Linear growth and decay

Given \(A(t)=a+9t\), substitute \(t=3\): \(A(3)=a+9\times3=a+27\). Since \(A(3)=57\), we get \(a+27=57\), so \(a=30\). Option 27 is the amount added over 3 time units, not the initial value \(a\). Exam tip: substitute the specified value of \(t\) first, then solve the resulting equation.

Given R(t)=b-11t and R(4)=76. Substituting t=4 gives b-11(4)=76, so b-44=76. Adding 44 to both sides gives b=120. Hence, 120 is the correct option. If b were 110, then R(4)=110-44=66, not 76. Exam tip: In a linear expression, first substitute the given value of the variable and then isolate the unknown term.

The coefficient \(b\) is the rate of change, or slope, per unit increase in \(x\). If \(b<0\), \(Q(x)\) decreases as \(x\) rises, so it represents decay. In contrast, \(b=0\) represents a constant quantity. Exam tip: use the sign of the slope to identify growth or decay.

In linear decay, a quantity decreases by a fixed amount in equal time intervals. Losing 5 litres every hour follows this pattern. Becoming half each hour is exponential decay. Exam tip: look for a fixed difference, not a fixed ratio.

Linear growth needs a positive rate and initial value (24). (y=24+7x) satisfies both conditions.

A linear rule has the form \(y=a+bx\), where the initial value is \(a\), found by putting \(x=0\). In \(y=150-12x\), the initial value is 150 and the coefficient of \(x\) is \(-12\), so the value decreases linearly. \(y=12-150x\) also represents decay, but its initial value is 12. Exam tip: put \(x=0\) to check the initial value, and look for a negative slope to identify decay.

Given \(M(t)=m+8t\). Substituting \(t=2\), \(M(2)=m+16=44\), so \(m=28\). Now, at \(t=7\), \(M(7)=28+8\times7=28+56=84\). Therefore, 84 is the correct answer. The value 76 can result from incorrectly counting the increase. Exam tip: first use the given condition to find the unknown constant \(m\), then substitute the required value of \(t\).

Given \(N(t)=n-6t\). At \(t=5\), \(N(5)=n-30=70\), so \(n=100\). Therefore, \(N(12)=100-6\times12=100-72=28\). Hence, 28 is correct. A value such as 34 can result from an error in subtraction. Exam tip: first find the unknown constant \(n\) from the given value, then substitute the required value of \(t\).

Substituting the given values into y=45+kx gives 81=45+6k. Hence, 6k=36 and k=6. Option 36 is only the difference 81-45; it must be divided by x=6 to find k. Exam tip: In a linear equation, substitute the given x and y values before solving for the unknown coefficient.

At \(x=8\), the value of \(y\) is 136. Substituting in the equation gives \(136=200-8k\). Hence, \(8k=64\), so \(k=8\). Note that 64 is the value of \(8k\), not of \(k\). Exam tip: Substitute the given values first, then isolate the unknown term carefully.

Substituting t=7 gives A(7)=20+5(7)=55 and B(7)=50+2(7)=64. Since 64 is greater than 55, the value of B(t) is greater. They are not equal because the two values are different. Exam tip: To compare linear expressions at a given value, substitute that value into both expressions first.

Substituting \(t=6\), \(C(6)=140-8\times6=92\) and \(D(6)=110-3\times6=92\). Hence, the two model values are equal, so \(C(6)=D(6)\). Options A and B are incorrect because neither value is greater than the other, and both values are not zero. Exam tip: evaluate each expression separately at the given value of \(t\) before comparing them.

Putting \(T(t)=102\) gives \(54+6t=102\). Hence, \(6t=48\), so \(t=8\). Therefore, option C is correct. For example, at \(t=7\), \(T(t)=54+42=96\), not 102. Exam tip: equate the function to the target value first, then isolate and solve for the variable.

Given \(U(t)=150-15t\), put \(U(t)=45\). This gives \(150-15t=45\), so \(-15t=-105\) and hence \(t=7\). If \(t=9\), the value is \(15\), not 45. Exam tip: In linear decay questions, substitute the target value into the function and solve the resulting equation for \(t\).

First, f(x+4)=25+6(x+4)=25+6x+24. Therefore, f(x+4)-f(x)=(25+6x+24)-(25+6x)=24. Hence, 24 is the correct option. 18 may seem tempting, but an increase of 4 units at a rate of 6 gives 6×4=24. Exam tip: For a linear function ax+b, increasing x by h always changes the function by ah.

Linear decay has the form \(a-bx\), where \(b>0\). In \(P(x)=80-6x\), the value decreases by 6 for every increase of 1 in \(x\). \(80+6x\) shows linear growth. In exams, check the sign of the \(x\)-coefficient.

In a linear expression \(y=a+bx\), \(a=y(0)\), while \(b\) gives the change in \(y\) for every increase of 1 in \(x\). Here, \(y(0)=18\) and the increase is 7, so \(y=18+7x\). In \(y=7+18x\), the initial value is 7 and the rate of increase is 18, so it is not correct. Exam tip: Substitute \(x=0\) to check the constant term quickly.

A linear relation has the form \(y=a+bx\), where \(a=y(0)\) and \(b\) is the change in \(y\) for each increase of 1 in \(x\). Here, \(y(0)=130\) and \(y\) decreases by 8 each time, so \(b=-8\). Hence, \(y=130-8x\). In option A, \(y\) increases, while option B has initial value 8. Exam tip: put \(x=0\) to identify the initial value from the constant term.

In a linear polynomial \(ax+b\), the value decreases as \(x\) increases when the coefficient \(a\) is negative. In \(-3x+7\), the coefficient of \(x\) is \(-3\). Exam tip: check the sign of the \(x\)-coefficient first.

Putting x=4 gives y=170-10(4)=130. Putting x=11 gives y=170-10(11)=60. Therefore, the difference between the two y-values is 130-60=70. Note that 60 is the value of y at x=11, not the difference. Exam tip: For a linear expression, substitute both x-values first and then subtract the resulting y-values.