Linear growth and decay

Given \(y=80-5x\) and the condition \(y=3x\), set the two expressions equal: \(80-5x=3x\). Adding \(5x\) to both sides gives \(80=8x\), so \(x=10\). If \(x=8\), then \(y=40\) while \(3x=24\), so it does not satisfy the condition. Exam tip: When two expressions represent the same variable, equate them and solve the resulting linear equation.

Given \(L(t)=2t+11\), we get \(L(3t)=2(3t)+11=6t+11\). Therefore, \(L(3t)-L(t)=(6t+11)-(2t+11)=4t\). \(6t\) is part of the value of \(L(3t)\), not the difference. Exam tip: substitute \(3t\) into the function first, then subtract the two expressions using brackets.

Given \(K(t)=90-4t\). Substituting \(2t\) for \(t\) gives \(K(2t)=90-4(2t)=90-8t\). Therefore, \(K(t)-K(2t)=(90-4t)-(90-8t)=90-4t-90+8t=4t\). Hence, \(4t\) is correct. The distractor \(-4t\) may result from not changing every sign while subtracting the second bracket. Exam tip: when finding a function at \(2t\), replace every \(t\) by \(2t\), then distribute the minus sign carefully.

Given \(Y(x)=a+5x\), we get \(Y(2)=a+10\) and \(Y(6)=a+30\). Therefore, \(Y(2)+Y(6)=2a+40=86\). Hence \(2a=46\), so \(a=23\). For example, choosing 28 would give a sum of 96, not 86. Exam tip: Substitute each given \(x\)-value into the function separately before adding.

Given \(Z(x)=b-6x\), we get \(Z(1)=b-6\) and \(Z(4)=b-24\). Hence, \(Z(1)+Z(4)=(b-6)+(b-24)=2b-30\). Using \(2b-30=90\), we obtain \(2b=120\), so \(b=60\). If 63 were used, the sum would be 96, not 90. Exam tip: Substitute each given value of \(x\) separately before adding the function values.

For equality, set \(p(x)=q(x)\): \(10+3x=4x-2\). Subtracting \(3x\) from both sides gives \(10=x-2\), so \(x=12\). On checking, \(p(12)=46\) and \(q(12)=46\). For example, at \(x=10\), their values are 40 and 38, so it is not correct. Exam tip: To find when two polynomials are equal, equate their expressions first and solve for \(x\).

Since \(p(x)=q(x)\), set \(120-7x=60+5x\). Subtracting 60 from both sides gives \(60-7x=5x\), so \(60=12x\). Hence, \(x=5\). If \(x=6\) is substituted, the two expressions do not have equal values. Exam tip: equate the two linear expressions and collect all \(x\)-terms on one side.

In \(y=18+(k-2)x\), the coefficient of \(x\), or slope, is \(k-2\). For linear growth, the slope must be positive, so \(k-2>0\). Hence, \(k>2\). When \(k=2\), \(y=18\) is constant, not increasing. Exam tip: in \(y=mx+c\), check \(m>0\) for linear growth.

In \(y=100-(k+3)x\), the coefficient of \(x\), or slope, is \(-(k+3)\). For linear decay, the slope must be negative, so \(-(k+3)<0\). This gives \(k+3>0\), hence \(k>-3\). When \(k=-3\), the slope is zero and \(y=100\) is constant, not decay. Exam tip: check the sign of the coefficient of \(x\) to identify linear growth or decay.

\(S(2)=s-4(2)=s-8\) and \(S(9)=s-4(9)=s-36\). Therefore, \(S(2)-S(9)=(s-8)-(s-36)=28\). The terms containing \(s\) cancel, so the condition is true for every real \(s\). Neither \(s=28\) nor \(s=4\) is a specially required value. Exam tip: In differences of function values, write both values first and check whether the parameter terms cancel.

Find the difference: (B(t)-A(t))=(30+5t)-(18+2t)=12+3t. The coefficient of t is 3, so the difference increases by 3 for every increase of 1 in t. Therefore, option A is correct. The value 12 is the initial difference, not the rate of increase, so option D is incorrect. Exam tip: when subtracting linear expressions, subtract the constant terms and the coefficients of t separately.

Subtracting gives \(C(t)-D(t)=(150-3t)-(120-8t)=30+5t\). Thus, for every increase of 1 in \(t\), the difference increases by 5. Option B is incorrect because \(D(t)\) decreases 5 units faster than \(C(t)\), so the gap between them increases. Exam tip: When subtracting linear expressions, change the signs of every term in the second bracket carefully.

In \(y=80-5x\), every increase of 1 in x decreases y by a constant 5, so it shows linear decay. Option B has a positive slope and represents growth. Exam tip: linear decay always has a negative coefficient of x.

In \(P(x)=240-8x\), the constant term 240 is the initial value and the coefficient \(-8\) shows a decrease as x increases. \(240+8x\) represents growth. Exam tip: a negative slope indicates decay.

Given \(f(x)=5x+20\), we get \(f(2x)=5(2x)+20=10x+20\). Therefore, \(f(2x)-f(x)=(10x+20)-(5x+20)=5x\). Setting \(5x=30\) gives \(x=6\). Hence, option C is correct. For instance, when \(x=5\), the difference is only \(25\), not \(30\). Exam tip: substitute \(2x\) for every \(x\) in \(f(x)\) before simplifying the subtraction.

Given \(g(x)=96-6x\), we get \(g(2x)=96-6(2x)=96-12x\). Therefore, \(g(x)-g(2x)=(96-6x)-(96-12x)=6x\). Setting \(6x=36\) gives \(x=6\), so option C is correct. For example, at \(x=5\), the difference would be only \(30\). Exam tip: while finding \(g(2x)\), replace every \(x\) in the function by \(2x\).

This is a linear function. The midpoint of 1 and 9 is \(\frac{1+9}{2}=5\), so \(Q(1)+Q(9)=2Q(5)\). Checking directly, \(Q(1)=96\) and \(Q(9)=64\), giving a sum of \(160\). Also, \(Q(5)=80\), hence \(2Q(5)=160\). \(Q(4)\) is not at the midpoint, so it is not correct. Exam tip: for a linear function, the sum of the values at two inputs equals twice the value at their midpoint.

The change in x is 9−4=5. In the linear expression y=70+ax, the change in y is a×5. Hence, 5a=35, so a=7. Option 5 represents the change in x, not the value of a. Exam tip: In y=mx+c, identify m as the rate of change.

The change in \(x\) is \(8-3=5\). In \(y=160-bx\), \(y\) decreases by \(b\) units for every 1-unit increase in \(x\). Hence, the total decrease is \(5b\). Given \(5b=45\), we get \(b=45/5=9\). Therefore, \(9\) is correct. If \(b=15\), the decrease would be \(5\times15=75\), not 45. Exam tip: For a linear relation, total change equals rate per unit times the change in the variable.

“\(A(t)\) is 4 more than \(B(t)\)” means \(A(t)=B(t)+4\). Thus, \(12+3t=18+2t+4\), which gives \(t=10\). Checking: \(A(10)=42\) and \(B(10)=38\), so the difference is 4. At \(t=9\), the difference is only 3. Exam tip: For “more than” questions, first write the difference as \(A(t)-B(t)\).

“\(C(t)\) is 12 less than \(D(t)\)” means \(C(t)=D(t)-12\). Thus, \(130-5t=70+3t-12\). Simplifying gives \(72=8t\), so \(t=9\). Option \(t=8\) is close, but the difference between the two values is not 12 there. Exam tip: For “less than” comparisons, subtract the stated amount from the larger expression before forming the equation.

Given f(t)=a+4t, we get f(1)=a+4, f(3)=a+12, and f(5)=a+20. Their sum is 3a+36. Hence, 3a+36=81, so 3a=45 and a=15. If a were 17, the sum would be 87, so it is not correct. Exam tip: substitute each given value of t into the function before adding the results.

Given \(g(t)=b-3t\), we get \(g(2)=b-6\), \(g(4)=b-12\), and \(g(6)=b-18\). Their sum is \((b-6)+(b-12)+(b-18)=3b-36\). Hence \(3b-36=144\), so \(3b=180\) and \(b=60\). If 54 is used instead, the sum becomes \(126\), not 144. Exam tip: In such questions, first evaluate the function at every given input and then form the sum equation.

In linear decay, the change over equal time intervals is constant. Here the change is always \(-7\), so the slope is fixed and the model has the form \(P=a-7t\). A 10% decrease depends on the current value, so it is not linear. Exam tip: check for equal first differences.

In \(Q(t)=Q_0-rt\), the quantity decreases by the same amount \(r\) in every unit of time, so it is a first-degree linear decay model. \(Q_0+rt\) shows growth, while \(e^{-rt}\) shows exponential decay. Exam tip: a constant rate of change indicates a linear relation.