Linear growth and decay

For the models to be equal, set \(32+7t=86-2t\). Moving \(2t\) to the left and 32 to the right gives \(9t=54\). Therefore, \(t=6\). At \(t=5\), the two model values are not equal, so it is a plausible but incorrect option. Exam tip: when equating linear models, collect all terms containing \(t\) on one side first.

For equality, write \(55+5t=135-3t\). Moving \(3t\) to the left and 55 to the right gives \(8t=80\), so \(t=10\). Substituting \(t=8\) does not make the two values equal, so it is a close but incorrect distractor. Exam tip: collect all terms containing \(t\) on one side and constant terms on the other.

The coefficient of x is 9, so f(x) increases by 9 whenever x increases by 1. From 3 to 8, x increases by 5; therefore, f(x) increases by \(5\times 9=45\). Hence, \(f(8)=f(3)+45=64+45=109\). Option 100 represents an increase of only 36, not the correct increase over 5 steps. Exam tip: for a linear function, multiply the slope by \(\Delta x\) to find the change in the function value.

Given (g(x)=b-8x), first use (g(4)=91): (b-8(4)=91), so (b-32=91). Hence, (b=123). Now substitute (x=10): (g(10)=123-8(10)=123-80=43). Therefore, 43 is correct. A choice such as 51 can result from an arithmetic error in subtraction. Exam tip: find the unknown constant first, then substitute the required value of x.

In \(y=18-2x\), the coefficient of x is \(-2\), so y falls by 2 for every increase of 1 in x. Option B represents growth. Exam tip: linear decay has a negative x-coefficient and degree 1.

In \(P(t)=120-5t\), the constant negative coefficient \(-5\) means the quantity falls by 5 for each unit of time. \(120-5t^2\) is not linear because the variable has power 2. Exam tip: check that the variable’s power is 1.

In \(18-3x\), the coefficient of x is \(-3\), so the quantity falls by 3 units for every 1-unit rise in x. \(18+3x\) shows growth instead. Exam tip: a linear polynomial has x to the power 1.

In linear decay, the quantity decreases by an equal amount for each unit increase, so its graph is a straight line with negative slope. For example, \(p(x)=a-bx\), where \(b>0\). Exam tip: a negative coefficient of \(x\) indicates decay.

Given \(C(t)=24+kt\), we have \(C(7)=24+7k\) and \(C(2)=24+2k\). Hence, \(C(7)-C(2)=(24+7k)-(24+2k)=5k\). Since this difference is 45, \(5k=45\), so \(k=9\). If \(k=8\), the difference would be \(5\times8=40\), not 45. Exam tip: when subtracting two values of a linear expression, the constant term 24 cancels out.

Given \(D(t)=160-kt\), we have \(D(4)=160-4k\) and \(D(11)=160-11k\). Therefore, \(D(4)-D(11)=(160-4k)-(160-11k)=7k\). Since this difference is \(63\), \(7k=63\), so \(k=9\). For example, \(k=8\) would give a difference of \(56\), not \(63\). Exam tip: Write the two function values separately before subtracting; the constant terms often cancel.

\(24-3x\) is linear, and its coefficient of x is \(-3\); therefore, the value falls by 3 for each one-unit increase in x. Check: \(P(x+1)-P(x)=-3\). In \(24-3x^2\), the rate of decrease is not constant. Exam tip: look for degree 1 with a negative coefficient.

The slope of \(P(t)=a-bt\) is the constant negative value \(-b\), so equal time intervals produce equal decreases. A changing slope indicates non-linear decay. Exam tip: identify \(-b\) as the constant decay rate.

Given \(P(t)=p+8t\), substitute \(t=4\): \(p+8\times4=70\), so \(p=38\). Next, \(P(s)=118\) gives \(38+8s=118\). Thus \(8s=80\), hence \(s=10\). For example, using 9 gives 110, not 118. Exam tip: find the constant \(p\) from the first condition before using the second condition.

Given \(R(t)=r-7t\), substitute \(R(3)=112\): \(r-7(3)=112\), so \(r=133\). Next, \(R(s)=63\) gives \(133-7s=63\), hence \(7s=70\) and \(s=10\). If \(s=9\), the model gives \(70\), not \(63\). Exam tip: first find the constant \(r\) using the known time, then use the second condition to solve for the variable.

Let the linear expression be \(y=a+5x\), since \(y\) must increase by 5 when \(x\) increases by 1. Using \(y(4)=37\) gives \(a+5(4)=37\), so \(a=17\). Therefore, \(y=17+5x\) is correct. Although \(y=57-5x\) gives \(y(4)=37\), its coefficient of \(x\) is \(-5\), so it represents a decrease, not an increase. Exam tip: substitute the given \(x\)- and \(y\)-values to find the constant term.

A decrease of 9 per step means that the coefficient of \(x\) must be \(-9\). So write the equation as \(y=a-9x\). Using \(y(6)=58\), we get \(a-9(6)=58\), hence \(a=112\). Therefore, \(y=112-9x\) is correct. Option A has slope \(-9\), but it gives \(y(6)=4\), not 58. Exam tip: substitute the given \(x\)-value into each option to check the condition quickly.

“9 more” means \(M(t)=N(t)+9\). Thus, \(42+4t=96-5t+9\), or \(42+4t=105-5t\). Hence \(9t=63\), so \(t=7\). At \(t=6\), the difference is only \(0\), so it is a close but incorrect option. Exam tip: For “more than” questions, equate the larger quantity to the smaller quantity plus the stated difference.

“16 more” means \(A(t)=B(t)+16\). So, \(140-6t=44+2t+16\), which gives \(140-6t=60+2t\) and then \(80=8t\). Hence, \(t=10\). At \(t=9\), the difference is \(24\), not \(16\). Exam tip: In “more than” questions, first write the difference as \(A(t)-B(t)\).

Given \(F(t)=32+rt\), we have \(F(3)=32+3r\) and \(F(8)=32+8r\). Substituting these into \(F(3)=F(8)-40\) gives \(32+3r=32+8r-40\). Hence, \(5r=40\), so \(r=8\). If \(r=10\), the change from \(t=3\) to \(t=8\) would be 50, not 40. Exam tip: for a linear function, the change over an interval equals \(r\) multiplied by the time interval.

Given \(G(2)=G(9)+56\), we have \(G(2)-G(9)=56\). Substituting \(G(2)=150-2r\) and \(G(9)=150-9r\) gives \((150-2r)-(150-9r)=7r\). Hence, \(7r=56\), so \(r=8\). If \(r=7\), the difference would be only \(49\), so it is not correct. Exam tip: In linear functions, subtracting values at two inputs eliminates the constant term.

Given \(P(n)=a+15n\). Using \(P(4)=140\), we get \(a+15\times4=140\), so \(a=80\). Therefore, \(P(9)=80+15\times9=80+135=215\). Option 230 does not apply the increase of 15 correctly. Exam tip: first find the constant \(a\) from the given value, then substitute the required value of \(n\).

Given V(y)=c-1800y, substitute V(3)=24600: c-1800(3)=24600, so c-5400=24600 and c=30000. Now, V(8)=30000-1800(8)=30000-14400=15600. Therefore, 15600 is correct. The value 17400 would correspond to 7 years, not 8 years. Exam tip: first find the unknown constant c from the given value, then substitute the required value of y.

We need both \(y=18+4x\) and \(y=7x\). Therefore, \(18+4x=7x\). Subtracting \(4x\) from both sides gives \(18=3x\), so \(x=6\). Check: for \(x=6\), \(y=18+24=42\) and \(7x=42\). For \(x=5\), the two values are 38 and 35, so it is not correct. Exam tip: equate the two expressions for y first, then solve for x.

Given \(y=96-6x\) and the condition \(y=2x\), we get \(96-6x=2x\). Adding \(6x\) to both sides gives \(96=8x\), so \(x=12\). For \(x=10\), \(y=36\) whereas \(2x=20\), so it is not correct. Exam tip: First equate the given expression for \(y\) to the required condition on \(y\).

Given \(L(t)=3t+14\), we get \(L(4t)=3(4t)+14=12t+14\). Therefore, \(L(4t)-L(t)=(12t+14)-(3t+14)=9t\). Hence, \(9t\) is correct. \(12t\) is only part of \(L(4t)\); it does not account for subtracting \(L(t)\). Exam tip: when evaluating a function at \(4t\), replace every occurrence of \(t\) with \(4t\).