Question 1

solve the system of equation by the subtraction methos
15x+5y=28
-3x=y+6

Accepted Answer

Subtract the last equation by y on both sides, then multiply by 5 to obtain:
15x+5y=28
-15x-5y=30
Then add the first equation to the second from which it follows that:
15x-15x+5y-5y=0x+0y=0=58.
The system of equation has no solutions.

Question 2

Are there worksheets on this topic?

Accepted Answer

yeah there is an exercise it's called systems of equations with simple elimination, however you don't need to multiply it is already set to be able to be eliminated without that ext a step.
Another thing you could do is you could ask your Math teacher to give you some examples or you could use a Math book.

Question 3

At 3:02, Sal says you could have solved this with substitution. How could you do that?

Accepted Answer

EpicOne,
The system of equations was
-3y+4x=11
y+2x=13
If you subtracted 2x from both sides of the second equation to get
y=13-2x  
Now you can substitute (13-2x) for the y in the first equation.
-3(13-2x)+4x=11  
Then you can solve for x.

After finding x, 
Then substitute that answer for x in either equation and solve for y.

The substitute both answers into the other equation to double check.  
If the equation is in balance, you have the correct answers.

I hope that helps.

Question 4

Why is it that when you only want to multiply a certain "term" in an equation, you have to multiply that entire side of that equation, rather than let's say for example, the term was a fraction and you needed to multiple it to make it NOT a fraction, you don't multiply that entire side of that equation?

For example, like the equation Khan gave us:
y+2x=13

Why is it when we need to multiply the "y" it to make it cancel out with the "y" in the other equation, -3y+4x=11, we multiply the entire left side of the equation?

That now gives us: 
3y+6x=39

Now, if the equation was, 6/3x+2x=13, and we wanted to take away that fraction, we would do this:

3*6/3x+2x=13*3

...right? We would NOT multiply the entire left side of the equation, only multiply that fraction (so we can "get rid" of it) and the other side of the equation because we have to do what we did on one side, to the other. Why is that? Why do we have the multiply the ENTIRE side of the equation in the one Khan gave us and not in one with a fraction?

I hope my question wasn't too lengthy!

Accepted Answer

Actually in your second example you would also need to multiply the entire left side, otherwise the equation doesn't hold anymore.
To see this, I'll use numbers instead of x or y.
As an example, take the simple equation 5/2 + 1/2 = 3 (can you see that this is correct?), and say you want to get rid of that first fraction.  If you multiply only the first fraction, what do you get?
2*5/2 + 1/2 = 2*3
=> 5 + 1/2 = 6
Hang on, that's definitely not correct.
Let's try again, this time multiplying the whole left side, and of course the right side as well:
2*5/2 + 2*1/2 = 2*3
=> 5 + 1 = 6.
Yep, that looks good.
It doesn't matter whether you use x and y, or just numbers (because x and y are just placeholders for numbers we don't know yet).  Every time you multiply or divide a term of an equation, you have to do the same to the other terms, and to the other side of the equation.

In your example (6/3)x+2x=13, we would do the same, multiplying both sides by 3:
3*(6/3)x + 3*2x = 3*13
=> 6x + 6x = 39
=> 12x = 39 = x = 39/12 = 13/4.
Multiplying only the first term gives a completely different, and incorrect, result of x=39/8 (try it!)

Question 5

couldn't x be 8 and y be -3?

Accepted Answer

Let's try using your x and y values for the second equation:
```
x = 8
y = -3
y + 2x = 13
-3 + 2 * 8 = 13
-3 + 16 = 13
13 = 13
```
Great! It works! Now let's try using them for the first equation:
```
x = 8
y = -3
-3y + 4x = 11
(-3 * -3) + (4 * 8) = 11
9 + 32 = 11
41 = 11
```
Uh-oh... Something went wrong there! As you can see, just because it works for one equation, doesn't mean it will work for both! That's the tricky part of these questions. In fact, you could find lots of different values that satisfy each equation alone, but there will only be one set of values that will work for both :)

Question 6

why did you multiply by three and why did you chose to turn it into three y

Accepted Answer

−3𝑦 + 4𝑥 = 11
𝑦 + 2𝑥 = 13

The reason Sal multiplies the second equation by 3 is that he wants the 𝑦-term in both equations to have the same coefficient (albeit different signs):
−3𝑦 + 4𝑥 = 11
3𝑦 + 6𝑥 = 39

The first equation tells us that adding 11 is the same thing as adding (−3𝑦 + 4𝑥).
This means that we can add 11 to the right-hand side of the second equation and (−3𝑦 + 4𝑥) to the left-hand side without changing the equality.
3𝑦 + 6𝑥 + (−3𝑦 + 4𝑥) = 39 + 11

*Since the two 𝑦-terms have the same coefficient they will cancel each other out*, and we are left with:
10𝑥 = 50 ⇔ 𝑥 = 5

Question 7

Without graphing complete parts a and b
X= -y
Y+x = 4
Is it identical lines, parallel or intersecting lines at a single point and ihow many solutions does this system have

Accepted Answer

Rewrite the two equations like this

x=-y --> y = -x
y+x=4 --> y = -x +4

*Looking at these equations* you can see that the slope of both lines are the same (-1) and that the y-intersections of the two lines are different (0 for the first line and 4 for the second line). Try to visualize this!

When you have the same slope (ex. -1) on two lines that intersect with the y-axis on different places (ex. 0 and 4), these lines will never intersect with eachother - they are parallel.
Since these two lines never intersect with eachother, there is no point on their lines that they have in common and therefore there is no solution to this system.

You could also see that the system does not have a solution by *solving the system*, you will get an inconsistent result (0=4 or 0=-4) - the lines are parallel.

8th grade (Eureka Math/EngageNY)

Course: 8th grade (Eureka Math/EngageNY) > Unit 4

Systems of equations with elimination: -3y+4x=11 & y+2x=13

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