# Need help solving linear equations with elimination and substiution method

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

• 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.) – layman May 17 '15 at 1:21
• @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 – ruler501 May 17 '15 at 1:24
• @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. – layman May 17 '15 at 1:25
• It's x = 3/2y = 1 – victoria May 17 '15 at 1:43
• I tried my best to edit it though. – victoria May 17 '15 at 1:43

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.

• i think coefficient matrix method is too advanced for OP – Taylor Ted May 17 '15 at 2:16
• 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. – user137731 May 17 '15 at 2:17
• That was a lot, but I got it now. Thank you! – victoria May 17 '15 at 2:51
• @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. – user137731 May 20 '15 at 23:40