Systems of equations with graphing: 5x+3y=7 & 3x-2y=8

Solve the system of linear equations by graphing, and they give us two equations here. 5x plus 3y is equal to 7, and 3x minus 2y is equal to 8. When they say, "Solve the system of linear equations," they're really just saying find an x and a y that satisfies both of these equations. And when they say to do it by graphing, we're essentially going to graph this first equation. Remember, the graph is really just depicting all of the x's and y's that satisfy this first equation, and then we graph the second equation that's depicting all of the x's and y's that satisfy that one. So if we were looking for an x and a y that satisfies both, that point needs to be on both equations or it has to be on both graphs. So it'll be the intersection of the two graphs. So let's try to see if we can do that. So let's focus on this first equation, and I want to graph it. So I have 5x plus 3y is equal to 7. There's a couple of ways we could graph this. We could put this in slope-intercept form, or we could just pick some points here. You just really need two points to graph a line. So let me just set some points over here. Let's say x and y. When x is equal to 0, what does y need to be equal to? So when x is equal to 0, we have-- let me do it over here-- we have 5 times 0, plus 3 times y, is equal to 7. That's just 0 over there. So you have 3y is equal to 7. Divide both sides by 3, you get y is equal to 7/3, which is the same thing as 2 and 1/3 if we want to write it as a mixed number. Now let's set y equal to 0. So if we set y as equal to 0, we get 5x, plus 3 times 0, is equal to 7. Or in this part right over here, it just becomes 0. So we have 5x is equal to 7. Divide both sides by 5, and we get x equals 7/5, which is the same thing as 1 and 2/5. So let's graph both of these points, and then we should be able to graph this line, or at least a pretty good approximation of that line. So we have the point, 0, 2 and 1/3. So that's that point right over there. So I'll call it 0, 7/3 right over there, and then we have the point, 7/5, 0, or 1 and 2/5, 0. So 1 and 2/5. 2/5 is a little less than a half. So 1 and 2/5, 0. So our line is going to look something like this. I just have to connect the dots. It's always hard to draw the straight line. I'll draw it as a dotted line. So it would look something like this. Normally, when you have to solve a system of equations by graphing, they normally give you a little bit cleaner numbers, but we'll try our best and see if we can see where these two lines intersect. So now, let's worry about the second one right over there. So we have 3x minus 2y is equal to 8. So I'll do the same thing. So 3x minus 2y is equal to 8. We'll just look at the x- and the y-intercepts. So first, the y-intercept. When x is equal to 0, this whole thing boils down to 3 times 0 minus 2y is equal to 8. That's just 0. So you have negative 2y is equal to 8. Divide both sides by negative 2, we get y is equal to negative 4. So the y-intercept is 0, negative 4. Right over here, and we mark it 0, negative 4. And then, let's set y equal to 0. So when y is equal to 0, this term right over here just becomes 0. So we get 3x minus 2 times 0, so that's just 0. So 3x is equal to 8. Divide both sides by 3, you get x is equal to 8/3. And 8/3 is the same thing as 2 and 2/3. So it puts it right about there. That's the point, 0, 8/3. Now let me try my best to graph it. Connect these two dots. So let me do my very best. I drew a dotted line there. It goes something like that. And just eyeballing it, it looks like these two lines intersect right over there. I'm hoping that this will give us a clean answer. And this is the point 2, negative 1. So that is the point, 2, negative 1. Right? The x value here is 2. y value is negative 1. Now, that's what we got just by eyeballing it, and clearly these are hand-drawn graphs. Not very precise. Let's verify or let's see if 2, negative 1 does satisfy both of these equations, if it's an x and y value that satisfies both and lies on both the graphs. So if you put 2, negative 1 in this first equation, you get 5 times 2, plus 3 times negative 1, and we're going to see if that is equal to 7. So this is 10 plus negative 3. So that's the same thing as 10 minus 3. Does that equal 7? And yeah, it does. 10 minus 3 is equal to 7. So 2, negative 1 is definitely on that graph or definitely satisfies that equation. And then, let's do it with the other one. So if we do 2, negative 1, you have 3 times 2, minus 2 times negative 1. And we're testing to see if that is equal to 8. So 3 times 2. 3 times 2 is 6. And then 2 times negative 1 is negative 2. But then, we're subtracting that, so it's 6 plus 2. 6 plus 2 is equal to 8. And it definitely does equal 8. So we have the coordinate or we have the point, 2, negative 1 satisfying both equations. So we've solved the system of linear equations by graphing.