# Need help applying substition/addition method to linear equations

I'm having a very difficult time solving these equations. For some odd reason I keep getting a graph answer. For example, when I try to solve $3x + 2y = 14$; it keeps telling me the answer is, $m=−32$, $b=7$ where $m$ is the slope and $b$ is the $y$-intercept. Which doesn't make sense to me since I have to apply the addition method which I did.

If anyone could help me figure out what I did wrong, so I could go from there, that would be great.

I need to Apply Substitution Method For: $$x + 2y = 4$$ $$2x - y = – 7$$

And I also need to Apply Addition (Elimination) Method For: $$3x + 2y = 14$$ $$5x – 2y = 18$$

• What are you referring to when you say "it" keeps telling you the answer. – Gregory Grant May 9 '15 at 15:25
• The work I'm doing is giving me a plot answer, and I'm not sure why. – Larry May 9 '15 at 15:26
• You don't need a computer to solve this, these are very simple systems. Solve the first equation for $x$, so that $x=4-2y$. Then substitute that value for $x$ into the second equation $2(4-2y)-y=-7$. Then solve that for $y$ and then substitute that value of $y$ into either equation to find $x$. – Gregory Grant May 9 '15 at 15:27
• For the second system you add the two equations, thereby cancelling out the term with $y$ in it, you get $8x=32$. That's easy then to solve for $x$, and you substitute back to get $y$. – Gregory Grant May 9 '15 at 15:29
• Could you explain who(m)/what is giving a plot answer ? Whatever it is, I suspect you asked for in some manner. Thanks. – Claude Leibovici May 9 '15 at 15:36

$$\begin{matrix} x+2y=4 & \quad &2x-y=-7 \\ x=4-2y & \quad & \\ & \quad & 2(4-2y)-y=-7 \\ & \quad & 8-4y-y=-7 \\ & \quad & 8-5y=-7 \\ & \quad & -5y=-15\\ & \quad & y=3\\ x=4-2(3) & \quad & \\ x=-2 & \quad & \\ \end{matrix}$$
$$\begin{matrix} \text{Given (a):} & 3x+2y=14\\ \text{Given (b):} & 5x-2y=18\\ \text{Add (a) and (b):} & 8x=32\\ \text{Solve:} & x=4\\ \text{Multiply (a) by 5 getting (c):} & 15x+10y=70\\ \text{Multiply (b) by -3 getting (d):} & -15x+6y=-54\\ \text{Add (c) and (d):} & 16y=16\\ \text{Solve:} & y=1 \end{matrix}$$
That last is actually easier to solve for $y$ using substitution after finding $x$, but I thought I would show you how to do the entire thing by elimination.