Linear growth and decay

Given \(F(t)=65+rt\), we have \(F(4)=65+4r\) and \(F(12)=65+12r\). Substituting these into \(F(4)=F(12)-96\) gives \(65+4r=65+12r-96\). Hence, \(8r=96\), so \(r=12\). For example, if \(r=14\), then \(F(12)-F(4)=8\times14=112\), not 96. Exam tip: when subtracting two values of a linear function, the constant term 65 cancels out.

Given \(G(t)=260-rt\), we have \(G(3)=260-3r\) and \(G(11)=260-11r\). Hence, \(G(3)-G(11)=8r\). The question states that this difference is \(104\), so \(8r=104\), giving \(r=13\). If \(r=16\), the difference would be \(8\times16=128\), so it is not correct. Exam tip: when subtracting values of a linear function at two times, the constant term \(260\) cancels out.

Given P(n)=a+25n and P(4)=210, we get a+25(4)=210. Thus, a+100=210, so a=110. Now P(12)=110+25(12)=110+300=410. Therefore, 410 is the correct answer. The value 385 may result from incorrectly counting 7 steps from 4 to 12; the correct difference is 12−4=8 steps. Exam tip: In a linear expression, first use the given value to find the constant a.

Given \(V(y)=c-3200y\). Substituting \(V(5)=44000\) gives \(c-3200\times5=44000\), so \(c=60000\). Hence, \(V(11)=60000-3200\times11=60000-35200=24800\). Option B, 28000, would result from using an incorrect value in place of 11. Exam tip: first find \(c\) using the given condition, then substitute the required value of \(y\).

Given \(y=42+7x\), the required condition is \(y=13x\). Therefore, \(42+7x=13x\). Subtracting \(7x\) from both sides gives \(42=6x\), so \(x=7\). Option \(x=6\) may seem close, but it gives \(y=84\) and \(13x=78\), which are not equal. Exam tip: First equate the two expressions, then collect like terms on one side.

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.

From (260-15x=68+3x), (192=18x), so (x=\frac{32}{3}). None of the given options is valid.

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 \(m<0\).

Here (R(0)=110) is directly the initial value. The difference gives (m=15), but the initial value remains the same.

\(S(5)=s-55\) 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.