Linear growth and decay

Given \(y=210-12x\) and the condition \(y=3x\), substitute the expression for \(y\): \(210-12x=3x\). Adding \(12x\) to both sides gives \(210=15x\), so \(x=14\). Option \(x=15\) may seem close, but then \(y=30\) while \(3x=45\), so they are not equal. Exam tip: when a condition is given for a variable, substitute the provided expression before solving.

Given \(L(t)=6t+31\), we get \(L(5t)=6(5t)+31=30t+31\) and \(L(2t)=6(2t)+31=12t+31\). Hence, \(L(5t)-L(2t)=(30t+31)-(12t+31)=18t\). \(30t\) is only the linear term in \(L(5t)\); \(L(2t)\) must also be subtracted. Exam tip: substitute each input first, then use brackets while subtracting the complete expressions.

Given \(K(t)=280-8t\), we get \(K(2t)=280-8(2t)=280-16t\) and \(K(6t)=280-8(6t)=280-48t\). Therefore, \(K(2t)-K(6t)=(280-16t)-(280-48t)=32t\). \(48t\) is only the magnitude of the linear term in \(K(6t)\), not the difference between the two function values. Exam tip: when substituting \(2t\) or \(6t\), replace every occurrence of \(t\) by the complete expression.

Given \(Y(x)=a+9x\), we get \(Y(4)=a+36\) and \(Y(10)=a+90\). Hence, \(Y(4)+Y(10)=2a+126=218\). Therefore, \(2a=92\), so \(a=46\). If 52 were used, the sum would be 230, so it is not correct. Exam tip: First substitute each given \(x\)-value in the function, then add the resulting expressions.

Given \(Z(x)=b-12x\), we have \(Z(3)=b-36\) and \(Z(8)=b-96\). Thus, \(Z(3)+Z(8)=2b-132=148\). Hence \(2b=280\), so \(b=140\). If 144 were used, the sum would be \(156\), not 148. Exam tip: Substitute each given \(x\)-value into the function separately before adding the results.

For the two polynomials to be equal, \(44+8x=12x-8\). Adding 8 to both sides gives \(52+8x=12x\), and then \(52=4x\). Hence, \(x=13\). At \(x=13\), both expressions have the value 148. \(x=12\) is a close distractor, but the two values are not equal there. Exam tip: In equality questions, first equate the two expressions and collect all \(x\)-terms on one side.

In a linear equation, the coefficient of \(x\) is the slope or rate of change. Here, the slope is \(k-9\). For linear growth, it must be positive: \(k-9>0\), so \(k>9\). If \(k=9\), the slope is zero and \(y=72\) is a constant line, not growth. Exam tip: check whether the slope is positive for growth and negative for decay.

The coefficient of \(x\), or slope, in \(y=320-(k-6)x\) is \(-(k-6)=6-k\). For linear decay, \(y\) must decrease as \(x\) increases, so the slope must be negative: \(6-k<0\). Hence, \(k>6\). When \(k=6\), the slope is zero and the value remains constant, not decaying. Exam tip: in \(y=a+mx\), decay occurs when \(m<0\).

\(S(5)=s-55\) and \(S(13)=s-143\). Therefore, \(S(5)-S(13)=(s-55)-(s-143)=88\). The terms containing \(s\) cancel, so the given condition is true for every value of \(s\). Neither \(s=88\) nor \(s=11\) is required. Exam tip: When subtracting expressions, use brackets and change the sign of every term in the second expression.

First find the difference: (B(t)-A(t))=(91+13t)-(58+7t)=33+6t. The coefficient of t is 6, so the difference increases by 6 whenever t increases by 1. Here, 33 is the initial difference, not the rate of increase, so option D is incorrect. Exam tip: the rate of the difference of two linear expressions is the difference of their t-coefficients.

Find the difference: \(C(t)-D(t)=(330-9t)-(230-18t)=100+9t\). The coefficient of \(t\) is 9, so the difference increases by 9 for every one-step increase in \(t\). Option B would be correct only if the coefficient of \(t\) in the difference were \(-9\). Exam tip: When subtracting linear expressions, change the signs of every term in the second bracket.

In \(y=250-6x\), the coefficient \(-6\) means that when \(x\) increases by 1, \(y\) decreases by 6 every time. In \(250-6x^2\), the rate of decrease is not constant. Exam tip: linear decay has a negative coefficient and power of \(x\) equal to 1.

In linear decay, y falls by an equal amount over equal intervals of x, so the graph has a constant negative slope. A constant ratio of decrease indicates exponential decay instead. Exam tip: look for a constant difference, not a constant ratio.

Given \(f(x)=10x+35\), we get \(f(4x)=10(4x)+35=40x+35\). Therefore, \(f(4x)-f(x)=(40x+35)-(10x+35)=30x\). Setting \(30x=150\) gives \(x=5\), so option B is correct. For instance, if \(x=4\), the difference is only \(120\), not \(150\). Exam tip: while finding \(f(4x)\), replace every \(x\) in the function by \(4x\).

Given \(g(x)=300-15x\), we get \(g(5x)=300-15(5x)=300-75x\). Hence, \(g(x)-g(5x)=(300-15x)-(300-75x)=60x\). Setting \(60x=180\) gives \(x=3\). Therefore, option B is correct. For example, \(x=2\) gives a difference of only \(120\), not \(180\). Exam tip: while finding \(g(5x)\), replace every \(x\) in the rule by \(5x\).

This is a linear function, so the sum of its values at two inputs equals twice its value at the midpoint of those inputs. The midpoint of 6 and 18 is \(\frac{6+18}{2}=12\). Hence, \(P(6)+P(18)=2P(12)\). Checking directly, \(P(6)=111\) and \(P(18)=183\), so their sum is 294, while \(2P(12)=2\times147=294\). Exam tip: for a linear function, use \(P(a)+P(b)=2P\left(\frac{a+b}{2}\right)\).

The function \(Q(t)=320-8t\) is linear. The midpoint of \(5\) and \(17\) is \(\frac{5+17}{2}=11\), so for a linear function, \(Q(5)+Q(17)=2Q(11)\). Hence, \(Q(11)\) is correct. \(Q(10)\) is a close distractor, but 10 is not the midpoint of 5 and 17. Exam tip: In such questions, find the average of the two inputs to determine \(k\).

In \(y=500-12x\), the coefficient of x is the constant negative number \(-12\), so y decreases at a constant rate. The \(x^2\) relation is not linear. Exam tip: linear decay has a negative coefficient of x.

The change in x is 15 − 7 = 8. In y = 360 − bx, y decreases by b units for every 1-unit increase in x. Hence, the total decrease is 8b = 136. Therefore, b = 136/8 = 17. If b were 16, the decrease would be 8 × 16 = 128, not 136. Exam tip: For a linear expression, multiply the change in x by its coefficient to find the total change in y.

The condition is \(A(t)=B(t)+6\). So, \(50+7t=92+4t+6\), which gives \(3t=48\) and hence \(t=16\). Therefore, the correct option is \(t=16\). At \(t=14\), the difference is only \(50+98-(92+56)=0\), not 6. Exam tip: translate “6 more than” as \(A=B+6\).

The condition is (310-12t=70+6t-6). This gives (246=18t), so (t=\frac{41}{3}), and none of the options is correct.

The sum is ((a+16)+(a+48)+(a+80)=3a+144). From (3a+144=252), (a=36), which is not in the options.

Given \(g(t)=b-7t\), we get \(g(4)=b-28\), \(g(8)=b-56\), and \(g(12)=b-84\). Their sum is \(3b-(28+56+84)=3b-168\). Hence, \(3b-168=258\), so \(3b=426\) and \(b=142\). If 138 were used, the sum would be 246, not 258. Exam tip: First write the function value for each given input, then combine like terms carefully.

In \(P(t)=80-5t\), the power of \(t\) is 1, so it is linear. The coefficient \(-5\) means the quantity falls by 5 per time unit. \(80+5t\) represents growth, not decay. Exam tip: for linear decay, check for a negative coefficient of \(t\).

In linear decay, the quantity decreases by an equal amount in equal time intervals, so its graph has a constant negative slope. Option B shows exponential decay because the quantity is halved each time. Exam tip: a constant negative slope indicates linear decay.