Linear growth and decay

For the two polynomials to be equal, \(18+5x=7x-4\). Subtracting \(5x\) from both sides gives \(18=2x-4\), and adding 4 gives \(22=2x\). Hence, \(x=11\). Substituting \(x=10\) does not give equal values, so it is a close but incorrect option. Exam tip: Equate the expressions first, then collect the variable terms on one side.

Put p(x)=q(x): 150-8x=54+4x. Subtracting 54 from both sides gives 96-8x=4x, so 96=12x. Hence, x=8. If x=9 is substituted, the two expressions do not have equal values. Exam tip: First equate the two polynomials, then collect all x-terms on one side.

In the linear equation \(y=25+(k-4)x\), the coefficient of \(x\), or slope, is \(k-4\). For linear growth, the slope must be positive. Thus, \(k-4>0\), which gives \(k>4\). When \(k=4\), the slope is zero and \(y=25\) is a constant line, not growth. Exam tip: check whether the coefficient of \(x\) is positive for growth and negative for decay.

In the linear form \(y=a+mx\), the value of \(y\) decreases as \(x\) increases when the slope \(m\) is negative; this represents linear decay. Here, the slope is \(-(k-5)=5-k\). For decay, \(5-k<0\), which gives \(k>5\). If \(k<5\), the slope is positive and the relation shows growth, while for \(k=5\), the slope is zero and the value remains constant. Exam tip: To identify growth or decay, check the sign of the coefficient of \(x\).

Here (R(0)=65) is directly the initial value. The difference gives (m=11), but the initial value remains the same.

\(S(3)=s-18\) and \(S(10)=s-60\). Therefore, \(S(3)-S(10)=(s-18)-(s-60)=42\). The parameter \(s\) cancels out, so the given condition is true for every real \(s\). Values such as \(s=42\) or \(s=6\) are merely particular cases, not required values. Exam tip: In a difference of function values, check whether a common constant parameter cancels.

Find the difference: \(B(t)-A(t)=(40+7t)-(28+3t)=12+4t\). The coefficient of \(t\) is 4, so the difference increases by 4 for every increase of 1 in \(t\). The value 12 is only the initial difference, not the rate of increase. Exam tip: When subtracting linear expressions, change the signs of all terms in the expression being subtracted.

Find the difference: \(C(t)-D(t)=(170-4t)-(110-9t)=60+5t\). Since the coefficient of \(t\) is \(5\), the difference increases by \(5\) for every one-unit increase in \(t\). The number \(60\) is the initial difference, not the rate of change. Exam tip: When subtracting an expression, change the signs of all terms in the expression being subtracted.

In \(y=100-5x\), the coefficient \(-5\) means that \(y\) decreases by 5 units for every 1-unit increase in \(x\). Relations containing \(x^2\) or \(1/x\) are not linear. Exam tip: linear decay has a negative coefficient of \(x\).

In \(y=18-3x\), the coefficient of x is the constant \(-3\), so y falls at the same rate as x increases. A relation containing \(x^2\) is not linear. Exam tip: for linear decay, look for x to power 1 with a negative coefficient.

Given \(f(x)=7x+18\), we get \(f(2x)=7(2x)+18=14x+18\). Hence, \(f(2x)-f(x)=(14x+18)-(7x+18)=7x\). Setting \(7x=42\) gives \(x=6\). Therefore, option C is correct. For \(x=5\), the difference would be \(35\), not \(42\). Exam tip: while finding \(f(2x)\), substitute the entire \(2x\) in place of \(x\).

Given \(g(x)=144-8x\), we get \(g(2x)=144-8(2x)=144-16x\). Therefore, \(g(x)-g(2x)=(144-8x)-(144-16x)=8x\). Setting \(8x=64\) gives \(x=8\), so option C is correct. For \(x=7\), the difference is \(56\), not 64. Exam tip: while finding \(g(2x)\), substitute the entire expression \(2x\) for \(x\).

This is a linear function. Here, \(P(3)=42+6\times3=60\) and \(P(11)=42+6\times11=108\), so \(P(3)+P(11)=168\). Also, \(P(7)=42+6\times7=84\); hence \(2P(7)=2\times84=168\). Therefore, the correct answer is \(P(7)\). \(P(6)\) is a close distractor, but \(2P(6)=156\), not 168. Exam tip: for a linear function, the function value at the midpoint of two inputs equals the average of their function values.

Since this is a linear function, the average of the function values at two inputs equals the function value at their midpoint. The midpoint of 2 and 12 is 7. Checking directly, \(Q(2)=150\) and \(Q(12)=100\), so \(Q(2)+Q(12)=250\). Also, \(Q(7)=125\), hence \(2Q(7)=250\). Therefore, \(Q(7)\) is correct. \(Q(6)\) is close, but \(2Q(6)=260\). Exam tip: for a linear function, first find the average of the two input values in questions of this form.

The change in x is 12 - 5 = 7. In the linear expression y = 90 + ax, the change in y is aΔx. Hence, 7a = 63, so a = 9. The constant term 90 does not affect the change in y. Exam tip: First find Δx, then use Δy = aΔx in such questions.

The change in \(x\) is \(12-4=8\). In \(y=220-bx\), every increase of 1 in \(x\) decreases \(y\) by \(b\). Hence, the total decrease is \(8b=96\), giving \(b=12\). If \(b=11\), the decrease would be \(8\times11=88\), not 96. Exam tip: for a linear relation, check the rate using \(\frac{\text{total change}}{\text{change in input}}\).

“\(A(t)\) is 5 more than \(B(t)\)” means \(A(t)=B(t)+5\). Thus, \(20+4t=35+3t+5\), which gives \(t=20\). At \(t=18\), the difference is only \(3\), so it is close but not correct. Exam tip: In “more than” questions, first write the difference as \(A(t)-B(t)\).

“12 less than (D(t))” means (C(t)=D(t)-12). Therefore, (180-7t=60+5t-12). Simplifying gives (180-7t=48+5t), so (132=12t) and hence (t=11). Therefore, option C is correct. At (t=10), the difference between the two functions is not 12. Exam tip: In “less than” questions, subtract the stated difference from the reference expression first.

The sum is ((a+10)+(a+20)+(a+40)=3a+70). From (3a+70=126), (a=\frac{56}{3}), so none of the listed options is correct.

Given \(g(t)=b-4t\), we get \(g(1)=b-4\), \(g(5)=b-20\), and \(g(9)=b-36\). Their sum is \((b-4)+(b-20)+(b-36)=3b-60\). Hence, \(3b-60=156\), so \(3b=216\) and \(b=72\). Therefore, option D is correct. If \(b=68\), the sum would be \(144\), not 156. Exam tip: Substitute each input value separately before adding the function values.