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3 Responses
Nice question-worth double-checking! Step by step: in direct proportion we expect y = kx, which means the graph goes through the origin and the ratio y/x stays constant (equal to k). For y = 3x + 5, the line doesn’t pass through the origin, and y/x isn’t constant: for x = 1 we get 8, for x = 2 we get 11/2, and so on. Also, doubling x wouldn’t double y (x = 2 gives y = 11, but x = 4 gives y = 17, not 22). So, strictly speaking, y = 3x + 5 is not direct proportion; it’s a linear relationship with slope 3 and intercept 5. That said, you can rewrite it as y − 5 = 3x and say “the amount of y above the baseline 5 is directly proportional to x,” and some people still call this “direct proportion” because the gradient 3 acts like the constant of proportionality-so in many contexts it’s treated as directly proportional once you account for the fixed start at 5. Hope this helps!
No-the +5 breaks direct proportion. “Directly proportional” means y = kx for some constant k, so the ratio y/x is constant and the graph passes through the origin. Your equation y = 3x + 5 is linear but not proportional because y/x = 3 + 5/x depends on x. A quick check: if x = 1 then y = 8; doubling x to 2 gives y = 11, which is not double 8, so the “double in, double out” test fails. However, the adjusted variable y − 5 is directly proportional to x, since y − 5 = 3x.
Nope-the +5 spoils pure direct proportion, which must be y = kx (pass through the origin), because y/x isn’t constant. For example, x=2 gives y=11, but doubling x to 4 gives y=17 (not double 11), so it’s linear-but-not-proportional.