoying fractions, but we’ve found y! Now let’s plug equation 4 into equation 1:

3x + 2y = 11

3x + 2(^{47}/_{26}) = 11

3x + ^{47}/_{13} = 11

3x = ^{143}/_{13} – ^{47}/_{13}

3x = ^{96}/_{13}

x = ^{33}/_{13}

And that’s it. If you want to make sure it’s right, we can plug in our answers into the original equations 1 and 2.

1) 3x + 2y = 11

2) 7x – 4y = 10

1) 3(^{33}/_{13}) + 2(^{47}/_{26}) = 11

2) 7(^{33}/_{13}) – 4(^{47}/_{26}) = 10

If you find that this doesn’t hold true, it means we made an error somewhere.

The best way to learn is by practicing, so try out these examples on your own:

What is the value of x and y for 2x + 4y = 22 and 7x – 3y = 26?

Answer: x = 5, y = 3

What is the value of x and y for ^{1}/_{2}x – 6y = -1 and 8x + 3y = 17?

Answer: x = ^{1}/_{2}, y = ^{1}/_{3}

If you can solve these, you’re all set for solving system of equations by substitution!

If you prefer to watch a video version of this tutorial:

You can also check out more great math tutorials.

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]]>The post The Birthday Problem appeared first on Toronto Tutor Team.

]]>Just for some context, in a random group of 23 people there is a 50% probability that two people will have the same birthday. In a group of 70 people there is a 99% probability!

Let P_{n} represent the probability that some pair will have the same birthday in a group of “n” random people

We know that there are 365 days in a year, so there are 365 possible birthdays (ignoring leap years). Given this, the chance that any 2 people have the same birthday is ^{1}/_{365}

This means that P_{2} = ^{1}/_{365}, the probability of 2 people having the same birthday in a group of 2 people

The solution space for this problem can be divided into 2 distinct possibilities, either a pair of matching birthdays exists, or there are no matching birthdays. Given this we can say:

P(birthday match exists) + P(no birthday matches) = 1

Rearranging:

P(birthday match exists) = 1 – P(no birthday matches)

So

P_{n} = 1 – P(no birthday matches amongst n people)

We established that P_{2} = ^{1}/_{365}

Put another way, if you have two people, there are 364 possible days where they do not match

P(no birthday matches, n=2) = ^{364}/_{365}

So P_{2} = 1 – ^{364}/_{365}

Now if we add in a third person, there are 363 possibilities that the third person’s birthday does not match with the first 2.

So the probability that the third person does not match with either of the first 2 is ^{363}/_{365}

In order for there to be no matches between all 3 people, the first 2 must not match, AND the third person must not match with the first 2. In probability, if A **and** B must happen, we take the product of their probabilities, so:

P(no birthday matches, n=3) = ^{364}/_{365} * ^{363}/_{365}

Which means: P_{3} = 1 – ^{364}/_{365} * ^{363}/_{365}

Following the same logic, if we introduce a 4th person who does not match with the first 3

P_{4} = 1 – ^{364}/_{365} * ^{363}/_{365} * ^{362}/_{365}

Following this trend:

P_{n} = 1 – ^{365-1}/_{365} * ^{365-2}/_{365} * ^{365-3}/_{365} … * ^{365-n+1}/_{365}

The reason it is -n+1 (instead of -n) is because we begin comparing when there are 2 people, meaning if there are 2 people then there is 1 potential day they could match. The number of days that could match is always one less than the number of people there are.

Before we can continue with the birthday problem we need to quickly cover factorials. Factorials are denoted by an exclamation mark, and is the product of all of the integers up to and including any integer.

n! = 1 x 2 x 3 x … x n

Our goal now is to express Pn in terms of factorials.

Consider ^{365!}/_{(}_{365-n)!} = ^{1*2*…*(365-n)*(365-n+1)*…*(365-1)*365}/_{1*2*…*(365-n)}

Notice how the numerator has all the same terms as the denominator, and more,

So we can cancel these out: ^{1*2*…*(365-n)*(365-n+1)*…*(365-1)*365}/_{1*2*…*(365-n)}

And we are left with ^{365!}/_{(}_{365-n)!} = (365 – n + 1) * (365 – n + 2) * … * (365 – 2) * (365 – 1) * **365**

And if we just change the order of the terms: ^{365!}/_{(}_{365-n)!} = (365 – 1) * (365 – 2) * … * (365 – n + 2) * (365 – n + 1) * **365**

Remember that P_{n} = 1 – ^{365-1}/_{365} * ^{365-2}/_{365} * ^{365-3}/_{365} … * ^{365-n+1}/_{365}

And we can see that the numerators of P_{n} are practically the same as ^{365!}/_{(}_{365-n)!} except for the bold 365 at the end.

Now if we look at the denominator of the product in P_{n}, it’s simply 365 multiplied by itself n-1 times, so 365^{n-1}. But remember we have an extra 365 in the numerator, so if we multiply the denominator by an extra 365 we can cancel it out. So we will use a denominator of 365^{n}

The final step is to combine all the pieces we’ve put together:

P_{n} = 1 – ^{365!}/_{(}_{365-n)! * 365n}

By taking the factorial terms and dividing it by 365^{n} we have recreated the long string of products in the original P_{n} formula, and this is the general solution to the birthday problem!

So if we return to the example of 23 people, all we do if plug in 23

P_{23} = 1 – ^{365!}/_{(}_{365-23)! * 36523} = 1 – ^{365!}/_{342! * 36523} = 1 – 0.493 = 0.507

In a group of 70 people

P_{70} = 1 – ^{365!}/_{(}_{365-70)! * 36570} = 1 – ^{365!}/_{295! * 36570} = 1 – 0.00084 = 0.99916

That’s the solution to the birthday problem! All of the fractions can get a bit confusing, so it really helps to write it out.

If you prefer to watch a video version of this tutorial:

You can also check out more great math tutorials.

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]]>The post Simple Divisibility Tests 1 to 10 appeared first on Toronto Tutor Team.

]]>It starts super simple.

Any number that divides by 1 is divisible by 1. Pretty obvious.

Any number that ends with the digits 2, 4, 6, 8 or 0 is divisible by 2.

Essentially any even number. For example: 12, 24, 26, 88, 100

A number is divisible by 3 if the sum of the digits of the number is divisible by 3.

For example, is 4325 divisible by 3?

Add the digits 4 + 3 + 2 + 5 = 14

14 is clearly not divisible by 3 (we can even use the same rule to check: 1 + 4 = 5, which is not divisible by 3!)

Is 654 divisible by 3?

6 + 5 + 4 = 15, which is divisible by 3, so yes! (just to confirm the rule, 1 + 5 = 6, which is definitely divisible by 3)

A number is divisible by 4 is the last 2 digits are divisible by 4.

For example, is 5469428 divisible by 4?

That’s a huge number, but we just have to look at the last 2 digits, 28.

Since 28 / 4 = 7, 5469428 is divisible by 4.

Consider 8826, is that divisible by 4?

Looking at the last 2 digits, 26, we see 26 / 4 = 6.5, so 8826 is not divisible by 4.

This is another simple one, any number ending in 0 or 5 is divisible by 5.

So 10, 105, 12345, 267548760, etc., all divisible by 5.

A number is divisible by 6 if it is divisible by 2 AND 3.

So basically we have to apply both of the earlier strategies.

For example, is 444 divisible by 6?

The last digit is a 4, and it’s even, so it’s divisible by 2.

Next, 4 + 4 + 4 = 12, since 12 is divisible by 3, 444 is divisible by 3.

Since 444 is divisible by 2 and 3, it’s divisible by 6.

If we add another 4, and get 4444, it’s still divisible by 2, but 4 + 4 + 4 + 4 = 16, which is not divisible by 3.

Therefore, 4444 is not divisible by 6.

You can try another example, 12345678 (it’s not divisible by 6, check it yourself!)

This one is a little tricky.

First we take the number’s last digit and multiply by 2.

Then subtract this result from the remaining number.

If the difference is divisible by 7, then the number is divisible by 7.

The problem is that if the number is too big for you to easily tell if the difference is divisible by 7, you may need to repeat this several times.

For example, is 5291 divisible by 7?

The last digit is 1, which we double to 2.

With the last digit removed, the number is now 529.

529 – 2 = 527

It’s not really obvious if 527 is divisible by 7, so we do this again.

The last digit is 7, double it to 14

The remaining number is 52

52 – 14 = 38, which is a manageable number

We know 38 doesn’t divide cleanly by 7, so we know 5291 is not divisible by 7

Let’s try another example: 54320

Last digit it 0, so we double it and subtract it from the remaining number, which gives us 5432 – (2 * 0) = 5432

We do it again, last digit 2, double it, subtract: 543 – (2 * 2) = 539

Once again, last digit 9, so 54 – (2 * 9) = 54 – 18 = 35

We know 35 / 7 = 5, therefore we know that 54320 is divisible by 7

If you want to practice, show that 687239 is divisible by 7

For a number to be divisible by 8, the last 3 digits need to be divisible by 8.

For example: 2347240

The last 3 digits are 240

Since 240 / 8 = 30, 2347240 is divisible by 8

It’s not always so clear that a 3 digit number is divisible by 8, but it is much easier to check using long division or some other method, especially when dealing with huge numbers.

For practice, show that 9329108 is not divisible by 8

A number is divisible by 9 if the sum of the digits is divisible by 9. Pretty similar to the method for 3.

For example, 29819

2 + 9 + 8 + 1 + 9 = 21

21 does not divide by 9, so 29819 is not divisible by 9

Try showing that 1233 is divisibly by 9!

This is an easy one, a number is divisible by 10 if the last digit it 0.

Examples: 540, 5230, 301290

Pretty straightforward.

Hopefully you find these techniques useful! Things are easy enough to check with a calculator, but having strong mental math is a surprisingly useful skill to have. It’s a great way to do quick sanity checks when you are solving bigger problems, and helps you avoid silly mistakes.

If you prefer video:

You can also check out more great math tutorials.

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]]>The post Why are there Infinite Primes? appeared first on Toronto Tutor Team.

]]>First to lay the groundwork, a number p is prime if:

a) p > 1

b) p has no positive divisors except 1 and p (itself)

Now let’s get into the infinite number of primes proof.

Let’s suppose q is the product of all primes up to p_{n} plus 1

Expressed as a formula:

q = p_{1}*p_{2}*p_{3}*…*p_{n} + 1, where all of the p_{i}‘s are prime

If we start plugging in numbers:

q = 2*3*5*…*p_{n} + 1

Given this, we can say q is not divisible by any prime numbers ≤ p_{n}

Why is that? It’s because of the “+ 1” we added to the end of q. Let’s say we didn’t have the + 1, then q would be divisible by every prime number up to pn, simply because it is a product of those numbers. The + 1 essentially forces a remainder, because 1 is smaller than the smallest prime, so if q is divided by any of the prime numbers used to make it, there will always be a remainder of 1.

Now let’s look at an example:

Let n = 3, then q = 2*3*5 + 1 = 31

if we divide 31 by 2, 3, or 5 we always get a remainder of 1

You can try this for any n, but you will always see that the product q is not divisible by any of the n primes used in the product.

This means that either q itself is a prime number,

or if it is not a prime number there must exist a prime number p_{k} that q is divisible by, where p_{n} < p_{k} ≤ q

One the above has to be true, and in either case, it requires there to exist a prime number, either q or pk, that is greater than p_{n}. Since n can be anything, there must be infinite primes, because we can always find a prime larger than the n^{th} prime.

If you prefer video:

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]]>The post Greatest Common Divisor appeared first on Toronto Tutor Team.

]]>The greatest common divisor is expressed as:

gcd (a, b)

Where a and b are the two numbers you want to find the greatest common divisor for.

The first step is the division algorithm, which states:

a = qb + r

Where a, b, q, and r are integers, and |a| > |b| > 0

q is the quotient and r is the remainder

For example, let

a = 51 and b = 10

q = 5 and r = 1, so 51 = 5*(10) + 1

Now the algorithm is as follows:

a = q_{1}b + r_{1 }

if r_{1} = 0, then gcd (a, b) = b

if r_{1} ≠ 0, b = q_{2}r_{1} + r_{2}

if r_{2} = 0, then gcd (a, b) = r_{1}

if r_{2} ≠ 0, r_{1} = q_{3}r_{2} + r_{3}

.

.

.

if r_{n} = 0, r_{n-2} = q_{n}r_{n-1} + r_{n}, then gcd (a, b) = r_{n-1}

The key is that at each step if the remainder is not zero, the divisor becomes the new dividend, and the remainder becomes the new divisor.

Let’s look at an example:

gcd (1234, 5678)

5678 = 4*(1234) + 742

1234 = 1*(742) + 492

742 = 1*(492) + 250

492 = 1*(250) + 242

250 = 1*(242) + 8

242 = 30*(8) + 2

8 = 4*(2) = 0

We finally got a remainder of 0! This means rn = 0, which means rn-1, which is 2, is the greatest common divisor.

Now consider 3 numbers:

gcd (2, 6, 17)

= gcd (gcd (2, 6), 17) The way to do this is to simply break the problem into gcd’s between a pair of numbers

= gcd (2, gcd (6, 17)) We can either initially consider the first pair (2, 6), or the second pair (6, 17)

= gcd (2, 1)

= 1

Here are some more examples you can try on your own:

gcd (12345, 1745)

Answer = 5

gcd (21, 39, 1044)

Answer = 3

If you prefer video:

You can also check out more great math tutorials.

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