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## Algebra (all content)

### Course: Algebra (all content)>Unit 10

Lesson 13: Evaluating expressions with unknown variables

# Worked example: evaluating expressions using structure (more examples)

A few more examples where Sal evaluates expressions with multiple unknown variables. Created by Sal Khan.

## Want to join the conversation?

• I think there is solution to the equation being talked about at . 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 • 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!
• 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 • How do I evaluate an equation like y+7/x-5=2/3? • 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?
(1 vote) • 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

Is this true or did I do something wrong? • 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.
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).
• 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)?
(1 vote) • 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.
• At , why did sal cross out the expression divided by 3?
(1 vote) • 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!
(1 vote) • 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
• 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?
(1 vote) • 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. 