2 problems have me stuck and factoring is involved in one

[maris]

(x-y)^2=40
x^2+y^2=60
what is xy?
How do you get the answer

next problem
a=3b
6b=12c
a=6c (how do you get this answer?)

 

[tkhunny]

Re: 2 problems have me stuck and factoring is involved in on

1) multiply out (x-y)^2

2) Divide the second equation by 2

3) Why are you stuck? What have you tried?

 

[maris]

Re: 2 problems have me stuck and factoring is involved in on

I have forgotten how to do these problems, so even when you say multiply out and divide by 2 I do not understand the logic/reasoning. Please forgive me, but if you could would you explain the steps to me?
Thanks for responding

 

[tkhunny]

Re: 2 problems have me stuck and factoring is involved in on

You can prove it out by multiplication, but stick these in your head.

(x+y)^2 = x^2 + 2xy + y^2

(x-y)^2 = x^2 - 2xy + y^2

In this case, once you make the transformation, the first will look ALMOST like the second. You should be able to do something with that.

 

[JeffM]

Re: 2 problems have me stuck and factoring is involved in on

I have forgotten how to do these problems, so even when you say multiply out and divide by 2 I do not understand the logic/reasoning. Please forgive me, but if you could would you explain the steps to me?
Thanks for responding

Hi Maris
In the Read before Posting entry, it explains that the volunteers here do not know what you are doing, whether you are in grade school, high school, or college or even whether or not you are a student. The volunteers need to have some information about you in order to give you an answer that meets your needs. If you are a student, most volunteers here will not do your homework for you; they will only give you hints, and their hints are most likely to be helpful only if they know what you have tried and where you are stuck. You need to do something like say, "I am in tenth grade" or "I am trying to help my fourth grade child" or "I am taking a review course in algebra, but have not studied it for fifteen years and have forgotten so much." You also need to explain what you yourself have done in trying to solve your problem. That way we do not waste time repeating what you have already done.

Do you remember ONE WAY to solve a system of linear equations like
u + v = 7
u - v = 1.

It is easy to ELIMINATE the v by adding the two equations together and get
2u = 8.
So u = 4
So v = 3.

Now that was a very simple elimination, but it's always helpful in math to wonder whether techniques that you already know in one context may help in another context. Now what tkhunny suggested as a hint was to multiply out (x - y)[sup:wucekc38]2[/sup:wucekc38], getting x[sup:wucekc38]2[/sup:wucekc38] - 2xy + y[sup:wucekc38]2[/sup:wucekc38]. Why is that helpful. First, it gets xy into the picture, which is what the problem is asking about. Second, it gets x[sup:wucekc38]2[/sup:wucekc38] and y[sup:wucekc38]2[/sup:wucekc38] into BOTH equations, which means that an elimination MAY BE possible.

What tkhunny meant in his original post about dividing by 2 was to divide the second equation in the second problem by 2. What do YOU get when you do that? Now do you see a way to get to the third equation?

 

[maris]

Re: 2 problems have me stuck and factoring is involved in on

I have forgotten how to do these problems, so even when you say multiply out and divide by 2 I do not understand the logic/reasoning. Please forgive me, but if you could would you explain the steps to me?
Thanks for responding

Hi Maris
In the Read before Posting entry, it explains that the volunteers here do not know what you are doing, whether you are in grade school, high school, or college or even whether or not you are a student. The volunteers need to have some information about you in order to give you an answer that meets your needs. If you are a student, most volunteers here will not do your homework for you; they will only give you hints, and their hints are most likely to be helpful only if they know what you have tried and where you are stuck. You need to do something like say, "I am in tenth grade" or "I am trying to help my fourth grade child" or "I am taking a review course in algebra, but have not studied it for fifteen years and have forgotten so much." You also need to explain what you yourself have done in trying to solve your problem. That way we do not waste time repeating what you have already done.

Do you remember ONE WAY to solve a system of linear equations like
u + v = 7
u - v = 1.

It is easy to ELIMINATE the v by adding the two equations together and get
2u = 8.
So u = 4
So v = 3.

Now that was a very simple elimination, but it's always helpful in math to wonder whether techniques that you already know in one context may help in another context. Now what tkhunny suggested as a hint was to multiply out (x - y)[sup:47ijur4l]2[/sup:47ijur4l], getting x[sup:47ijur4l]2[/sup:47ijur4l] - 2xy + y[sup:47ijur4l]2[/sup:47ijur4l]. Why is that helpful. First, it gets xy into the picture, which is what the problem is asking about. Second, it gets x[sup:47ijur4l]2[/sup:47ijur4l] and y[sup:47ijur4l]2[/sup:47ijur4l] into BOTH equations, which means that an elimination MAY BE possible.

What tkhunny meant in his original post about dividing by 2 was to divide the second equation in the second problem by 2. What do YOU get when you do that? Now do you see a way to get to the third equation?

I am a college Grad and I am studying for a standardized test I am taking. I am not great in math and I have not had a math course in 4 years. This question is on a practice test and I am having a hard time still understanding it. Question " If a=3b and 6b=12c, then a=" Is there a simple way you can explain how to do this problem? It looks easy and it probably is but it is racking my brain. The answer is 6c?

 

[masters]

Re: 2 problems have me stuck and factoring is involved in on

next problem
a=3b
6b=12c
a=6c (how do you get this answer?)

Given:

[1] \(\displaystyle a=3b\)
[2] \(\displaystyle 6b=12c\)

Solve [2] for \(\displaystyle b\).

\(\displaystyle 6b=12c\)
\(\displaystyle b=\frac{12c}{6}\)
\(\displaystyle b=2c\)

Substitute results for \(\displaystyle b\) into [1].

[1]\(\displaystyle a=3b\)

\(\displaystyle a=3(2c)\)

\(\displaystyle \boxed{a=6c}\)

 

[maris]

Re: 2 problems have me stuck and factoring is involved in on

next problem
a=3b
6b=12c
a=6c (how do you get this answer?)

Given:

[1] \(\displaystyle a=3b\)
[2] \(\displaystyle 6b=12c\)

Solve [2] for \(\displaystyle b\).

\(\displaystyle 6b=12c\)
\(\displaystyle b=\frac{12c}{6}\)
\(\displaystyle b=2c\)

Substitute results for \(\displaystyle b\) into [1].

[1]\(\displaystyle a=3b\)

\(\displaystyle a=3(2c)\)

\(\displaystyle \boxed{a=6c}\)

ahhh Thank you very much masters!!

 

[Denis]

Re: 2 problems have me stuck and factoring is involved in on

(x-y)^2=40 [1]
x^2+y^2=60 [2]
what is xy?

Keep it simple:
[1]: x^2 - 2xy + y^2 = 40 ; x^2 + y^2 = 2xy + 40
[2]: x^2 + y^2 = 60
SO: 2xy + 40 = 60
2xy = 20
xy = 10

 

[maris]

Re: 2 problems have me stuck and factoring is involved in on

(x-y)^2=40 [1]
x^2+y^2=60 [2]
what is xy?

Keep it simple:
[1]: x^2 - 2xy + y^2 = 40 ; x^2 + y^2 = 2xy + 40
[2]: x^2 + y^2 = 60
SO: 2xy + 40 = 60
2xy = 20
xy = 10

Ok great thanks Denis!