Linear growth and decay

For the two polynomials to be equal, \(60+9x=14x-10\). Rearranging gives \(70=5x\), so \(x=14\). Substituting \(x=14\) gives \(186\) for both polynomials. \(x=13\) is a close distractor, but the two values are not equal there. Exam tip: In equality questions, equate the expressions and collect the variable terms on one side.

Putting p(x)=q(x) gives 330-18x=78+3x. Subtracting 78 from both sides gives 252-18x=3x, so 252=21x. Hence, x=12. If \(x=14\), the values of the two expressions are not equal. Exam tip: collect the x-terms on one side and constants on the other before solving.

In a linear equation, the coefficient of \(x\) gives the rate of change (slope). Here the slope is \(k-11\). For linear growth, this slope must be positive: \(k-11>0\), so \(k>11\). If \(k=11\), the slope is zero and \(y=90\) remains constant, so there is no growth. Exam tip: check the sign of the coefficient of \(x\) to identify growth, decay, or a constant value.

The expression can be written as \(y=410-(k-8)x=410+(8-k)x\). For linear decay, the coefficient of \(x\) (the slope) must be negative. Thus, \(8-k<0\), which gives \(k>8\). When \(k=8\), the slope is zero and the value remains constant; when \(k<8\), the slope is positive, giving linear growth. Exam tip: check the sign of the coefficient of \(x\) to identify linear growth or decay.

Here (R(0)=140) is directly the initial value. The difference gives (m=18), but the initial value remains the same.

\(S(6)=s-13\times6=s-78\) and \(S(15)=s-13\times15=s-195\). Therefore, \(S(6)-S(15)=(s-78)-(s-195)=117\). The terms containing \(s\) cancel, so the condition is true for every value of \(s\). No particular value such as \(s=117\) or \(s=13\) is required. Exam tip: when subtracting function values, check whether common parameter terms cancel.

Find the difference: \(B(t)-A(t)=(111+15t)-(72+8t)=39+7t\). The coefficient of \(t\) is 7, so the difference increases by 7 whenever \(t\) increases by 1. Here, 39 is the initial difference, not the rate of increase; therefore, the option involving 39 is incorrect. Exam tip: The rate of change of the difference of two linear expressions equals the difference of their \(t\)-coefficients.

\(C(t)-D(t)=(390-11t)-(270-22t)=120+11t\). Thus, for every increase of 1 in \(t\), the difference increases by \(11\), so option A is correct. Option B incorrectly treats the resulting rate as negative; \(-11-(-22)=+11\). Exam tip: while subtracting linear expressions, change the sign of every term in the second bracket.

In \(y=50-3x\), the degree of \(x\) is 1 and its coefficient is \(-3\), so \(y\) decreases by 3 for every 1-unit increase in \(x\). \(y=50-3x^2\) is quadratic, not linear. Exam tip: identify degree 1 with a negative coefficient.

In \(y=120-8x\), the coefficient of x is \(-8\), so y falls by 8 when x rises by 1. \(120+8x\) represents growth. Exam tip: in \(y=a+bx\), a negative b indicates linear decay.

Given \(f(x)=12x+50\), we get \(f(5x)=60x+50\) and \(f(2x)=24x+50\). Therefore, \(f(5x)-f(2x)=(60x+50)-(24x+50)=36x\). Setting this equal to \(180\) gives \(36x=180\), so \(x=5\). If \(x=4\), the value is only \(144\), so it is not correct. Exam tip: substitute each input into the function first, then subtract carefully; identical constant terms cancel out.

Given \(g(x)=420-14x\), we get \(g(2x)=420-28x\) and \(g(6x)=420-84x\). Hence, \(g(2x)-g(6x)=(420-28x)-(420-84x)=56x\). Setting \(56x=224\) gives \(x=4\). For \(x=3\), the expression equals only \(168\), so the nearest distractor is not correct. Exam tip: when substituting \(2x\) or \(6x\) into a function, multiply every occurrence of \(x\) by that input.

This is a linear function. The midpoint of 7 and 19 is \(\frac{7+19}{2}=13\), so \(P(7)+P(19)=2P(13)\). Checking directly, \(P(7)=146\) and \(P(19)=242\), giving a sum of \(388\); also, \(P(13)=194\) and \(2\times194=388\). \(P(12)\) is a close distractor, but 12 is not the midpoint of the two given inputs. Exam tip: for a linear function, the average of two function values equals the function value at the midpoint of their inputs.

This is a linear function. The midpoint of 6 and 18 is 12, so for a linear function, Q(6)+Q(18)=2Q(12). Checking directly, Q(6)=348 and Q(18)=204, giving a sum of 552. Also, Q(12)=276, so 2Q(12)=552. Q(11) is not the midpoint, so it cannot be 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 17 - 9 = 8. In the linear expression y = 220 + ax, the change in y is aΔx, so 8a = 144. Hence, a = 144/8 = 18, making option C correct. If a were 16, the increase in y would be only 128. Exam tip: the constant term 220 does not affect the change in y.

The change in x is \(16-8=8\). In \(y=420-bx\), y decreases by b units for every 1-unit increase in x. Therefore, over 8 units, the total decrease is \(8b\). Given \(8b=152\), we get \(b=152/8=19\). If b were 18, the decrease would be \(8\times18=144\), not 152. Exam tip: divide the total change in y by the change in x to find the rate b.

The condition is \(A(t)=B(t)+8\). Thus, \(70+9t=118+5t+8\), which gives \(4t=56\) and hence \(t=14\). Therefore, \(t=14\) is correct. At the closest distractor, \(t=13\), the difference between the functions is only \(4\), not \(8\). Exam tip: Translate “8 more than” as adding \(+8\) to the second expression before forming the equation.

The statement “\(C(t)\) is 18 less than \(D(t)\)” means \(C(t)=D(t)-18\). Thus, \(350-14t=80+4t-18\), giving \(288=18t\) and hence \(t=16\). Checking, \(C(16)=126\) and \(D(16)=144\), so \(C(16)\) is 18 less than \(D(16)\). At \(t=15\), the two values are equal, so it is not correct. Exam tip: for “less than by” questions, subtract the stated difference from the larger quantity.

The sum is ((a+27)+(a+63)+(a+99)=3a+189). From (3a+189=279), (a=30), so option (A) is correct.

The sum is ((b-40)+(b-72)+(b-104)=3b-216). From (3b-216=288), (b=168), so option (D) is correct.