2
$\begingroup$

I need help solving system of equations. I never worked with these equations before, so I'm not sure how this works. If anyone could show me the steps so I could go from there, that would be great. Anyways, I need to use substitution method to solve x - 3/2y = 1 2x -7y = 10 Also, I need to use addition method for x - 2y =2 5x + 2y = 22

Thank you in advance,

$\endgroup$
  • 1
    $\begingroup$ Please edit your question and add parentheses where they are needed in your equations so that the problem is clear for us :). For example, you wrote x - 3/2y = 1. Do you mean $\dfrac{x- 3}{2y} =1$? If so, you should write it as (x - 3)/2y =1. Did you mean $x - \dfrac{3}{2y} = 1$? If so, you should write it as x - (3/2y) = 1. (I think that's the only equation I was having problems interpreting.) $\endgroup$ – layman May 17 '15 at 1:21
  • 1
    $\begingroup$ @user46944 my guess is he meant $x- \frac{3}{2}y = 1$ or x - (3/2)y = 1 since it seems to be system of linear equations $\endgroup$ – ruler501 May 17 '15 at 1:24
  • 1
    $\begingroup$ @ruler501 Wow, good point. I think your guess is probably right, but this should be a lesson to the OP about parentheses -- there are three different ways to interpret the equation when it is given without parentheses. $\endgroup$ – layman May 17 '15 at 1:25
  • $\begingroup$ It's x = 3/2y = 1 $\endgroup$ – victoria May 17 '15 at 1:43
  • $\begingroup$ I tried my best to edit it though. $\endgroup$ – victoria May 17 '15 at 1:43
2
$\begingroup$

Instead of solving your homework for you, I'll show you how each of the methods work with a slightly harder system of linear equations.

Here's the system of linear equations that'll I'll be working on:

$$\begin{cases} x-3y+\frac 43z = 2 \\ 2x- z =1 \\ 3y-4z=0\end{cases}$$

What does solving this system of linear equations mean? If means that we find an $x$, a $y$, and a $z$ that works in all $3$ equations.

So let's try a couple of different approaches:


Substitution Method:

For this method we want to solve for on variable in terms of other variables and then plug that into a different equation. This can help us reduce the number of variables in a given equation. Let's see how it works:

I notice that the second and third equations each only have $2$ variables so let's start with one of them. I'll start with the second equation and I'll isolate one variable on one side and everything else on the other side:

$$2x-z=1 \implies z=2x-1$$

Now that we have this, we can plug it in to the other two equations. Then we have the new -- and equivalent system:

$$\begin{cases} x-3y+\frac 43z = x-3y+\frac 43(2x-1) = \frac {11}3x -3y -\frac 43 = 2 \\ z=2x-1 \\ 3y-4z= 3y-4(2x-1) = -8x+3y+4=0 \end{cases} \\ \iff \begin{cases} \frac {11}3x -3y = \frac {10}3 \\ z=2x-1 \\ -8x + 3y = -4\end{cases}$$

Notice now that the first and third equations have only two variables -- and importantly they're the same two variables. So let's solve for one variable in one of them and then plug that expression into the other one:

$$-8x + 3y = -4 \implies x = \frac {3y+4}{8}$$

Now let's plug this expression into the first equation to get a new set of equations:

$$\begin{cases} \frac {11}3x -3y = \frac {11}3(\frac {3y+4}{8}) - 3y = \frac {11}8 y + \frac {11}6 -3y=\frac {10}3 \\ z=2x-1 \\ -8x + 3y = -4\end{cases} \\ \iff \begin{cases} y=-\frac {12}{13} \\ z=2x-1 \\ -8x + 3y = -4\end{cases}$$

Now that we've solved for one, we're just about done. Plug $y=-\frac {12}{13}$ into the equation $-8x + 3y = -4$ to get $x=\frac 2{13}$, then plug that into $z=2x-1$ to get $z=-\frac 9{13}$. And we're done!


Elimination Method:

For this method, we'll be adding/ subtracting multiples of one equation from another to get rid of variables. Let's see how it works:

I notice that if I added the third equation to the first, the $y$ term would cancel out of each. So let's do that:

$$x-3y+\frac 43z =2 \\ \underline{+\ \ \ \ \ 3y-4z = 0} \\ \ \ \ \ \ \ \ x-\frac 83z = 2$$

If we replace our first equation with this new one, we get a new set of equations:

$$\begin{cases} x-\frac 83z = 2 \\ 2x- z =1 \\ 3y-4z=0\end{cases}$$

Now notice that if we subtracted $2$ times the first equation from the second we could get rid of the $x$'s. So let's do that:

$$ 2x- z =1\\ \underline{-\ \ \ (2x-\frac {16}3z =4)} \\ \frac {13}3z =-3$$

Solving that we get $z = -\frac 9{13}$. Plugging that into the first equation then gets us $x$ and into the third equation gets us $y$. Done.


Coefficient Matrix Method:

Notice that we really only need $x$, $y$, and $z$ as placeholders in our calculations until the very end -- all of the math ends up just affecting the coefficients. So let's rewrite our system without those $x$'s, $y$'s, and $z$'s getting in the way. We'll write it as a matrix:

$$\left[\begin{array}{ccc|c} 1 & -3 & \frac 43 & 2 \\ 2 & 0 & -1 & 1 \\ 0 & 3 & -4 & 0 \end{array}\right]$$

Make sure you understand how I built this matrix.

Now remember that this is just a more convenient way of writing a system of linear equations. Also remember that there are a few things we can do to a system of equations without changing the solutions. We can

  1. Multiply both sides of an equation by a nonzero constant.
  2. Rearrange the order of the equations.
  3. Add a multiple of one equation to another.

How do we perform these three operations on our matrices? Each of these operations is equivalent to the following operations on our matrix:

  1. Multiply any row by a nonzero constant.
  2. Reorder the rows of our matrix.
  3. Add a multiple of a row to another row.

Now that we have these operations, what should our goal be? It'd be nice if we could get the matrix into the form

$$\left[\begin{array}{ccc|c} 1 & 0 & 0 & a \\ 0 & 1 & 0 & b \\ 0 & 0 & 1 & c \end{array}\right]$$

because then if we converted this matrix back into equation form we'd just have

$$\begin{cases} x + 0y + 0z = x = a \\ 0x + y + 0z = y = b \\ 0x + 0y + z = c\end{cases}$$

and then we could just read off our answers. So now that we have our operations and we have a goal, let's see if we can achieve that goal:

$$\left[\begin{array}{ccc|c} 1 & -3 & \frac 43 & 2 \\ 2 & 0 & -1 & 1 \\ 0 & 3 & -4 & 0 \end{array}\right]_{R_2 \to R_2 - 2\cdot R_1} \\ \sim \left[\begin{array}{ccc|c} 1 & -3 & \frac 43 & 2 \\ 0 & 6 & -\frac {11}3 & -3 \\ 0 & 3 & -4 & 0 \end{array}\right]_{R_2 \to \frac 16R_2} \\ \sim \left[\begin{array}{ccc|c} 1 & -3 & \frac 43 & 2 \\ 0 & 1 & -\frac {11}{18} & -\frac 12 \\ 0 & 3 & -4 & 0 \end{array}\right]_{R_3 \to R_3-3\cdot R_2} \\ \sim \left[\begin{array}{ccc|c} 1 & -3 & \frac 43 & 2 \\ 0 & 1 & -\frac {11}{18} & -\frac 12 \\ 0 & 0 & -\frac {13}6 & \frac 32 \end{array}\right]_{R_3 \to -\frac 6{13}\cdot R_3} \\ \sim \left[\begin{array}{ccc|c} 1 & -3 & \frac 43 & 2 \\ 0 & 1 & -\frac {11}{18} & -\frac 12 \\ 0 & 0 & 1 & -\frac 9{13} \end{array}\right]_{R_2 \to R_2 + \frac {11}{18}\cdot R_3} \\ \sim \left[\begin{array}{ccc|c} 1 & -3 & \frac 43 & 2 \\ 0 & 1 & 0 & -\frac {12}{13} \\ 0 & 0 & 1 & -\frac 9{13} \end{array}\right]_{R_1 \to R_1 +3\cdot R_2 -\frac 43\cdot R_3} \\ \sim \left[\begin{array}{ccc|c} 1 & 0 & 0 & \frac 2{13} \\ 0 & 1 & 0 & -\frac {12}{13} \\ 0 & 0 & 1 & -\frac 9{13} \end{array}\right]$$

Which gives us the answer $x=\frac 2{13}$, $y=-\frac {12}{13}$, $z=-\frac 9{13}$ -- exactly what we found above.

$\endgroup$
  • $\begingroup$ i think coefficient matrix method is too advanced for OP $\endgroup$ – Taylor Ted May 17 '15 at 2:16
  • $\begingroup$ I just wanted to add it for fun. OP didn't ask for it and doesn't have to use it, but given a little thought (s)he should be able to figure out how it works. $\endgroup$ – user137731 May 17 '15 at 2:17
  • $\begingroup$ That was a lot, but I got it now. Thank you! $\endgroup$ – victoria May 17 '15 at 2:51
  • $\begingroup$ @victoria If I answered your question, you can accept it by clicking on the green arrow left of my answer. That way others will know it has already been answered. $\endgroup$ – user137731 May 20 '15 at 23:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.