Question 1

I think there is solution to the equation being talked about at 1:35. 3a+5b=2. 15a+15b=?
I am not too sure I double checked and it seems correct.
= 3(5a+5b)
= 3(2a+3b+5b)
= 3(2a+2)
= 6a+6
a = -6/6
a = -1

Therefore,  
-3+5b=2
5b=5
b=5/5
b=1

Hence, 
15(-1)+15(1) = ?
-15+15 = 0

Accepted Answer

You did some really clever factorisation but there are some pretty clever errors as well. Lets review the first part where you got `a = -1`.
At the start you write `= 3(5a+5a)`, but what does it equal? The biggest error comes from this statement.

Lets replace the question mark in the equation with `c`, this seems proper, so:
`15a + 15b = c`, this is how the first line should look.
`3(5a + 5b) = c`
`3(2a + 3a + 5b) = c`, you did make a typo here but this step was cool.
`3(2a + 2) = c`
`6a + 6 = c`
In the last 2 steps of your calculation of `a` you put `c` as 0, but we can't make that assumption out of the blue, `c` can be anything. In the last equation I wrote we see that if we know `a` we can calculate `c` or vice versa. But to know `a` we need `b`.

At this point we know that to solve this problem we need more information!

Question 2

The 'insoluble' 3a + 5b = 2 question.  Why not divide both sides by the common denominator of 3 and 5, i.e. 15, giving after simplification, a +b  = 2/15

Accepted Answer

how? now 15/3=5 so when we divide LHS by 15 then 5 comes below 'a' and similarly 3 comes below 'b' . how does that cancel 5 or 3 . maybe you are taking LCM of expression after dividing .. :-)

Question 3

How do I evaluate an equation like y+7/x-5=2/3?

Accepted Answer

you mean searching for x and y? I've tried like, expressing x via y, and then you can plug in the values you get, and solve for y, and eventually for x. but this one seems to be insolvable..

it's possible I guess. I'm tired now but that's my best idea, good luck

Question 4

3a+5b=c
3a=-5b+c
a=-5/3b+c/3
3(-5/3b+c/3)+5b=c
-5b+c+5b=c
c=c

It led to infinite solution. Hence, not enough information, am I correct?

Accepted Answer

It really does not help to substitute a question back into itself, it will not ever give a solution.  If you have 3 variables, you have to have three independent equations to find a solution.

Question 5

Whilst doing this exercise, I got the question "If 8a+2b+6c= -6, what is 16a+4b+12c?." Can I factor out "8a+2b+6c= -6" by doing 8+2+6(a+b+c)= -6, eventually getting -2 2/3 for (a+b+c)?

Accepted Answer

I don't think your method will work. The way I do these is I compare: 8a+2b+6c to 16a+4b+12c. What can you multiply with 8a+2b+6c to get to 16a+4b+12c? Well, 8a can become 16a if we multiply by 2. Does that work for all the terms? Let's see. 2b can become 4b also by multiplying by 2. And, 6c becomes 12c by multiplying by 2.

We now know: 2 (8a+2b+6c) = 16a+4b+12c. Thus, 2(-6) must also = 16a+4b+12c. Answer is: -12.

We now know:  2 (8a+2b+6c) = 16a+4b+12c.  Thus, 2(-6) must also = 16a+4b+12c.  Answer is:  -12.

Question 6

At 1:10, why did sal cross out the expression divided by 3?

Accepted Answer

He's dividing both sides of the equation by 3.  3 divided by 3 = 1.  And, if you multiply anything by 1, it doesn't change the other value (5*1 = 5;  x*1 = x; etc).

You can also look at it as he is reducing the fraction.  This process of cancelling common factors is done all the time when reducing fractions.

Question 7

Um, I was just wondering couldn't you do the first problem:3x+3y+3z=1 and 12x+12y+12z=?
on a different way?
I solved the problem by dividing equation 2 by equation 1 so i got um... 4x+4y+4z=4
Or you could of just 4(3x+3y+3z=1)===== 12x+12y+12z=4!! Therfore the 4 would replace the question mark!!
I got the same answer as Sal but i did it in a different way so i  was just wondering if this could work?Does it work all the time? Sorry if this confused you!

Accepted Answer

You can solve this problem using System of linear equations.
Say that w is equal to 1 and the answer to 3x + 3y + 3z = 1w, so you create a matrix for both equations
3     3     3   |  1
12   12   12  |  ?w

So you can multiple the first row for 4 and then it will be equal to the second row, so you will know that w must be multiplied for 4 to get the right answer.

Knowing that w is equal to 1, 4 * 1 = 4

Question 8

in the first problem I didn't understand why sal divided 3 from both sides of 3(x+y+z) equals 1. can someone please explain?

Accepted Answer

Sal sees a common structure to both equations. Getting them to something in common let him use information from one equation to solve the other. The common structure in the two equations is "a + b + c". If he can both equations to that term with whatever else remains getting it there, he can use one equation to solve the other.

In this example he takes the approach of getting to a common "a + b + c" in both equations. Again, then he can use one of them to substitute in the other.

3x + 3y + 3z = 1    [ Original equation. Want to get it to "x + y + z" so it can be substituted in the second equation ]
3(x + y + z) = 1     [ Factored out a common 3 from each of the first here terms. ]
(x + y + z = 1/3   [ Divide both sides by 3 to get x + y +z by itself. ]

Now he can replace "x + y + z" by 1/3 in the second equation to end with 4.

I hope this helps.

Question 9

How do you do the addition -8 + 30? I like to think of the difference in absolute value, and then I look at what number is smaller and what number is bigger and try to think of it as number line and figure out if I will go to the right of zero or to the left of zero.

I recently heard Sal say in a video that he likes to rewrite these, I think he distributed out the negative sign somehow and he had two numbers in the parentheses. But I got wrong answer because of this.

-8+30 is not - (30-8) I know that now. The only way I can think of is -(-30+8).

Accepted Answer

I see... so do you do this in your head or do you write it down somehow? It would seem easier maybe to do it in your head, but writing it down might help the thinking process and avoid mistakes. And 30 - 8 is not 24 by the way, that's 22.

Question 10

I think I solved the last one, but I might be wrong:
For the last question I was able to do the following

3a + 5b = 2
15a + 15b = ?

consider the following:

x(3a+5b) = 15a +15b 
(I'm asking what number times the first equation is equal to the second equation)

x(3a + 5b) = 15a +15b
x = 15a +15b / 3a + 5b
x = 5a +3b / 2

so....

(3a + 5b) x ( (5a + 3b) / 2 ) = 15a + 15b

intuition tells me that:
 3a+5b = 5a+3b  
because of the rules of multiplication and addition, so.....
3a+5b = 2
5a+3b = 2
if we substitute these values we get:

2 x (2/2) = 15a + 15b
15a + 15b = 2

so the answer is "2"
Is this true or did I do something wrong?

Accepted Answer

The problem with your assessment is that 3a+5b = 5a+3b is not really true. You are probably thinking of the Commutative Property which says you can move around numbers when adding them. So 3a+5b = 5b+3a, the variables stay with their coefficients, you cannot move the variables around as if they are separate from the numbers they are being multiplied by.
The reason why we can't solve for 3a + 5b = 2 and 15a + 15b = ? is because the ? mark should be another variable as x. The equations would become 3a + 5b = 2 and 15a + 15b = x, and there is a little rule in algebra that you need at least the same number of equations as the number of variables we are solving for. Hence two equations are not enough for three variables.
Final note, your writing of
x(3a + 5b) = 15a +15b
x = 15a +15b / 3a + 5b
x = 5a +3b / 2 
has a little error where changing 15a +15b into 5a +3b doesn't work because you can't divide the coefficients of each variable directly (15 divided by 3 and then by 5). In fact the equation should have be written as
x = (15a +15b) / (3a + 5b)
x = (15a +15b) / (2)
x = (15/2)a + (15/2)b
x = (15/2)(a + b) 
and we are stuck in the same situation as in the video with x = (15)(a + b).
Hope this was helpful.

Course: Algebra (all content) > Unit 10

Worked example: evaluating expressions using structure (more examples)

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