How Do I Solve Linear Equations with Variables and Coefficients?

[I'm_Learning]

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

Homework Statement

Solve for x and y:

7x-y=2
2x+2y=5

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.

ax+by=1
cx+dy=2

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

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.

14x-2y=4
2x+2y=5
16x=9

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

Thanks.

 

[adjacent]

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.

Welcome to PF
Is that really $!$?
What is your education level?

 

[Office_Shredder]

Is that really $!$?

!= is a common symbol for not equal to

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.

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).

 

[I'm_Learning]

Welcome to PF
Is that really $!$?
What is your education level?

I am in year 8

!= 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).

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.

 

[Office_Shredder]

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?

 

[I'm_Learning]

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?

-1?

-ax-by=-1
cx+dy=2

-ax-b=-1
cx+d=2

I really have no idea. There are too many variables I'm getting confused.

 

[Tanya Sharma]

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 .

 

[I'm_Learning]

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 .

ok now this makes sense and I can easily do it.

a=2,b=3,c=3,d=5

2x+3y=1
4x+5y=2
times top equation by 2 and change the signs

-4x-6y=-2
4x+5y=2

-y=0
y=0

but I just don't get how to do it without numbers.

 

[Office_Shredder]

ok now this makes sense and I can easily do it.

a=2,b=3,c=3,d=5

2x+3y=1
4x+5y=2
times top equation by 2 and change the signs

-4x-6y=-2
4x+5y=2

-y=0
y=0

but I just don't get how to do it without numbers.

Why did you multiply the top equation by 2, and what similar process can you do to the equation ax+by = 1?

 

[I'm_Learning]

Why did you multiply the top equation by 2, and what similar process can you do to the equation ax+by = 1?

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:

 

[HallsofIvy]

The original equations were
ax+by=1
cx+dy=2

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?

 

[Office_Shredder]

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:

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?

 

[I'm_Learning]

The original equations were
ax+by=1
cx+dy=2

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?

If I multiply the top by d and the bottom by b and change the signs I get:

-dax-dby=-d
bcx+dby=2b

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

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?

what is this thing θ?

 

[Ray Vickson]

If I multiply the top by d and the bottom by b and change the signs I get:

-dax-dby=-d
bcx+dby=2b

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



what is this thing θ?

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 $y$:
$$ax + by = 1 \Longrightarrow ax = 1 - by \Longrightarrow x = \frac{1}{a} - \frac{b}{a} y$$ Substitute that value of $x$ into the second equation:
$$cx + dy = 2 \Longrightarrow c\left(\frac{1}{a} - \frac{b}{a} y \right) + dy = 2 \\ = \frac{c}{a} + \left(d - \frac{bc}{a} \right) y = \frac{c}{a} + \frac{(ad-bc)}{a}y$$
Therefore we have
$$\frac{ad-bc}{a} y = 2 - \frac{c}{a}$$
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.

 

[I'm_Learning]

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 $y$:
$$ax + by = 1 \Longrightarrow ax = 1 - by \Longrightarrow x = \frac{1}{a} - \frac{b}{a} y$$ Substitute that value of $x$ into the second equation:
$$cx + dy = 2 \Longrightarrow c\left(\frac{1}{a} - \frac{b}{a} y \right) + dy = 2 \\ = \frac{c}{a} + \left(d - \frac{bc}{a} \right) y = \frac{c}{a} + \frac{(ad-bc)}{a}y$$
Therefore we have
$$\frac{ad-bc}{a} y = 2 - \frac{c}{a}$$
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.

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 :)

 

[vanceEE]

This confuses me even more.

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

 

[I'm_Learning]

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

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:

-4x-7y=5
2x+y=6

3x+4y=-2
-2x-3y=1

ect ect it is SO simple but when I don't know the values it really messes me up :(

 

[vanceEE]

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.

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.

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

$$(ax+by)+(cx+dx) = 3$$
$$ y(a+b) = 3-x(c+d) $$
$$∴ y = \frac{3-x(c+d)}{a+b} $$
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.

 

[Dick]

If I multiply the top by d and the bottom by b and change the signs I get:

-dax-dby=-d
bcx+dby=2b

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

You were close on part of it. But if you add the two don't you get -dax+bcx=(-d)+2b?

 

[I'm_Learning]

You were close on part of it. But if you add the two don't you get -dax+bcx=(-d)+2b?

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.

 

[Dick]

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.

You are getting there. Yes, just rearrange -dax+bcx=-d+2b. Factor out the x on the left side. Solve for x.

 

[I'm_Learning]

You are getting there. Yes, just rearrange -dax+bcx=-d+2b. Factor out the x on the left side. Solve for x.

Factoring out x gives me

x(a+c)=b
x=b/(a+c)

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

 

[vanceEE]

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.

 

[Dick]

Factoring out x gives me

x(a+c)=b
x=b/(a+c)

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

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?

 

[I'm_Learning]

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.

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.

$ax+by=3$
$cx+dy=-4$

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

 

[Dick]

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.

$ax+by=3$
$cx+dy=-4$

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

Right. You've got it.

 

[vanceEE]

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

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

Yes, this is correct.

 

[Ray Vickson]

Factoring out x gives me

x(a+c)=b
x=b/(a+c)

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

That "level of math" is easier than the way you are trying to do it! What I showed you was the way it was taught when I was a bit older than you are now. It is called "Gaussian elimination" and has been in use for more than 300 years; it was devised by Isaac Newton in 1670 and was clarified by Gauss in about 1810. It was taught in the 1950s onward as "Gaussian elimination", but should probably been called "Newton's Method" (one of many Newton methods!).

However, as I said in my post, you can take your choice and use whichever method you find easiest.

BTW: there is a difference between methods that are easiest for people vs. easiest for computers. The method I showed you is easiest for people; the method you want to use is easiest for computers.

 

[vanceEE]

That "level of math" is easier than the way you are trying to do it! What I showed you was the way it was taught when I was a bit older than you are now. It is called "Gaussian elimination" and has been in use for more than 300 years; it was devised by Isaac Newton in 1670 and was clarified by Gauss in about 1810. It was taught in the 1950s onward as "Gaussian elimination", but should probably been called "Newton's Method" (one of many Newton methods!).

However, as I said in my post, you can take your choice and use whichever method you find easiest.

BTW: there is a difference between methods that are easiest for people vs. easiest for computers. The method I showed you is easiest for people; the method you want to use is easiest for computers.

Vick's right, you are performing matrix operations in a sense. But whatever's easiest AND fastest for you is what matters. I myself find it less time consuming to do direct substitution. I find matrices to be more concise when dealing with a large amount of equations. Glad you understand these problems now!

 

[I'm_Learning]

That "level of math" is easier than the way you are trying to do it! What I showed you was the way it was taught when I was a bit older than you are now. It is called "Gaussian elimination" and has been in use for more than 300 years; it was devised by Isaac Newton in 1670 and was clarified by Gauss in about 1810. It was taught in the 1950s onward as "Gaussian elimination", but should probably been called "Newton's Method" (one of many Newton methods!).

However, as I said in my post, you can take your choice and use whichever method you find easiest.

BTW: there is a difference between methods that are easiest for people vs. easiest for computers. The method I showed you is easiest for people; the method you want to use is easiest for computers.

ok maybe i'll try and learn that method. It just seems so much longer and more confusing than 3 or 4 simple manipulations like in the method I did. Anyway thanks for helping :)

 

[I'm_Learning]

I wanted to just add an extra post to this. Sorry for the double post but I don't think it is worth making another thread.

Solve the following system of linear equations. Assume the same rules apply as the previous questions.

$ax+by=u$
$cx+dy=v$

I got the following solution of: $y=\frac{uc-va}{cb-ad}$ and $x=\frac{du-bv}{da-bc}$ and I'm told to verify that the answers I get are actually valid solutions. How would I go about this? Thanks.

 

[Dick]

I wanted to just add an extra post to this. Sorry for the double post but I don't think it is worth making another thread.

Solve the following system of linear equations. Assume the same rules apply as the previous questions.

$ax+by=u$
$cx+dy=v$

I got the following solution of: $y=\frac{uc-va}{cb-ad}$ and $x=\frac{du-bv}{da-bc}$ and I'm told to verify that the answers I get are actually valid solutions. How would I go about this? Thanks.

Substitute your solution for $x$ and $y$ into $ax+by$ and make sure the result is $u$. Same thing for the other equation.

 

[vanceEE]

I wanted to just add an extra post to this. Sorry for the double post but I don't think it is worth making another thread.

Solve the following system of linear equations. Assume the same rules apply as the previous questions.

$ax+by=u$
$cx+dy=v$

I got the following solution of: $y=\frac{uc-va}{cb-ad}$ and $x=\frac{du-bv}{da-bc}$ and I'm told to verify that the answers I get are actually valid solutions. How would I go about this? Thanks.

Also, since a,b,c,d,u, and v are arbitrary constants, you can choose any real numbers for a,b,c,d, u, and v (as long as they satisfy any initial conditions) to prove that $y=\frac{uc-va}{cb-ad}$ and $x=\frac{du-bv}{da-bc}$ are solutions that satisfy the system of equations.
you would say:
"Let a = 4, b = 3, c = 5, d = 8, u = 2, and v = 9 where bc-ad ≠ 0
Then $$4x+3y = 2$$
$$5x+8y = 9$$
By our solution: $$y = \frac{(2*5)-(9*4)}{(5*3)-(4*8)} = \frac{26}{17}$$
$$ x=\frac{(8*2)-(3*9)}{(8*4)-(3*5)} = \frac{-11}{17} $$
$$ 4(\frac{-11}{17})+3(\frac{26}{17}) = 3 $$
$$ 5(\frac{-11}{17})+8(\frac{26}{17}) = 9$$
Therefore, $y=\frac{uc-va}{cb-ad}$ and $x=\frac{du-bv}{da-bc}$ are solutions that satisfy the system of equations for any real numbers a,b,c,d,u, and v where bc-ad ≠ 0."

 

[Ray Vickson]

ok maybe i'll try and learn that method. It just seems so much longer and more confusing than 3 or 4 simple manipulations like in the method I did. Anyway thanks for helping :)

It is a bit longer, but MUCH less confusing. Look how long it took you to finally get your desired method right. The longer method takes more steps, but each step is straightforward and hardly requires any thinking at all. I will go through the method once more, using vanCEE's example. The steps rely on one simple property: if you do the same thing to both sides of an equation, you get an equation again.
$$4x + 3y = 2\;\;\; (1)\\ 5x + 8y = 9 \;\; \;(2)$$
Isolate $x$ in eq.(1); that is, re-write the equation so that only $x$ terms are on the left. Do this by subtracting the term $3y$ from both sides:
$$4x + 3y - 3y = 2 - 3y, \text{ or}\\ 4x = 2 - 3y \;\;\;\;\;\;\;(3)$$
Turn (3) into an equation of the form $x = \ldots$ by dividing both sides of (3) by 4:
$$4x/4 = (2 - 3y)/4, \text{ or}\\ x =\frac{2 - 3y}{4} = \frac{2}{4}-\frac{3}{4} y = \frac{1}{2} - \frac{3}{4} y\;\;\;(4)$$
Now, wherever you see $x$ in equation (2), put in the above value instead:
$$9 = 8y + 5x = 8y + 5\left(\frac{1}{2} - \frac{3}{4} y \right) = \frac{5}{2} + \left( 8 - 5 \frac{3}{4} \right) y = \frac{5}{2} +\frac{4 \cdot 8 - 5 \cdot 3}{4} y = \frac{5}{2} + \frac{17}{4} y$$
In other words, we have
$$\frac{5}{2} + \frac{17}{4} y = 9, \text{ or}\\ \frac{17}{4} y = 9 - \frac{5}{2} = \frac{18 - 5}{2} = \frac{13}{2}$$
Thus
$$y =\frac{13}{2} \cdot \frac{4}{17} = \frac{26}{17}$$
Substitute this value of $y$ into (4), to get $x$:
$$x = \frac{1}{2} - \frac{3}{4}\cdot \frac{16}{17} = -\frac{11}{17}$$

 

[I'm_Learning]

...

That is definitely going to take some getting used to. My book is called Basic Mathematics by Lang and he teaches the shorter way. I'm on to solving linear equations using 3 variables and instead of having 2 equations, we now have 3. I can solve them using my method but how does the Gaussian method handle 3 equations? While it may be simpler I find it faster to use my method and I can easily deal with 3 equations without much work at all.