I’m trying to make friends with straight lines, but they keep slipping through my fingers like buttered toast. I get the idea that y = mx + c has a gradient (tilt) and a y‑intercept (where it bonks into the y‑axis), but when I actually do the algebra, my signs do backflips.
Example 1: If I have 5x − 3y = 9, I tried to rearrange it to get y = mx + c. My (probably wrong) attempt was: divide everything by 3 and say y = (−5/3)x + 3. Then I sketched it and the line seemed to cross the y‑axis in the opposite place I expected. I think I messed up with the negatives, but I can’t see exactly where.
Example 2: If the gradient is m = 1/2 and the line passes through (4, −1), I wanted the y‑intercept. I did a very silly thing: I said b = m × x × y = (1/2) × 4 × (−1) = −2, so the y‑intercept is −2. That feels like I multiplied apples by doorknobs.
Analogy attempt: If the gradient is like the tilt of a skateboard ramp and the y‑intercept is the spot where the ramp’s base kisses the floor at x = 0, how do I stop my ramp from teleporting to the wrong wall?
Could someone explain a reliable, step‑by‑step way to get the gradient and the y‑intercept in cases like these-and a quick sanity check so I can tell if my signs are the right way up? I keep tripping over the minus signs and mixing up when to plug in x = 0 versus x = something else.
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3 Responses
For 5x − 3y = 9, move 5x over (−3y = 9 − 5x) and divide by −3 to get y = (5/3)x − 3, so m = 5/3 and b = −3; quick sanity check: set x = 0 in the original and you’ll get y = −3. For m = 1/2 through (4, −1), plug into y = mx + b (not b = mxy-I used to do that too!) so −1 = 2 + b ⇒ b = −3; nice refresher here: https://www.khanacademy.org/math/algebra/two-variable-linear-equations/slope-intercept-form/v/slope-intercept-form.
I love this question-straight lines are like tidy little machines once the signs stop doing gymnastics! The safest recipe is: to get y = mx + b from Ax + By = C, first move the x-term to the other side, then divide by the full coefficient of y (including its sign). In Example 1: 5x − 3y = 9 → −3y = −5x + 9 → divide by −3 → y = (5/3)x − 3. Your slip was dividing by +3 instead of −3; a super-quick sanity check is to set x = 0 in the original: 5·0 − 3y = 9 → y = −3, so the y-intercept must be −3 (that instantly tells you which side of the axis the line “bonks”). Bonus memory: from Ax + By = C, slope m = −A/B and intercept b = C/B (here m = 5/3, b = −3). In Example 2, with m = 1/2 through (4, −1), use b = y − mx: b = −1 − (1/2)·4 = −1 − 2 = −3, so y = (1/2)x − 3. Another way is point–slope form: y − (−1) = (1/2)(x − 4) → same result. Final sanity checks: (i) intercept is always the value of y when x = 0, (ii) increasing x by 1 should change y by exactly m (rise over run), and (iii) positive m slopes up to the right, negative m slopes down-if your graph disagrees, a sign probably flipped. Want to try one more: what are the slope and intercept of −2x + 4y = 12, and where does your quick x = 0 check put the y-intercept?
A clean rule: from Ax + By = C, the slope is m = −A/B and the y‑intercept is b = C/B, so 5x − 3y = 9 gives m = 5/3 and b = −3, i.e., y = (5/3)x − 3 (your sign flipped because isolating y means dividing by −3, not 3). With m = 1/2 through (4, −1), use y = mx + b ⇒ −1 = (1/2)·4 + b ⇒ b = −3; sanity check is “slide to x = 0 to find where the ramp’s base touches the floor,” never multiply x and y-see a short review at https://www.khanacademy.org/math/algebra/linear-equations-and-inequalities/slope-intercept-form/a/slope-intercept-form-review.