Alissa Fong

**MA, Stanford University**

Teaching in the San Francisco Bay Area

Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts

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To solve a system of equations by graphing when the lines are not in slope-intercept (y=mx+b) form, start by rewriting the equations in slope-intercept form. Do this by solving the equations for y, meaning use inverse operations so that y is on one side of the equal sign and everything else (the x and the constant) is on the other side. Once you have the equations in slope-intercept form, graph the lines by first plotting the y-intercepts, then using the slope to find a second point. After graphing the two lines, identify the point that the lines intersect. This point is the solution to the system of equations. Check you answer by plugging in the point to both equations. If the equations are true, then the point is the correct solution.

Here I’m given 2 equations, and asked to find where they cross which means find the solution by graphing them and I’m kind of bummed out because these lines are not in y equals mx plus b form, so that means before I do any graphing, I’m going to have to do some solving for y. Let’s choose to designate each different equation with a colour. That will help me to keep my work organized and remember what I’m doing in the first place.

Okay, so with this red equation, I want to get y all by itself. First thing I’m going to do is add x to both sides, and then I’m going to go through and divide both sides by 2. So now I have y equals 1/2x plus 2. I’ll graph it in just a second.

My other equation I’m going to write in blue so again I can keep my work straight x plus y equals 5, when we solve for y by subtracting x from both sides, we’ll see y equals -x plus 5. Now these are the two equations that are the same as those guys, only written in different form and I did that because these are easier to graph. Okay my first dot is going to go at the y intercept. From there I’ll count the slope. Let’s go over to the graph. My first dot will go at the y intercept of 2, and from there I’ll count the slope of 1/2, up 1 over 2, up 1 over 2. I’m going to go in both directions and then use a ruler to show my line.

All right my next point, not my next point. My next line I’m going to do in blue and I want to show my y intercept of +5 from there my slope is -1, so down 1 right 1, down 1 right 1. Starting at +5, I’m going to go down 1, right 1, down 1 okay like that. Use my ruler to connect those points and I’m looking for where these two lines intersect or cross because that’s going to be the solution to my system of equations. Okay there it is right there, the point where those two lines cross is at 2 for my x value, 2, 3 for my y value.

I’m going to have to go back and double check using Algebra that that’s the correct solution, but I think that’s going to be pretty right because I did a pretty goo graph using a ruler and stuff. So let’s go check I’m going to erase this stuff and what I need to do to verify is make sure that the point that I think is the solution (2,3) does indeed work in both equations.

Let’s try it in the red one. 2 times my y number take away my x number needs to be equal to 4. So let’s see 2 times 3, that’s my y value, take away my x number needs to be equal to 4, oops 6 take away 2 is 4, good okay. So it works in the first one. I’m almost done checking. I’m not done yet. Checking means you have to look at both original equations.

My x value plus my y value should hopefully equal to 5 and good 2 plus 3 does equal to 5, that means I got the right point. If you got that your point only worked in one equation and maybe didn’t work in either one that would mean that you didn’t graph precisely and you need to do the whole problem again.

So when you see a problem that says solve by graphing, don’t play around. Get out that ruler, get out that graph paper and a pencil because you might make errors. You have to do it precisely the first time around.