# Linear Systems with Multiplication

## Key Questions

I also need to assume that, like me, you count discovering that a system has no solutions (is inconsistent) counts as "solving the system"
Third, I assume that the number of variables equals the number of equations.
Finally, I assume that the "multiplication method" is the same as the method I was taught to call the "addition/subtraction method".

With these four assumptions, the answer is,

Yes. Any linear system of $n$ equations in $n$ variables can be solved by this method:

$a x + b y = c$
$\mathrm{dx} + e y = f$

Mutiply the first equation by $d$ and the second by $- a$ to get:

$a \mathrm{dx} + b \mathrm{dy} = c d$
$- a \mathrm{dx} - a e y = - a f$

$b \mathrm{dy} - a e y = c d - a f$

$\left(b d - a e\right) y = c d - a f$

So, as long as we don't have $0$ on the left and non-zero on tlhe right (which would indicate an inconsistent system), and we also don't have $0 = 0$ (infinitely many solutions) we get:

$y = \frac{c d - a f}{b d - a e}$

Using the number we get for $y$, we can go back and find $x$.

• One method is L division

LCM can be found using this technique where you divide the numbers using common divisors. For e.g If you need LCM for 2,3,4,5 then you can get that by
$2 \lfloor 2 , 3 , 4 , 5$

$\lfloor 1 , 3 , 2 , 5$
As all of the reminders are prime numbers you multiply them along with the number you used in division. i.e. $L C M = 2 \cdot 1 \cdot 3 \cdot 2 \cdot 5$.
Which is $L C M = 60$. Similarly you can do the same for any number. One of the other method is to multiply all the numbers and divide by GCD. But finding GCD also uses the above method.

• There are more than one way to solve the system of equations

The most utilised methods are elimination and substitution methods. I prefer substitution than elimination.

Other methods like Cramer's rule and other matrix methods such as Gauss elimination, Gauss - Jacobi are available. These are pretty advanced and can solve any number of linear equations.

A comparison of substitution and elimination methods is given below.

Example

$6 x + 4 y = 2$---------->Eqn 1
$x - 2 y = 3$---------->Eqn 2

Elimination method
Multiply Eqn 2 by '2' an add with Eqn 1.

$6 x + 4 y = 2$
$2 x - 4 y = 6$
______+
$8 x = 8$
$x = 1$

Substitute in one of the equations. Using Eqn 1 we have

$6 \cdot 1 + 4 y = 2$
$4 y = 2 - 6$
$y = - 1$

Hence the solution is $x = 1 , y = - 1$

Substitution Method
From Eqn 2 we have

$x = 3 + 2 y$ --> Eqn 3

Substitute in Eqn 1
$6 \cdot \left(3 + 2 y\right) + 4 y = 2$
$18 + 12 y + 4 y = 2$
$16 y = 2 - 18$
$16 y = - 16$
$y = - 1$
Use in eqn 3
$x = 3 + 2. - 1$
$x = 1$
So we get $x = 1 , y = - 1$.

• There are certain steps to follow before finding the solution.

Let's say we have two equations: $3 x - y = 15$ and
$6 x + 3 y = 20$.

The first thing you have to do is determine which of the variables that, when added, can be canceled to just get one term. In this case, we don't have any variables to be canceled out, so we need to first multiply one system. Let's begin with the first one, where the whole system is multiplied by -2

$3 x - y = 15$
$- 2 \left(3 x - y\right) = - 2 \left(15\right)$
$- 6 x + 2 y = - 30$

Notice that with $- 6 x$, we can add this part to the second equation that has $6 x$ to to cancel the $x$-terms from each other:

$- 6 x + 2 y = - 30$
$6 x + 3 y = 20$
. . . . . . . . . . . . . . . . . . .
$5 y = - 10$

By doing so, we get a sum of equations in terms of $y$ only. Thus, we can solve for $y$:

$y = - 2$

Then you would substitute the $y$-value into one of the equations and solve of $x$. Since the $y$-value is solved, we can use the original equations; for that I will use $3 x - y = 15$ instead of $- 6 x + 2 y = - 30$.

Let's do both equations and see if we get the same answer:

$3 x - y = 15$
$3 x + 2 = 15$
$3 x = 13$
$x = \frac{13}{3}$

$6 x + 3 y = 20$
$6 x + 3 \left(- 2\right) = 20$
$6 x = 26$
$x = \frac{26}{6} \to x = \frac{13}{3}$ when simplified.

Huzzah! We got the same x-value! Therefore,

x=13/3; y =-2

In most of these problems, there are other ways that can be multiplied easily to get the answer. For the system of equations above, we could also solve for $x$ first by multiplying the first equation by 3, which will cancel the $y$'s by the $3 y$ in the second equation.

Sometimes there will be cases where one of the variables does not equal the other (if you follow the steps carefully). This means that the system of equations cannot equal each other, thus there is no solution.

Be careful when doing some fractional equations; I would first multiply each one by the most common denominator to get whole integers, thus allowing the equations to be easier to use elimination. For more information on how to do this, see this link:

http://www.regentsprep.org/regents/math/algebra/AV5/Fequations.htm

And finally, here's another example:

Hopefully all of these help and good luck with algebra!