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Homework Help: Linear equations need some help

  1. Jan 31, 2014 #1
    Hello all, this is my first post here and I hope to get some help. I am not as good at math as most of you so sorry if this is a really easy question

    1. The problem statement, all variables and given/known data
    Solve for x and y:


    I can do these types really easy but on the next page of my book it says:

    Let a,b,c,d be numbers such that ad-bc != 0. Solve the follow for x and y in terms of a,b,c,d.


    I am confused how to answer this. Can you give me a hint? Thank you.

    3. The attempt at a solution

    I can't really give an attempt because I don't know where to start. For example the other linear equation I can times the top one by 2 to get -2y so I can take it away from the second equation.


    x=9/16 answer but I am stuck on ax+by=1, cx+dy=2

  2. jcsd
  3. Jan 31, 2014 #2


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    Welcome to PF :wink:
    Is that really ##!##?
    What is your education level?
    Last edited: Jan 31, 2014
  4. Jan 31, 2014 #3


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    != is a common symbol for not equal to

    Try doing the same thing with your general system - what do you need to multiply the first equation by in order to cancel the y's when you add the second equation?(don't worry about whether things are equal to zero or not right now).
  5. Jan 31, 2014 #4
    I am in year 8

    I'm not sure. I'm looking to see what I could times the top equation by in order to get rid of the y's but the only thing I can think of is -y but I don't see how to do it.
  6. Jan 31, 2014 #5


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    In the solution you gave you multiplied the top equation by -2 so that the coefficient of y in the first equation was -2, and the coefficient in the second equation was 2.

    In your new set of equations, the coefficient in the second equation of y is d. What do you have to multiply the first equation by to get a coefficient of -d for y?
  7. Jan 31, 2014 #6



    I really have no idea. There are too many variables I'm getting confused.
  8. Jan 31, 2014 #7
    No need to get confused .There are only two variables 'x' and 'y' .

    'a' , 'b' , 'c' , 'd' are constants (numbers) just like 2,5,9 . Solve the equations just as you would do normally .

    The condition given to you ad-bc != 0 ensures that the the set of equations has a unique solution .Do not worry about the given condition .

    For a clearer understanding give any arbitrary values to a,b,c,d and solve them .For example put a=1,b=2,c=3,d=4 and solve the given set of equations .
  9. Jan 31, 2014 #8
    ok now this makes sense and I can easily do it.


    times top equation by 2 and change the signs



    but I just don't get how to do it without numbers.
  10. Jan 31, 2014 #9


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    Why did you multiply the top equation by 2, and what similar process can you do to the equation ax+by = 1?
  11. Jan 31, 2014 #10
    I multipled the top eqaution by 2 so that -2y and 2y will cancel. So do I multiply the top equation by d? Sorry if I am just stupid I am trying to understand what you are saying D:
  12. Jan 31, 2014 #11


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    The original equations were

    Okay, in your example, you multiplied the first equation by 2 in order that the two coefficients of y would be the same. Yes, if you multiply the first equation by d and the second equation by b, the two equations will both have "bd" as the coefficient of y and subtracting the two equations will eliminate "y", leaving a single equation in x. Of course, multiplying the first equation by d will change the coefficient of x and the constant term, just as multiplying the second equation by b will change that equation.

    What do you get when you multiply the first equation by d and the second equation by b? What do you get when you subtract those two equations?
  13. Jan 31, 2014 #12


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    Actually you did it so that the x's would cancel in this latest example, but let's focus on the y term. Your second equation has cx+dy = 2. Your first equation is ax+by = 1. If I multiply the top equation by θ, I get θax+ θby = θ and cx+dy = 2. When I add the two equations together I get
    θax+cx+ θby+dy = θ+2.

    If I want the y terms to cancel, I need θby+dy = 0. What does θ have to be then?
  14. Jan 31, 2014 #13
    If I multiply the top by d and the bottom by b and change the signs I get:


    so now I see the -dby and dby cancel leaving -dax+bcx=-d2b is this right so far?

    what is this thing θ?
  15. Jan 31, 2014 #14

    Ray Vickson

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    Instead of doing row operations, I prefer to use direct elimination (explained below). Mathematically, they are completely equivalent, but some people find direct elimination easier to use and understand; others do not, so take your choice.

    From ##ax + by = 1##, we cannot have both ##a## and ##b## = 0, so at least one of them is non-zero: say ##a \neq 0##. That means we can divide by ##a## and so can solve for ##x## in terms of ##y##:
    [tex] ax + by = 1 \Longrightarrow ax = 1 - by \Longrightarrow x = \frac{1}{a} - \frac{b}{a} y[/tex] Substitute that value of ##x## into the second equation:
    [tex] cx + dy = 2 \Longrightarrow
    c\left(\frac{1}{a} - \frac{b}{a} y \right) + dy = 2 \\
    = \frac{c}{a} + \left(d - \frac{bc}{a} \right) y = \frac{c}{a} + \frac{(ad-bc)}{a}y[/tex]
    Therefore we have
    [tex] \frac{ad-bc}{a} y = 2 - \frac{c}{a}[/tex]
    Since we are assuming ##ad-bc \neq 0## we are allowed to divide through by it, so we can get a formula for ##y## in terms of ##a,b,c,d##. Then we can put that into the expression for ####x in terms of ##y##, to finally arrive at a formula for ##x##. You can see exactly where the assumption ##ad-bc \neq 0## comes in.
  16. Jan 31, 2014 #15
    This confuses me even more. I am just learning about linear equations today from my book. Can someone please just show me how to work this one out? Then I go begin to work on the other 19 questions :)
  17. Jan 31, 2014 #16
    Since we have the equation $$\frac{ad-bc}{a}y = 2 - \frac{c}{a}$$
    we can solve for y to get $$ y = (2-\frac{c}{a})(\frac{a}{ad-bc}) $$
    and from here we can simply use our y equation to find an equation for x in terms of a,b,c, and d by plugging
    $$ y = (2-\frac{c}{a})(\frac{a}{ad-bc}) $$ into either equation ax+by = 1 or cx+ dy = 2 to solve for x. We are going to leave our solutions in terms of a,b,c, and d since the question asks us to simply find formulas for the variables x and y in terms of the undetermined constants a,b,c, and d.

    ex. If someone says, "Dude find me an x and a y where 10x + 20y = 1 and where 90x + 15y = 2" you can say ok then $$ y = (2-\frac{90}{10})(\frac{10}{10*15-20*90}) = \frac{7}{165}$$
    and since ax+by = 1, then $$ x = \frac{1-by}{a} = \frac{1-b(2-\frac{c}{a})(\frac{a}{ad-bc})}{a} $$
    $$ = \frac{1-b(7/165)}{a} = \frac{1-20(7/165)}{10} = \frac{1}{66} $$
    Therefore, you would say $$ x = \frac{1}{66}$$ and $$ y = \frac{7}{165} $$ So if you are given ANY a,b,c, and a d you can find a y and an x where ax + by = 1 and cx + dy = 2 for any real numbers a,b,c, and d as long as ad-bc doesn't equal zero
    Last edited: Jan 31, 2014
  18. Jan 31, 2014 #17
    I am 12 years old I don't know what any of this stuff means. Can you just please help me in an easy way? Why must it be so complicated lol.

    I can solve ANY linear equation like:



    ect ect it is SO simple but when I don't know the values it really messes me up :(
  19. Jan 31, 2014 #18
    Yes, there is a slightly simpler way conceptually where we can use the system provided to initially have our function in terms of a,b,c, and d. Let us start with our a,b,c,and d as any number we want. I will choose a = 7, b = 21, c = 19 = d= 38. We can easily solve the system
    7x + 21y = 1
    19x + 38y = 2

    $$ \left( \begin{array}{ccc}
    7 & 21 & 1 \\
    19 & 38 & 2 \\ \end{array} \right) $$

    $$ \left( \begin{array}{ccc}
    7 & 21 & 1 \\
    0 & -19 & 5/7 \\ \end{array} \right) $$
    therefore, y = 5/133 and x = 4/133.

    $$ ax + by = 1$$
    $$ cx + dx = 2$$

    $$(ax+by)+(cx+dx) = 3$$
    $$ y(a+b) = 3-x(c+d) $$
    $$∴ y = \frac{3-x(c+d)}{a+b} $$
    Now since we have y in terms of a,b,c, and d, we need to solve for a single variable x or y, whichever is more convenient, and since we have values of x and y with respect to a,b,c, and d, we can easily check our answers.

    Since we have an equation for y in terms of a,b,c,d, and x, it would be more convenient for us to solve for x.
    $$ ax + by = 1$$
    then, $$ ax + b(\frac{3-x(c+d)}{a+b}) = 1$$
    $$ \frac{ax(b+d)+3b-(ab+bc)x}{b+d} = 1$$
    Which simplifies to
    $$\frac{x(ad-bc)+3b}{b+d} = 1$$
    $$ x(ad-bc)+3b = b+d $$
    $$ x(ad-bc) = d-2b $$
    $$∴ x = \frac{d-2b}{ad-bc} $$
    Now we can solve for y, and can plug in our arbitrary constants to check our work.
    Last edited: Jan 31, 2014
  20. Jan 31, 2014 #19


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    You were close on part of it. But if you add the two don't you get -dax+bcx=(-d)+2b?
  21. Jan 31, 2014 #20
    Yes I notice the mistake I made. So now I have: -dax+bcx=-d+2b so now I just rearrange the equation?

    ax+cx=b? Please I am really trying to do it! I can't do the other problems in my book until I figure out how to do these types... where you just have variables with no numbers.
  22. Jan 31, 2014 #21


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    You are getting there. Yes, just rearrange -dax+bcx=-d+2b. Factor out the x on the left side. Solve for x.
  23. Jan 31, 2014 #22
    Factoring out x gives me


    but I don't have any numbers to work with! It's like me saying that a + b = c there is no way to deduce what c is because we don't have any values just variables.

    So the final answer is x=b/(a+c) ? Or is the problem asking me for actual values?

    P.S Vance and Ray thank you for the help but you are really just making it more complicated for me. Sorry I don't understand that level of math. If you can help me that is great but please keep it simple thanks
  24. Feb 1, 2014 #23
    ax + by = 1
    cx + dy = 2

    c(ax+by=1) → acx + bcy = c
    a(cx + dy = 2) → acx + ady = 2a

    acx + bcy = c

    acx + ady = 2a

    (acx+bcy)-(acx+ady) = (c-2a)

    acx+bcy-acx-ady = c-2a (the acx's cancel out, now we are left with our y 's and a,b,c,d)

    bcy-ady = c-2a (factor out the y)

    y(bc-ad) = c-2a (divide by ad-bc)

    ∴y = (c-2a)/(bc-ad) = (2a-c)/(ad-bc) ← this is why you were provided the condition that ad-bc ≠ 0.

    The same can be done for x by multiplying by b and -d to cancel out the y's. A way to check your work is to make sure that you have y and x in terms a,b,c, and d. If you're missing one, you know you probably made a small error somewhere along the line. Note: This is simply an algebraic question asking you to find an x and a y in terms of a,b,c, and d. You don't need to provide any values, you simply need to have your answer in the form y = (c-2a)/(ad-bc) where there are a's,b's,c's, and d's on the right side and a single y on the left.
    Last edited: Feb 1, 2014
  25. Feb 1, 2014 #24


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    The answer should just be x is some combination of a,b,c and d. You aren't supposed to deduce what c is. Just x and y. But I don't know how you got x=b/(a+c) from -dax+bcx=-d+2b.There've been some algebra mistakes. Can you show what you did?
    Last edited: Feb 1, 2014
  26. Feb 1, 2014 #25
    Now this makes perfect sense :) thank you! Can you check the answer on this question for me?

    I took the time to learn latex so hopefully it is easier for you to read.


    Then I times the top by c and the bottom by a giving:

    ##(cax+cby)-(acx+ady)=3c+4a## I think this is right because before when the value was positive you changed the sign so I did the same and -- becomes +?

    ##cby=ady=3c+4a## and factor out the y ##y(cb-ad)=3c+4a##

    ##y=\frac{3c+4a}{cb-ad}## as my final answer?

    and likewise: ##x=\frac{3d+4b}{da-bc}## ?
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