Addition of simultaneous equations

[Probability]

I am just struggling getting to grips with addition of these simultaineous equations?

\(\displaystyle 2x+y=4\)
\(\displaystyle x-y=7\)

So I multiply the second equation by 2x to get;

\(\displaystyle 2x+y=4\)
\(\displaystyle 2x-2y=14\)

Now I can add the two equations

\(\displaystyle 2x+y=4\)
\(\displaystyle 2x-2y=14\)
\(\displaystyle 3y = 18\)
\(\displaystyle y = 6\)

So I can plug in the value of y

\(\displaystyle 2x+6=4\)
\(\displaystyle 2x = 4 -6\)
\(\displaystyle x = -1\)

Substitute into equation 1

\(\displaystyle 2(-1) + 6 = -2 + 6 = 4\)

However,

\(\displaystyle -1 - 6 = -7??\)

The original equation = 7

Please advise where have I gone wrong?

 

[pka]

I am just struggling getting to grips with addition of these simultaineous equations?
\(\displaystyle 2x+y=4\)
\(\displaystyle x-y=7\)

Just add these two:
\(\displaystyle 2x+y=4\)
\(\displaystyle x-y=7\) Add

\(\displaystyle 3x=11\) SO \(\displaystyle x=\frac{11}{3}\)

 

[Probability]

Just add these two:
\(\displaystyle 2x+y=4\)
\(\displaystyle x-y=7\) Add

\(\displaystyle 3x=11\) SO \(\displaystyle x=\frac{11}{3}\)

Thanks for that, I had that on my first attempt, however when I try to solve y as a fraction I get \(\displaystyle 1/3\), so the final answer = 3.33 and not 7?

So I am still getting something wrong?

 

[pka]

Thanks for that, I had that on my first attempt, however when I try to solve y as a fraction I get \(\displaystyle 1/3\), so the final answer = 3.33 and not 7? So I am still getting something wrong?

Yes that is completely wrong.
What you have is \(\displaystyle \frac{11}{3}-y=\frac{21}{3}\).
So what is \(\displaystyle y~?\)

Now you are in disparate need of a grade school review of basic mathematics.

BTW. If the answer were \(\displaystyle \frac{1}{3}\) and you turned in \(\displaystyle 0.33\), ​I would mark it wrong!

 

[Probability]

Thanks for that, I had that on my first attempt, however when I try to solve y as a fraction I get \(\displaystyle 1/3\), so the final answer = 3.33 and not 7?

So I am still getting something wrong?

I think I have got it now, my problem seems to be arithmetic I think?

\(\displaystyle 2x+y=4\)
\(\displaystyle x-y =7\)

add the equations

\(\displaystyle 3x = 11\)
\(\displaystyle x = 3(2/3) i.e. 11\)
\(\displaystyle 2 x 11 + y = 4\)
\(\displaystyle 22 + y = 4\)
\(\displaystyle y = 4 - 22\)
\(\displaystyle y = -18\)

Substitute back into the first equation

2x + y = 4

22 - 18 = 4

substitute into second equation

x - y = 7

11 - 18 = -7

I just can't seem to get a +7 ??

 

[pka]

I think I have got it now, my problem seems to be arithmetic I think?
\(\displaystyle 2x+y=4\)
\(\displaystyle x-y =7\)
add the equations
\(\displaystyle 3x = 11\)
\(\displaystyle x = 3(2/3) i.e. 11\)
\(\displaystyle 2 x 11 + y = 4\)
\(\displaystyle 22 + y = 4\)
\(\displaystyle y = 4 - 22\)
\(\displaystyle y = -18\)
Substitute back into the first equation
2x + y = 4
22 - 18 = 4
substitute into second equation
x - y = 7
11 - 18 = -7
I just can't seem to get a +7 ??

You lack the basic arithmetic skills to do these questions.
Please, please go have a face-to-face with a live tutor.

 

[Probability]

You lack the basic arithmetic skills to do these questions.
Please, please go have a face-to-face with a live tutor.

Thank you, but you have only told me what I already know.

 

[Mrspi]

Thank you, but you have only told me what I already know.

There are situations where a student is lacking SO many skills that it is virtually impossible to cover the gaps in this kind of message-board situation. When there are so many gaps, the only real remedy is for the student to work intensively, one on one, with a tutor.

So...I'll repeat what you've been told already: you need much more help than we can give you here.

 

[Probability]

I understand what has been said above by both members, and yes I am learning just as you did at the beginning, also I will have a word with my tutor when I go to the lesson on Wednesday, however we all must appreciate that this forum is marketed/advertised as the title suggests A MATHS HELP FORUM, and I appreciate that some people are very good at maths, but when it comes to explaining it to others, they become useless to the learner, and just tell them to go and get help

A help forum is there to help people not to go and tell them to get help

So let's start again.

\(\displaystyle 2x + y = 4\)
\(\displaystyle x - y = 7\)

I am now going to multiply the second equation by 2

\(\displaystyle 2x + y = 4\)
\(\displaystyle 2x - 2y = 14\)

Now I am going to subtract the equations above

\(\displaystyle 2x + y = 4\)
\(\displaystyle 2x - 2y = 14\)

3y = 18

y = 18/3

Putting the y value back into equation 1

\(\displaystyle 2x + 18/3 = 4\)
\(\displaystyle 2x = 4 - 18/3\)

x = -1

Putting both values back into the first equation

2(-1) + (18/3) = 4

In the second equation

x - y = 7

I think multiply -1 by 1 to get 1 then

1 - (-6) = 7

so both equations by this method get the correct answers

Constructive comments only please.

 

[Probability]

No one was trying to put you down. What they are saying is that the resources of this site are limited. Everyone here is a volunteer.

First, there is a simpler method to solve your system than yours. (Your method is not wrong although your application of it is). Let's try simpler method first.

\(\displaystyle 2x + y = 4\)

\(\displaystyle x - y = 7\)

Add the two equations

\(\displaystyle 2x + y + x - y = 4+7\)

\(\displaystyle 3x = 11\)

\(\displaystyle x = \dfrac{11}{3}\)

Substitute to find y in one of the equations. I'll take the simpler equation, thank you.

\(\displaystyle 7 = x - y = \dfrac{11}{3} - y\)

\(\displaystyle y = \dfrac{11}{3} - 7 = \dfrac{11}{3} - \dfrac{21}{3} = -\dfrac{10}{3}\)

Check your work

\(\displaystyle 2x + y = 2*\dfrac{11}{3} + (-\dfrac{10}{3}) = \dfrac{22 - 10}{3} = \dfrac{12}{3} = 4\)

\(\displaystyle x - y = \dfrac{11}{3} - (-\dfrac{10}{3}) = \dfrac{11+10}{3} = \dfrac{21}{3} = 7\)

Now your method works too if you do it right

\(\displaystyle 2x + y = 4\)

\(\displaystyle x - y = 7\)

Multiply the second equation by 2

\(\displaystyle 2x - 2y = 14\)

Subtract the multiplied equation from the first

\(\displaystyle 2x + y - 2x - (-2y) = 4 - 14\)

\(\displaystyle y + 2y = - 10\)

\(\displaystyle 3y = - 10\)

\(\displaystyle y = -\dfrac{10}{3}\)

Substitute

\(\displaystyle 7 = x - y = x - (-\dfrac{10}{3}), or\ x = 7 + (-\dfrac{10}{3}) = \dfrac{21}{3} - \dfrac{10}{3} = \dfrac{11}{3}\)

Then you check, but we have already checked these answers and know they are correct.

Jeff thank you for taking the time to explain the above equations to me, now I can follow your method of working which I can see works brilliant. Turning the pair of equations into a linear fashion and cancelling down I never thought of, and the way you have substituted into the equation without examples I would never have got that far, so thank you very much for your clear explanation.