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So I have a College Admission test tomorrow and I am hoping that you could help me understand how to arrive at the solution to this:

1.) Given the following equations: $$3x-y=30\\ 5x-3y=10$$ What are the values of $x$ and $y$?

$\quad$ a. $x=20, \; y=30$
$\quad$ b. $x=30, \; y=20$
$\quad$ c. $x=20, \; y=40$
$\quad$ d. $x=10, \; y=30$
$\quad$ e. $x=30, \; y=30$

When we solved this our teacher said that the correct answer is

$\quad a. x=20, \;y=30$

Could you please explain me why is that the answer please... and please tell me the title of the lesson so that I could study further in this field.. Thank you.

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    $\begingroup$ +1 This is an honest question, and the OP is showing that they want to know what sort of topic this is so that they can study further. Why so many downvotes? Yes, I know we to want to see work shown for the good of the site (and I'm typically a stickler for that), but I think it is also important to take into account the OP's attitude, too. :) $\endgroup$ – apnorton Aug 3 '13 at 3:29
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You have what we call a system of two linear equations in two unknowns. A corresponding "lesson" might be called solving simultaneous systems of equations.

$$\begin{align} 3x-y& =30 \tag{1}\\ 5x-3y &=10\tag{2}\end{align}$$

There are a number of ways you can solve for the $x, y$ values that satisfy both equations. One way you can approach this is by substitution: expressing $y$ in equation $(1)$ as a function of $x$, and "plugging" that function into into "$y$" in equation $2$:

$$ 3x - y = 30 \iff \color{blue}{\bf y = 3x - 30}\tag{1}$$

$$\begin{align} 5x - 3\color{blue}{\bf y} & = 10 \tag{2}\\ 5x - 3(\color{blue}{\bf 3x - 30}) & = 10 \\ 5x - 9x + 90 & = 10 \\ -4x & = -80 \\ \bf x & = \bf 20\end{align}$$

Now, go back to $(1)$ and "plug in" $x = 20$ to solve for $y$: $$\begin{align} (1)\quad x = 20,\quad 3x - y = 30 & \implies 3(20) - y =30 \\ &\iff -y = -30\\ & \iff {\bf y = 30}\end{align}$$

Viola!: We have a unique solution to the system given by equations $(1)$ and $(2)$, namely, $x = 20, \;y = 30$, and this corresponds to option $(a)$.


If you explore the link (Khan Academy) posted at the start of this answer, you'll find corresponding videos and practice problems for solving systems of linear equations by substitution (as we did just now) and also by elimination.

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  • $\begingroup$ In this case, I would prefer multiplying the first row by -3, adding the result to the second row, get x = 20, substitute back into the first equation and get y = 30. It is easier than substitution. $\endgroup$ – Matthias Aug 3 '13 at 9:22
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    $\begingroup$ @amWhy: needs a nice question badge! +1 $\endgroup$ – Amzoti Aug 4 '13 at 1:12
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Hint:

Just substitute in your equations $x$ and $y$ by the given values and check the answer.

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  • $\begingroup$ Short but helpful.:) $\endgroup$ – mrs Dec 16 '13 at 9:45
  • $\begingroup$ Voila! There you have it! $\endgroup$ – amWhy Mar 4 '14 at 13:23
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When given the answers in multiple choice format, I've always found it quicker to just substitute the values to see which works out but the proper way to do it is the following. You have these two equations:

$$3x-y = 30,$$

$$5x-3y = 10.$$

Let's multiply the top equation by $(-3)$ and see what happens.

$$(-3)(3x - y) = -9x + 3y = (-3)(30) = -90$$

That is to say that we get

$$-9x + 3y = -90.$$

note that in this new equation we have a $+3y$ and our other equation has $-3y$. If we add them, it will get rid of the $y$ dependence since they will cancel. Let's do this. Adding them we get:

$$-9x + 3y + 5x - 3y = -90 + 10 $$

Or

$$-4x = -80 \Rightarrow x = 20.$$

Can you see how to get $y$ from this? (Think: substitute the value of $x$ we just found.)

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... and please tell me the title of the lesson so that I could study further in this field..

This is the study of "simultaneous equations". In this linear simultaneous equations, because all the equations used describe straight lines. Sometimes there will be no solution to these types of problems. Sometimes there will be many solutions. Its best to learn matrices if you intend to solve systems of equations with more than three variables.

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Well, if you are taking a multiple choice test, then hopefully one (and only one) of the four answers works. This leads to a pragmatic (if slightly underhanded) approach to solving these problems, namely test all the values. This is likely to be faster than actually calculating $x$ and $y$ by hand if the number of choices is not too large.

Here's how you could do this problem with this approach. First, it seems easier to calculate $3x - y$ than $5x-3y$, so let's plug the 5 cases into the first equation.

a. $3x-y=3\times20-30=60-30=30$ works

b. $3x-y=3\times30-20=90-20=70$ no

c. $3x-y=3\times20-40=60-40=20$ no

d. $3x-y=3\times10-30=30-30=0$ no

e. $3x-y=3\times30-30=90-30=60$ no

So we've already ruled out all the answers other than (a), and hence you can just pick (a). We don't need the second equation at all.

Another possible closely related method (if you're pressed for time) would be after seeing that (a) works for the first equation, immediately check (a) against the second equation. If it works there too, you've found a correct answer, and there should only be one correct answer if you trust whoever wrote the exam, so you can just mark down (a). This is likely to save you time in a case like this where you quickly find one that satisfies the first equation.

As for understanding how to solve this outside such a controlled environment, amWhy's answer is very good as a basic treatment, so I won't repeat that. The basic idea is that you want to find a way to eliminate one variable from the equations by manipulating to solve for one variable and substituting into the other equation. Then solve that equation for the second variable, and go back to find the first variable after that.

If you want to do further study, this is a system of linear equations, and finding the solutions to such systems is a significant part of linear algebra. You probably don't want to try to just pick up a linear algebra textbook without a good understanding of solving systems like the above one though. I recommend trying the techniques above on a system with 3 equations and 3 unknowns to check that you understand how it works first.

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  • $\begingroup$ OMG thank you guys for the help that you gave me.. thank you very much... $\endgroup$ – user84466 Aug 3 '13 at 4:43
  • $\begingroup$ @user84466 You're welcome. If you found this helpful, you can click the upvote icon at the top of the post, and if you think it was the most helpful for you out of all of the answers you can click the check-mark to mark it as accepted. $\endgroup$ – user88377 Aug 3 '13 at 4:57
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For this type of problems,proceed with the given options below.

Take any one equation from this given problem. Substitute the values in the question.

Lets take 3x−y=30 ---- (1)

a. x=20,y=30

Substitute this values in (1)

3(20)-30 = 30 ,you got the answer.

Similarly check with other options.

simple.These kinda problems we can solve from answers.

Good luck

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