How Do I Solve Linear Equations with Variables and Coefficients?

[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)\\<br /> 5x + 8y = 9 \;\; \;(2)<br />
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}\\<br /> 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}\\<br /> x =\frac{2 - 3y}{4} = <br /> \frac{2}{4}-\frac{3}{4} y = \frac{1}{2} - \frac{3}{4} y\;\;\;(4)<br />
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)<br /> = \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}\\<br /> \frac{17}{4} y = 9 - \frac{5}{2} = \frac{18 - 5}{2} = \frac{13}{2}
Thus
<br /> 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.

 

[Ray Vickson]

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.

The Gaussian method handles 1000 equations in 1000 variables in just the same way as 2 equations in 2 variables. You just use the equations to eliminate variables one-by-one.

I have already pointed out to you that the Gaussian elimination method is equivalent to yours. It just arranges things in a slightly different way.

 

[vanceEE]

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.

You would handle them the same way; by eliminating variables by substitution. I suggest that you take the time to practice a few problems using this method. Being familiar with using direct substitution is essential for dealing with linear equations and any other higher maths you may encounter in the future.