I’m prepping for a test and practicing solving simultaneous equations by graph, and I keep getting tripped up when the slopes are fractions and the lines don’t cross at a neat grid point.
The system I’m working on is:
– y = (2/5)x – 1
– y = (-3/2)x + 5
Here’s what I tried:
– I started with the y-intercepts: (0, -1) for the first line and (0, 5) for the second.
– For y = (2/5)x – 1, I used the slope “rise 2, run 5” to go from (0, -1) to (5, 1). I wasn’t sure if it’s also okay to go “down 2, left 5” to get another point like (-5, -3), or if I should always move to the right.
– For y = (-3/2)x + 5, from (0, 5) I went down 3 and right 2 to (2, 2). I drew the lines with my very wobbly ruler (apparently my straightedge is allergic to being straight), and they seem to cross a little to the right of x = 3 and a little above y = 0. But I can’t tell exactly where.
I’m confused about two things:
1) Is my method for plotting from the fractional slopes actually correct?
2) When the intersection isn’t on a grid point, how do you read it cleanly for a test? Do you just estimate from the graph, or is there a sensible way to check that your estimate fits both equations without fully switching to algebra?
Bonus: Is two points per line enough here, or should I plot a third point to reduce my “shaky line” error?
Any help appreciated!
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3 Responses
Great instincts! Your slope moves are spot on: for y = (2/5)x − 1 you can go up 2 and right 5, or down 2 and left 5-both keep the ratio 2/5; for y = (−3/2)x + 5 you can go down 3 and right 2, or up 3 and left 2. To tame a wobbly straightedge, plot multiple lattice points using multiples of the slope steps: first line hits (0, −1), (5, 1), (10, 3); second line hits (0, 5), (2, 2), (4, −1). The exact intersection is where (2/5)x − 1 = (−3/2)x + 5, which gives x = 60/19 ≈ 3.16 and y = 5/19 ≈ 0.26-exactly “a bit right of 3 and a hair above 0.” When the crossing isn’t on a grid point and the task is “by graph,” estimate to the nearest tenth by drawing long lines through widely separated points; then sanity-check without “going full algebra” by plugging your guessed x into both y = (2/5)x − 1 and y = (−3/2)x + 5 and seeing if the y-values nearly match (small difference means your estimate is solid). Two points are enough in theory, but a third point (or using larger step multiples) is an excellent anti-wobble check and helps you draw a cleaner, more reliable line. If you want a refresher on both plotting from slope-intercept form and solving systems by graphing, this Khan Academy explainer is great: https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:systems-of-equations/solving-systems-graphically/v/solving-systems-graphically. Want to try another system together, or see a quick elimination walkthrough for how 60/19 and 5/19 pop out so neatly?
Yes-if the slope is a/b, from any plotted point you can go right b and up a or left b and down a; two points determine a line, but a third point can steady a wobbly line. For a non-grid intersection, estimate and check by plugging into both equations, or get it exactly by setting (2/5)x − 1 = (−3/2)x + 5 ⇒ (19/10)x = 6 ⇒ x = 60/19 and y = 5/19 ≈ (3.16, 0.26), which matches your sketch.
Your plotting method is correct. A slope of 2/5 means y increases by 2 when x increases by 5, and you can also go the opposite way: left 5 and down 2. Likewise, for −3/2 you can go right 2 and down 3, or left 2 and up 3. To reduce wobble, use larger, integer-friendly steps so your points are far apart: for y = (2/5)x − 1, points like (0, −1), (5, 1), (10, 3) work well; for y = (−3/2)x + 5, points like (0, 5), (2, 2), (4, −1). Two points determine a line, but a third point is a good check and helps you draw a straighter line.
When the intersection isn’t on a grid point, you read an estimate from the graph, then quickly check it by plugging the x-value into both equations and seeing if the y-values agree. For example, at x = 3 you get y1 = (2/5)·3 − 1 = 0.2 and y2 = (−3/2)·3 + 5 = 0.5, so the second line is above the first. At x = 3.2 you get y1 = 0.28 and y2 = 0.2, so now the first is above the second. The crossing is between 3.0 and 3.2; try x ≈ 3.16: y1 ≈ 0.264 and y2 ≈ 0.26, essentially matching. That fits your sketch: a bit right of x = 3 and a little above y = 0.
For completeness, solving exactly gives the intersection at x = 60/19 ≈ 3.158 and y = 5/19 ≈ 0.263. You won’t read those fractions off a graph, so a decimal to two places is usually fine on a “by graphing” question. Using well-spaced integer points and, if needed, a third point will keep your line straight and your estimate tight.